Past JEE Main Entrance Paper Set-I

Hint

(b) Let \(z=\frac{3+i \sin \theta}{4-i \cos \theta}\), after rationalising \(z=\frac{(3+i \sin \theta)}{(4-i \cos \theta)} \times \frac{(4+i \cos \theta)}{(4+i \cos \theta)}\)
As \(z\) is purely real
\(
\begin{aligned}
&\Rightarrow \quad 3 \cos \theta+4 \sin \theta=0 \quad \Rightarrow \tan \theta=-\frac{3}{4} \\
&\arg (\sin \theta+i \cos \theta)=\pi+\tan ^{-1}\left(\frac{\cos \theta}{\sin \theta}\right) \\
&=\pi+\tan ^{-1}\left(-\frac{4}{3}\right)=\pi-\tan ^{-1}\left(\frac{4}{3}\right)
\end{aligned}
\)

Hint

Let \(z=x+i y\)
Length of side \(=4\)
\(
\mathrm{AB}=4
\)
\(
\begin{aligned}
&|\mathrm{z}-\overline{\mathrm{z}}|=4 \\
&|2 \mathrm{y}|=4 ;|\mathrm{y}|=2  \\
&\mathrm{BC}=4
\end{aligned}
\)
\(
\mid \overline{\mathrm{z}}-(\overline{\mathrm{z}}-2 \operatorname{Re}(\overline{\mathrm{z}})) \mid=4
\)
\(
|2 x|=4 ;|x|=2
\)
\(
|z|=\sqrt{x^2+y^2}=\sqrt{4+4}=2 \sqrt{2}
\)

Hint

(c) \(\because-1+\sqrt{3} i=2 \cdot e^{\frac{2 \pi}{3} i}\) and \(1-i=\sqrt{2} \cdot e^{-\frac{i \pi}{4}}\)
\(
\therefore\left(\frac{-1+\sqrt{3} i}{1-i}\right)^{30}=\left(\sqrt{2} e^{\left(\frac{2 \pi}{3}+\frac{\pi}{4}\right) i}\right)^{30}
\)
\(
=2^{15} \cdot e^{-\frac{\pi}{2} i}=-2^{15} \cdot i
\)

Hint

(b) Let \(z_1=x_1+i y_1\) and \(z_2=x_2+i y_2\)
\(
\begin{aligned}
&\because\left|z_1-1\right|=\operatorname{Re}\left(z_1\right) \\
&\Rightarrow\left(x_1-1\right)^2+y_1^2=x_1^2 \\
&\Rightarrow y_1^2-2 x_1+1=0 \dots(i)\\
&\left|z_2-1\right|=\operatorname{Re}\left(z_2\right) \Rightarrow\left(x_2-1\right)^2+y_2^2=x_2^2 \\
&\Rightarrow y_2^2-2 x_2+1=0 \dots(ii)
\end{aligned}
\)
From eqn. (i) – (ii),
\(
\begin{aligned}
&y_1^2-y_2^2-2\left(x_1-x_2\right)=0 \\
&\Rightarrow y_1+y_2=2\left(\frac{x_1-x_2}{y_1-y_2}\right) \\
&\because \arg \left(z_1-z_2\right)=\frac{\pi}{6} \\
&\Rightarrow \tan ^{-1}\left(\frac{y_1-y_2}{x_1-x_2}\right)=\frac{\pi}{6} \\
&\Rightarrow \frac{y_1-y_2}{x_1-x_2}=\frac{1}{\sqrt{3}} \\
&\Rightarrow \frac{2}{y_1+y_2}=\frac{1}{\sqrt{3}} \quad\left[\text { From, } \frac{y_1-y_2}{x_1-x_2}=\frac{2}{y_1+y_2}\right] \\
&\therefore y_1+y_2=2 \sqrt{3} \Rightarrow \operatorname{lm}\left(z_1+z_2\right)=2 \sqrt{3}
\end{aligned}
\)

Hint

\(
\text { (c) Let } z=x+i y
\)
\(
\begin{aligned}
&\text { Then, }\left|\frac{z-i}{z+2 i}\right|=1 \Rightarrow x^2+(y-1)^2 \\
&=x^2+(y+2)^2 \Rightarrow-2 y+1=4 y+4 \\
&\Rightarrow \quad 6 y=-3 \Rightarrow y=-\frac{1}{2} \\
&\therefore \quad|z|=\frac{5}{2} \Rightarrow x^2+y^2=\frac{25}{4} \\
&\Rightarrow \quad x^2=\frac{24}{4}=6 \\
&\therefore \quad z=x+i y \Rightarrow z=\pm \sqrt{6}-\frac{i}{2} \\
&|\mathrm{z}+3 i|=\sqrt{6+\frac{25}{4}}=\sqrt{\frac{49}{4}} \\
&\Rightarrow \quad|z+3 i|=\frac{7}{2}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&z=x+i y \\
&\operatorname{Re}(z)=z \\
&\operatorname{Im}(z)=y
\end{aligned}
\)
\(
\begin{aligned}
&|\operatorname{Re}(z)|+|\operatorname{Im}(z)|=4 \\
&|x|+|y|=4 \\
&\text { If } x=0:|y|=4 \rightarrow y=\pm 4 \\
&y=0:|x|=4 \Rightarrow x=\pm 4
\end{aligned}
\)
\(
|z|=\sqrt{x^2+y^2}
\)
Minimum value of \(|z|=2 \sqrt{2}\)
Maximum value of \(|z|=4\)
\(
|z| \in[\sqrt{8}, \sqrt{16}]
\)
So, \(|z|\) can’t be \(\sqrt{7}\).

Hint

(b) Given equation is, \(|z-1|=|z-i|\)
\(\Rightarrow \quad(x-1)^2+y^2=x^2+(y-1)^2\)
\([\) Here, \(z=x+i y]\)
\(\Rightarrow \quad 1-2 x=1-2 y \Rightarrow x-y=0\)
Hence, locus is straight line with slope 1.

Hint

\(
\text { (a) } z=\frac{(1+i)^2}{a-i} \times \frac{a+i}{a+i}
\)
\(
\begin{aligned}
& z=\frac{(1-1+2 i)(a+i)}{a^2+1}=\frac{2 a i-2}{a^2+1} \\
|z|=& \sqrt{\left(\frac{-2}{a^2+1}\right)^2+\left(\frac{2 a}{a^2+1}\right)^2}=\sqrt{\frac{4+4 a^2}{\left(a^2+1\right)^2}} \\
& \Rightarrow|z|=\sqrt{\frac{4\left(1+a^2\right)}{\left(1+a^2\right)^2}}=\frac{2}{\sqrt{1+a^2}} \dots(i)
\end{aligned}
\)
Since, it is given that \(|z|=\sqrt{\frac{2}{5}}\)
Then, from equation (i),
\(
\sqrt{\frac{2}{5}}=\frac{2}{\sqrt{1+a^2}}
\)
Now, square on both side; we get
\(
\Rightarrow 1+a^2=10 \Rightarrow a=\pm 3
\)
since, it is given that \(a>0 \Rightarrow a=3\)
Then, \(z=\frac{(1+i)^2}{a-i}=\frac{1+i^2+2 i}{3-i}=\frac{2 i}{3-i}\)
\(
=\frac{2 i(3+i)}{10}=\frac{-1+3 i}{5}
\)
Hence, \(\bar{z}=\frac{-1}{5}-\frac{3}{5} i\)

Hint

\(
\begin{aligned}
&\text { (c) } \omega=\frac{5+3 z}{5-5 z} \Rightarrow 5 \omega-5 \omega z=5+3 z \\
&\Rightarrow 5 \omega-5=z(3+5 \omega) \Rightarrow z=\frac{5(\omega-1)}{3+5 \omega} \\
&\therefore|z|<1, \therefore 5|\omega-1|<|3+5 \omega| \\
&\Rightarrow 25(\omega \bar{\omega}-\omega-\bar{\omega}+1)<9+25 \omega \bar{\omega}+15 \omega+15 \bar{\omega} \\
&\quad(|z| = \quad z \bar{z})
\end{aligned}
\)
\(
\begin{aligned}
&16<40 \omega+40 \bar{\omega} \Rightarrow \omega+\bar{\omega}>\frac{2}{5} \Rightarrow 2 \operatorname{Re}(\omega)>\frac{2}{5} \\
&\operatorname{Re}(\omega)>\frac{1}{5}
\end{aligned}
\)

Hint

(a) Let \(t=\frac{z-\alpha}{z+\alpha}\)
\(\quad t\) is purely imaginary number.
\(
\begin{array}{ll}
\therefore t+\bar{t}=0 \\
\Rightarrow \quad \frac{z-\alpha}{z+\alpha}+\frac{\bar{z}-\alpha}{\bar{z}+\alpha}=0 \\
\Rightarrow \quad(z-\alpha)(\bar{z}+\alpha)+(\bar{z}-\alpha)(z+\alpha)=0 \\
\Rightarrow \quad z \bar{z}-\alpha^2+z \bar{z}-\alpha^2=0 \\
\Rightarrow \quad z \bar{z}-\alpha^2=0 \\
\Rightarrow \quad|z|^2-\alpha^2=0 \\
\Rightarrow \quad \alpha^2=4 \\
\Rightarrow \quad \alpha=\pm 2
\end{array}
\)

Hint

\(
\begin{aligned}
&|z|+z=3+i \\
&z=3-|z|+i \\
&\text { Let } 3-|z|=a \\
&\Rightarrow|z|=(3-a) \\
&\Rightarrow z=a+i \\
&\Rightarrow|z|=\sqrt{a^2+1} \\
&\Rightarrow 9+a^2-6 a=a^2+1 \\
&\Rightarrow a=\frac{8}{6}=\frac{4}{3} \\
&\Rightarrow|z|=3-\frac{4}{3}=\frac{5}{3}
\end{aligned}
\)

Hint

Answer is a \(\operatorname{Re}(z)=0\)
From the question, the condition given is:
\(
\begin{aligned}
&3\left|\mathrm{z}_1\right|=4\left|\mathrm{z}_2\right| \\
&\Rightarrow \frac{\left|z_1\right|}{\left|z_2\right|}=\frac{4}{3} \\
&\because\left[\mathrm{z}=|\mathrm{z}|(\cos \theta+i \sin \theta)=|\mathrm{z}| \mathrm{e}^{i \theta}\right] \\
&\Rightarrow \frac{z_1}{z_2}=\left|\frac{z_1}{z_2}\right| e^{i \theta} \text { and } \frac{z_2}{z_1}=\left|\frac{z_2}{z_1}\right| e^{-i \theta} \\
&\Rightarrow \frac{z_1}{z_2}=\frac{4}{3} e^{i \theta} \text { and } \frac{z_2}{z_1}=\frac{3}{4} e^{-i \theta} \\
&\Rightarrow \frac{3}{2} \frac{z_1}{z_2}=2 e^{i \theta} \text { and } \frac{2}{3} \frac{z_2}{z_1}=\frac{1}{2} e^{-i \theta}
\end{aligned}
\)
On adding these two,
\(
\begin{aligned}
&\Rightarrow z=\frac{3}{2} \frac{z_1}{z_2}+\frac{2}{3} \frac{z_2}{z_1}=2 e^{i \theta}+\frac{1}{2} e^{-i \theta} \\
&\because\left[\mathrm{e}^{\pm i \theta}=\cos \theta \pm i \sin \theta\right] \\
&\Rightarrow z=2 \cos \theta+2 i \sin \theta+\frac{1}{2} \cos \theta-\frac{1}{2} i \sin \theta \\
&\Rightarrow z=\frac{5}{2} \cos \theta+\frac{3}{2} i \sin \theta \\
&\Rightarrow|z|=\sqrt{\left(\frac{5}{2}\right)^2+\left(\frac{3}{2}\right)^2} \\
&\Rightarrow|z|=\sqrt{\frac{34}{4}} \\
&\therefore|z|=\sqrt{\frac{17}{2}}
\end{aligned}
\)
Note that ‘z’ is neither purely imaginary and nor purely real.

Hint

(d) Suppose \(z=\frac{3+2 i \sin \theta}{1-2 i \sin \theta}\)
Since, \(z\) is purely imaginary, then \(z+\bar{z}=0\)
\(
\begin{aligned}
&\Rightarrow \quad \frac{3+2 i \sin \theta}{1-2 i \sin \theta}+\frac{3-2 i \sin \theta}{1+2 i \sin \theta}=0 \\
&\Rightarrow \quad \frac{(3+2 i \sin \theta)(1+2 i \sin \theta)+(3-2 i \sin \theta)(1-2 i \sin \theta)}{1+4 \sin ^2 \theta} \\
&=0 \\
&\Rightarrow \quad \sin ^2 \theta=\frac{3}{4} \Rightarrow \sin \theta=\frac{\sqrt{3}}{2} \\
&\Rightarrow \quad \theta=-\frac{\pi}{3}, \frac{\pi}{3}, \frac{2 \pi}{3}
\end{aligned}
\)
Now, the sum of elements in \(A=-\frac{\pi}{3}+\frac{\pi}{3}+\frac{2 \pi}{3}=\frac{2 \pi}{3}\)

Hint

(a) \(|z|=1 \& \operatorname{Re} z \neq 1\)
Suppose \(z=x+i y \Rightarrow x^2+y^2=1\)…..(i)
Now, \(w=\frac{1+(1-8 \alpha) z}{1-z}\)
\(
\begin{aligned}
\Rightarrow & w=\frac{1+(1-8 \alpha)(x+i y)}{1-(x+i y)} \\
\Rightarrow \quad w &=\frac{1+(1-8 \alpha)(x+i y))((1-\mathrm{x})+\mathrm{iy})}{1-(x+i y))((1-\mathrm{x})+\mathrm{iy})} \\
\Rightarrow \quad & w=\frac{\left[\left(1+x(1-8 \alpha)(1-x)-(1-8 \alpha) y^2\right]\right.}{(1-x)^2+y^2} \\
&+i \frac{[(1+x(1-8 \alpha)) y-(1-8 \alpha) y(1-x)]}{(1-x)^2+y^2}
\end{aligned}
\)
As, \(w\) is purely imaginary. So,
\(
\begin{aligned}
&\operatorname{Re} w=\frac{\left[(1+x(1-8 \alpha))(1-x)-(1-8 \alpha) y^2\right]}{(1-x)^2+y^2}=0 \\
&\Rightarrow(1-x)+x(1-8 \alpha)(1-x)=(1-8 \alpha) y^2 \\
&\Rightarrow(1-x)+x(1-8 \alpha)-x^2(1-8 \alpha)=(1-8 x) y^2 \\
&\Rightarrow(1-x)+x(1-8 \alpha)=1-8 \alpha\left[\operatorname{From}(i), x^2+y^2=1\right] \\
&\Rightarrow 1-8 \alpha=1 \\
&\Rightarrow \alpha=0 \\
&\therefore \alpha \in\{0\}
\end{aligned}
\)

Hint

(b) Rationalizing the given expression
\(
\frac{(2+3 i \sin \theta)(1+2 i \sin \theta)}{1+4 \sin ^2 \theta}
\)
For the given expression to be purely imaginary, real part of the above expression should be equal to zero.
\(
\begin{aligned}
&\Rightarrow \frac{2-6 \sin ^2 \theta}{1+4 \sin ^2 \theta}=0 \quad \Rightarrow \sin ^2 \theta=\frac{1}{3} \\
&\Rightarrow \sin \theta=\pm \frac{1}{\sqrt{3}} \Rightarrow \theta=\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)
\end{aligned}
\)

Hint

Let, \(z=r(\cos \theta+i \sin \theta)\)
\(z^5=r^5(\cos 5 \theta+i \sin 5 \theta)\)
\(\operatorname{Im}\left((z)^5\right)=r^5 \sin 5 \theta\)
and \((\operatorname{Im} z)^5=r^5(\sin )^5 \theta\)
\(\frac{\operatorname{Im}\left(z^5\right)}{(\operatorname{Im} z)^5}=\frac{\sin 5 \theta}{(\sin )^5 \theta}=A\) (Let)
\(\Rightarrow \frac{d A}{d \theta}=\frac{5(\sin )^5 \theta \cos 5 \theta-5 \sin 5 \theta(\sin )^4 \theta \cos \theta}{\left((\sin )^5 \theta\right)^2}\)
\(\Rightarrow \frac{d A}{d \theta}=\frac{5(\sin )^4 \theta(\sin \theta \cos 5 \theta-\sin 5 \theta \cos \theta)}{(\sin )^{10} \theta}\)
\(\Rightarrow \frac{d A}{d \theta}=\frac{5}{(\sin )^6 \theta}[\sin (-4 \theta)]=0\)
\(\Rightarrow \theta=\frac{\pi}{4}\)
The minimum value of \(A\) will be at \(\theta=\frac{\pi}{4}\).
\(
\begin{aligned}
&\Rightarrow \frac{\sin 5 \frac{\pi}{4}}{\left(\sin \frac{\pi}{4}\right)^5} \\
&=\frac{\frac{-1}{\sqrt{2}}}{\left(\frac{1}{\sqrt{2}}\right)^5} \\
&=-(\sqrt{2})^4=-4
\end{aligned}
\)

Hint

(d) We know minimum value of \(\left|Z_1+Z_2\right|\) is ||\(Z_1|-| Z_2||\). Thus minimum value of \(\left|Z+\frac{1}{2}\right|\) is \(\left|Z\right|-\frac{1}{2}\) \(\leq\left|Z+\frac{1}{2}\right| \leq|Z|+\frac{1}{2}\)
Since, \(|Z| \geq 2\) therefore
\(
\begin{aligned}
& 2-\frac{1}{2}<\left|Z+\frac{1}{2}\right|<2+\frac{1}{2} \\
\Rightarrow \quad \frac{3}{2} &<\left|Z+\frac{1}{2}\right|<\frac{5}{2}
\Rightarrow \quad 1 <\left|Z\right|<2
\end{aligned}
\)

Hint

(c) \(\operatorname{Given}|\mathrm{z}|=1, \arg z=\theta\)
\(
\begin{aligned}
&\Rightarrow \bar{z}=\frac{1}{z} \\
&\therefore \arg \left(\frac{1+z}{1+\bar{z}}\right)=\arg \left(\frac{1+z}{1+\frac{1}{z}}\right)=\arg (z)=\theta
\end{aligned}
\)

Hint

\(
\text { (d) } \because \frac{w-\bar{w} z}{1-z} \text { is purely real }
\)
\(
\begin{aligned}
&\therefore \overline{\left(\frac{w-\bar{w} z}{1-z}\right)}=\left(\frac{w-\bar{w} z}{1-z}\right) \Rightarrow \frac{\bar{w}-w \bar{z}}{1-\bar{z}}=\frac{w-\bar{w} z}{1-z} \\
&\Rightarrow \bar{w}-\bar{w} z-w \bar{z}+w z \bar{z}=w-w \bar{z}-\bar{w} z+\bar{w} z \bar{z} \\
&\Rightarrow w-\bar{w}=(w-\bar{w})|z|^2 \\
&\Rightarrow|z|^2=1 \quad(\because w=\alpha+\mathrm{i} \beta \text { and } \beta \neq 0) \\
&\Rightarrow|z|=1 \text { and also given that } z \neq 1
\end{aligned}
\)

Hint

(b) \(\left|z_1\right|=12 \Rightarrow z_1\) lies on a circle with centre \((0,0)\) and radius 12 units.
And \(\left|z_2-3-4 i\right|=5 \Rightarrow z_2\) lies on a circle with centre \((3,4)\) and radius 5 units.
From figure, it is clear that \(\left|z_1-z_2\right|\) i.e., distance between \(z_1\) and \(z_2\) will be min when they lie at \(A\) and \(\mathrm{B}\) respectively i.e., \(O, C, B, A\) are collinear as shown.
Then \(z_1-z_2=A B=O A-O B=12-2(5)=2\). As above is the minimum value, we must have \(\left|z_1-z_2\right| \geq 2\).

Hint

(a) Given: \(\quad\left|z_1\right|=\left|z_2\right|=\left|z_3\right|=1\)
Now, \(\left|z_1\right|=1 \Rightarrow\left|z_1\right|^2=1 \Rightarrow z_1 \bar{z}_1=1\)
Similarly \(z_2 \bar{z}_2=1, z_3 \bar{z}_3=1\)
Now, \(\left|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}\right|=1 \Rightarrow\left|\bar{z}_1+\bar{z}_2+\bar{z}_3\right|=1\)
\(
\begin{aligned}
&\Rightarrow\left|\overline{z_1+z_2+z_3}\right|=1 \\
&\Rightarrow\left|z_1+z_2+z_3\right|=1
\end{aligned}
\)

Hint

\(
\text { (a) Given: } \arg (z)<0 \text { (given) } \Rightarrow \arg (z)=-\theta
\)

Now, \(z=r \cos (-\theta)+i \sin (-\theta)=r[\cos (\theta)-i \sin (\theta)]\)
Again \(-z=-r[\cos (\theta)-i \sin (\theta)]\)
\(
\begin{aligned}
&\quad=r[\cos (\pi-\theta)+i \sin (\pi-\theta)] \\
&\therefore \quad \arg (-z)=\pi-\theta ;
\end{aligned}
\)
Thus \(\arg (-z)-\arg (z)=\pi-\theta-(-\theta)=\pi-\theta+\theta=\pi\)

Hint

(d) \((1+i)^{n_1}+\left(1+i^3\right)^{n_1}+\left(1+i^5\right)^{n_2}+\left(1+i^7\right)^{n_2}\) \(=(1+i)^{n_1}+(1-i)^{n_1}+(1+i)^{n_2}+(1-i)^{n_2}\)
Using \(1+i=\sqrt{2}(\cos \pi / 4+i \sin \pi / 4)\)
and \(1-i=\sqrt{2}(\cos \pi / 4-i \sin \pi / 4)\)
We get the given expression as
\(=(\sqrt{2})^{n_1}\left[\cos \frac{n_1 \pi}{4}+i \sin \frac{n_1 \pi}{4}\right]\)
\(+(\sqrt{2})^{n_2}\left[\cos \frac{n_2 \pi}{4}+i \sin \frac{n_2 \pi}{4}\right]\)
\(+(\sqrt{2})^{n_2}\left[\cos \frac{n_2 \pi}{4}-i \sin \frac{n_2 \pi}{4}\right]\)
\(=(\sqrt{2})^{n_1}\left[2 \cos \frac{n_1 \pi}{4}\right]+(\sqrt{2})^{n_2}\left[2 \cos \frac{n_2 \pi}{4}\right]\)
\(=\) real number irrespective the values of \(n_1\) and \(n_2\)
\(\therefore(d)\) is the most appropriate answer.

Hint

(c) Given that \(|z+i \omega|=|z-i \bar{\omega}|\)
\(
\Rightarrow|z-(-i \omega)|=|z-(-\overline{i \omega})|
\)
\(\Rightarrow z\) lies on perpendicular bisector of the line segment joining \((-i \omega)\) and \((-\overline{i \omega})\), which is real axis, \((-i \omega)\) and \((-\overline{i \omega})\) being mirror images of each other.
\(\therefore \quad \operatorname{Im}(z)=0\).
If \(z=x\), then \(|z| \leq 1 \Rightarrow x^2 \leq 1 \Rightarrow-1 \leq x \leq 1\)
\(\therefore \quad\) (c) is the correct option.

Hint

(d) \(\because|z|=|\omega|\) and \(\arg z=\pi-\arg \omega\)
Let \(\omega=r e^{i \theta}\) then \(z=r e^{i(\pi-\theta)}\)
\(
\begin{gathered}
\Rightarrow z=r e^{i \pi} \cdot e^{-i \theta} \\
=\left(r e^{-i \theta}\right)(\cos \pi+i \sin \pi)=\bar{\omega}(-1)=-\bar{\omega}
\end{gathered}
\)

Hint

(d) \(\frac{1+i}{1-i}=\frac{(1+i)}{(1-i)(1+i)}=\frac{1-1+2 i}{2}=i\)
Now \(i^n=1 \Rightarrow\) the smallest positive integral value of \(n\) should be 4.

Hint

Given that \(\left(\frac{1+i}{1-i}\right)^{m / 2}=\left(\frac{1+i}{i-1}\right)^{n / 3}=1\)
\(
\begin{gathered}
\Rightarrow\left(\frac{(1+i)^2}{2}\right)^{m / 2}=\left(\frac{(1+i)^2}{-2}\right)^{n / 3}=1 \\
\Rightarrow i^{m / 2}=(-i)^{n / 3}=1 \\
m(\text { least })=8, n \text { (least) }=12 \\
\operatorname{GCD}(8,12)=4
\end{gathered}
\)

Hint

Given : \(\alpha_{\mathrm{k}}=\cos \frac{k \pi}{7}+i \sin \frac{k \pi}{7}=e^{\frac{i \pi k}{7}}\)
\(
\alpha_{k+1}-\alpha_k=e^{\frac{i \pi(k+1)}{7}}-e^{\frac{i \pi k}{7}}=e^{\frac{i \pi k}{7}}\left(e^{i \pi / 7}-1\right)
\)
\(
\left|\alpha_{k+1}-\alpha_k\right|=\left|e^{i \pi / 7}-1\right|
\)
\(
\begin{gathered}
\Rightarrow \sum_{k=1}^{12}\left|\alpha_{k+1}-\alpha_k\right|=12\left|e^{i \pi / 7}-1\right| \\
\text { Similarly, } \sum_{k=1}^3\left|\alpha_{4 k-1}-\alpha_{4 k-2}\right|=3\left|e^{i \pi / 7}-1\right| \\
\therefore \quad \frac{\sum_{k=1}^{12}\left|\alpha_{k+1}-\alpha_k\right|}{\sum_{k=1}^3\left|\alpha_{4 k-1}-\alpha_{4 k-2}\right|} \\
= 4
\end{gathered}
\)

Hint

Given : \(|z-3-2 i| \leq 2\),
which represents a circular region with centre \((3,2)\) and radius 2.
Now, \(|2 z-6+5 i|=2\left|z-\left(3-\frac{5}{2} i\right)\right|\)
\(=2 \times\) distance of \(z\) from \(P\)
(where \(z\) lies in or on the circle)
Also min distance of \(z\) from \(P=\frac{5}{2}\)
\(\therefore\) Minimum value of \(|2 z-6+5 i|=5\)

Hint

Let \(z=\frac{\sin x / 2+\cos x / 2+i \tan x}{1+2 i \sin x / 2}\) \(=\frac{(\sin x / 2+\cos x / 2+i \tan x)(1-2 i \sin x / 2)}{(1+2 i \sin x / 2)(1-2 i \sin x / 2)}\)
\(
=\frac{\begin{array}{l}
{[\sin x / 2+\cos x / 2+2 \sin x / 2 \tan x)} \\
\left.+i\left(\tan x-2 \sin ^2 x / 2-2 \sin x / 2 \cos x / 2\right)\right]
\end{array}}{\left(1+4 \sin ^2 x / 2\right)}
\)
But it is given that \(z\) is real.
\(
\begin{aligned}
&\therefore I_m(z)=0 \\
&\Rightarrow \tan x-2 \sin \frac{x}{2}\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)=0
\end{aligned}
\)
\(
\begin{aligned}
&\Rightarrow \frac{\sin x}{\cos x}-2 \sin ^2 x / 2-2 \sin x / 2 \cos x / 2=0 \\
&\Rightarrow \frac{\sin x}{\cos x}-(1-\cos x)-\sin x=0 \\
&\Rightarrow \sin x\left[\frac{1}{\cos x}-1\right]-[1-\cos x]=0 \\
&\Rightarrow \quad\left(\frac{1-\cos x}{\cos x}\right) \sin x-[1-\cos x]=0 \\
&\Rightarrow \quad(1-\cos x)\left(\frac{\sin x}{\cos x}-1\right)=0 \\
&\Rightarrow \quad \cos x=1 \quad \Rightarrow \quad x=2 n \pi \\
&\text { and } \tan x=1 \Rightarrow x=n \pi+\pi / 4 \\
&\therefore \quad x=2 n \pi, n \pi+\pi / 4
\end{aligned}
\)

Hint

(True) Let \(z=x+i y\), then \(1 \cap z \Rightarrow 1 \leq x \& 0 \leq y\) (by def.) Consider,
\(\frac{1-z}{1+z}=\frac{1-(x+i y)}{1+(x+i y)}=\frac{(1-x)-i y}{(1+x)+i y} \times \frac{(1+x)-i y}{(1+x)-i y}\)
\(=\frac{1-x^2-y^2}{(1+x)^2+y^2}-\frac{i y(1-x+1+x)}{(1+x)^2+y^2}\)
\(=\frac{1-x^2-y^2}{(1+x)^2+y^2}-\frac{2 i y}{(1+x)^2+y^2}\)
\(\frac{1-z}{1+z} \cap 0 \Rightarrow \frac{1-x^2-y^2}{(1+x)^2+y^2} \leq 0\)
and \(\frac{-2 y}{(1+x)^2+y^2} \leq 0\)
\(\Rightarrow 1-x^2-y^2 \leq 0\) and \(-2 y \leq 0\)
\(\Rightarrow x^2+y^2 \geq 1\) and \(y \geq 0\), which is true as
\(x \geq 1\) and \(y \geq 0\)
Hence, the given statement is true \(\forall z \in C\).

Hint

\(
\text { (b, c) } \quad\left|z^2+z+1\right|=1
\)
\(
\begin{aligned}
&\Rightarrow\left|\left(z+\frac{1}{2}\right)^2+\frac{3}{4}\right|=1\\
&\Rightarrow\left|\left(z+\frac{1}{2}\right)^2+\frac{3}{4}\right| \leq\left|z+\frac{1}{2}\right|^2+\frac{3}{4}\\
&\Rightarrow\left|\left(z+\frac{1}{2}\right)\right|^2 \geq \frac{1}{4}\\
&\Rightarrow\left|z+\frac{1}{2}\right| \geq \frac{1}{2}\\
&\text { also }\left|\left(z^2+z\right)+1\right|=1 \geq|| z^2+z|-1|\\
&\Rightarrow\left|z^2+z\right|-1 \leq 1\\
&\Rightarrow\left|z^2+z\right| \leq 2\\
&\Rightarrow|| z^2|-| z|| \leq\left|z^2+z\right| \leq 2\\
&\Rightarrow\left|r^2-r\right| \leq 2\\
&\Rightarrow r=|z| \leq 2 ; \quad \forall \quad z \in \mathrm{S}\\
&\text { Hence, set ‘ } \mathrm{S} \text { ‘ is infinite }
\end{aligned}
\)

Hint

We have,
\(
s z+t \bar{z}+r=0 \dots(i)
\)
On taking conjugate
\(
\overline{s z}+\overline{t z}+\bar{r}=0 \dots(ii)
\)
On solving Eqs. (i) and (ii), we get
\(
z=\frac{\bar{r} t-r \bar{s}}{|s|^2-|t|^2}
\)
(a) For unique solutions of \(z\)
\(
|s|^2-|t|^2 \neq 0 \Rightarrow|s| \neq|t|
\)
It is true.
(b) If \(|s|=|t|\), then \(\bar{r} t-r \bar{s}\) may or may not be zero.
So, \(z\) may have no solution.
\(\therefore L\) may be an empty set.
It is false.
(c) If elements of set \(L\) represents line, then this line and given circle intersect at maximum two-point.
Hence, it is true.
(d) In the case locus of \(z\) is a line, so \(L\) has infinite elements. Hence, it is true.

Hint

(a, b, d)
(a) \(\arg (-1-i)=\frac{-3 \pi}{4}\)
\(\therefore\) (a) is false
(b) \(f(t)=\arg (-1+i t)=\left[\begin{array}{l}\pi-\tan ^{-1}(t), t \geq 0 \\ -\pi+\tan ^{-1}(t), t<0\end{array}\right.\) \(\lim _{t \rightarrow 0^{-}} f(t)=-\pi\) and \(\lim _{t \rightarrow 0^{+}} f(t)=\pi\)
LHL \(\neq\) RHL \(\Rightarrow f\) is discontinuous at \(t=0\) \(\therefore\) (b) is false.
(c) \(\arg \left(\frac{z_1}{z_2}\right)-\arg z_1+\arg z_2\)
\(=2 n \pi+\arg z_1-\arg z_2-\arg z_1+\arg z_2\)
\(=2 n \pi\), multiple of \(2 \pi\)
\(\therefore\) (c) is true.
(d) \(\arg \left(\frac{\left(z-z_1\right)\left(z_2-z_3\right)}{\left(z-z_3\right)\left(z_2-z_1\right)}\right)=\pi \quad \Rightarrow \frac{\left(z-z_1\right)\left(z_2-z_3\right)}{\left(z-z_3\right)\left(z_2-z_1\right)}=k, \quad k \in R\)
\(
\Rightarrow\left(\frac{z-z_1}{z-z_3}\right)=k\left(\frac{z_2-z_1}{z_2-z_3}\right)
\)
\(\Rightarrow \quad z, z_1, z_2, z_3\) are concyclic. i.e. z lies on a circle. \(\therefore \quad\) (d) is false.

Hint

\((\mathbf{a}, \mathbf{b}) \quad \mathrm{a}-\mathrm{b}=1, \mathrm{y} \neq 0\)
\(\operatorname{Im}\left(\frac{a z+b}{z+1}\right)=y\)
\(
\Rightarrow \operatorname{Im}\left[\frac{a(x+i y)+b}{(x+1)+i y} \times \frac{(x+1)-i y}{(x+1)-i y}\right]=y
\)
\(
\begin{aligned}
&\Rightarrow \frac{-(a x+b) y+a y(x+1)}{(x+1)^2+y^2}=y \\
&\Rightarrow \frac{-a x y-b y+a x y+a y}{(x+1)^2+y^2}=y \\
&\Rightarrow \mathrm{a}-\mathrm{b}=(\mathrm{x}+1)^2+\mathrm{y}^2 \\
&\Rightarrow 1=(\mathrm{x}+1)^2+\mathrm{y}^2, \quad \therefore \mathrm{x}=-1 \pm \sqrt{1-\mathrm{y}^2}
\end{aligned}
\)

Hint

(a,c,d) Given : \(z=(1-t) z_1+t z_2\), where \(0<t<1\) \(\Rightarrow z=\frac{(1-t) z_1+t z_2}{(1-t)+t}\)
\(\Rightarrow \mathrm{Z}\) divides the join of \(z_1\) and \(z_2\) internally in the ratio \(t:(1-t)\).
\(
\therefore z_1, z \text { and } z_2 \text { are collinear }
\)
\(
\Rightarrow\left|z-z_1\right|+\left|z-z_2\right|=\left|z_1-z_2\right|
\)
Also \(z=(1-t) z_1+t z_2\)
\(
\begin{aligned}
&\Rightarrow \frac{z-z_1}{z_2-z_1}=t \text {, which is purely real number } \\
&\therefore \arg \left(\frac{z-z_1}{z_2-z_1}\right)=0 \Rightarrow \arg \left(z-z_1\right)=\arg \left(z_2-z_1\right)
\end{aligned}
\)
Also \(\frac{z-z_1}{z_2-z_1}=t \Rightarrow \frac{\bar{z}-\bar{z}_1}{\bar{z}_2-\bar{z}_1}=t\)
\(
\Rightarrow \frac{z-z_1}{z_2-z_1}=\frac{\bar{z}-\bar{z}_1}{\bar{z}_2-\bar{z}_1}
\)
\(
\Rightarrow\left(z-z_1\right)\left(\bar{z}_2-\bar{z}_1\right)=\left(\bar{z}-\bar{z}_1\right)\left(z_2-z_1\right)
\)
\(
\Rightarrow\left|\begin{array}{ll}
z-z_1 & \bar{z}-\bar{z}_1 \\
z_2-z_1 & \bar{z}_2-\bar{z}_1
\end{array}\right|=0
\)

Hint

\(
\text { (d) Taking }-3 i \text { common from } C_2 \text {, we get }
\)
\(
\begin{aligned}
&-3 i\left|\begin{array}{ccc}
6 i & 1 & 1 \\
4 & -1 & -1 \\
20 & i & i
\end{array}\right|=0 \quad\left(\because C_2 \equiv C_3\right) \\
&\Rightarrow x=0, \quad y=0
\end{aligned}
\)

Hint

(b) \(\sum_{i=1}^{13}\left(i^n+i^{n+1}\right)=\sum_{i=1}^{13} i^n(1+i)=(1+i) \sum_{i=1}^{13} i^n\),
Which forms a G.P.
Sum of G.P. \(=i(1+i) \frac{\left(1-i^{13}\right)}{1-i}=i-1\) as \(i^{13}=i\)

Hint

(d) Let \(z=\cos \frac{2 \pi}{7}+i \sin \frac{2 \pi}{7}\) By DeMoivre’s theorem,
\(
\begin{aligned}
z^k=\cos \frac{2 \pi k}{7}+i \sin \frac{2 \pi k}{7} \\
\text { Now, } & \sum_{k=1}^6\left(\sin \frac{2 \pi k}{7}-i \cos \frac{2 \pi k}{7}\right) \\
=& \sum_{k=1}^6(-i)\left(\cos \frac{2 \pi k}{7}+i \sin \frac{2 \pi k}{7}\right) \\
=&(-i) \sum_{k=1}^6 z^k=-i z \frac{\left(1-z^6\right)}{1-z}=-i\left(\frac{z-z^7}{1-z}\right) \\
=&(-i)\left(\frac{z-1}{1-z}\right)=\left[\because z^7=\cos 2 \pi+i \sin 2 \pi=1\right] \\
=& i\left(\frac{1-z}{1-z}\right)=i
\end{aligned}
\)

Hint

(c) Let \(z_1=r_1\left(\cos \theta_1+i \sin \theta_1\right)\)
and \(z_2=r_2\left(\cos \theta_2+i \sin \theta_2\right)\)
where \(r_1=\left|z_1\right|, r_2=\left|z_2\right|, \theta_1=\arg \left(z_1\right), \theta_2=\arg \left(z_2\right)\)
\(\therefore z_1+z_2=r_1\left(\cos \theta_1+i \sin \theta_1\right)+r_2\left(\cos \theta_2+i \sin \theta_2\right)\)
\(=\left(r_1 \cos \theta_1+r_2 \cos \theta_2\right)+i\left(r_1 \sin \theta_1+r_2 \sin \theta_2\right)\)
\(
\begin{aligned}
& \left|z_1+z_2\right|^2=r_1^2 \cos ^2 \theta_1+r_2^2 \cos ^2 \theta_2+2 r_1 r_2 \cos \theta_1 \cos \theta_2 \\
&+r_1^2 \sin ^2 \theta_1+r_2^2 \sin ^2 \theta_2+2 r_1 r_2 \sin \theta_1 \sin \theta_2 \\
&=r_1^2+r_2^2+2 r_1 r_2 \cos \left(\theta_1-\theta_2\right) \\
&\text { and }\left|z_1\right|+\left|z_2\right|=r_1+r_2 \\
&\text { Given }\left|z_1+z_2\right|=\left|z_1\right|+\left|z_2\right| \\
&\quad \Rightarrow\left|z_1+z_2\right|^2=\left|z_1\right|^2+\left|z_2\right|^2+2\left|z_1\right|\left|z_2\right| \\
&\Rightarrow r_1^2+r_2^2+2 r_1 r_2 \cos \left(\theta_1-\theta_2\right)=r_1^2+r_2^2+2 r_1 r_2 \\
&\Rightarrow \cos \left(\theta_1-\theta_2\right)=1 \Rightarrow \theta_1-\theta_2=0 \\
&\therefore \arg \left(z_1\right)=\arg \left(z_2\right)
\end{aligned}
\)

Hint

(a, d) Let \(z_1=a+i b, a>0\) and \(b \in R ; z_2=c+i d\), \(d<0, c \in R\), then
\(
\begin{gathered}
\left|z_1\right|^2=\left|z_2\right|^2 \Rightarrow a^2+b^2=c^2+d^2 \\
\Rightarrow a^2-c^2=d^2-b^2
\end{gathered}
\)
Now, \(\frac{z_1+z_2}{z_1-z_2}=\frac{(a+c)+i(b+d)}{(a-c)+i(b-d)}\)
\(
\begin{aligned}
&=\frac{\left[\left(a^2-c^2\right)+\left(b^2-d^2\right)\right]+i[(a-c)(b+d)-(a+c)(b-d)]}{(a-c)^2+(b-d)^2} \\
&=\frac{i[(a-c)(b+d)-(a+c)(b-d)]}{(a-c)^2+(b-d)^2} \quad \text { [Using (i)] }
\end{aligned}
\)
Which is purely imaginary number or zero in case
\(
a+c=b+d=0 .
\)

Hint

(a, b, c) \(z_1=a+i b\) and \(z_2=c+i d\).
Acc. to the ques, \(|z_1|^2=\left|z_2\right|^2=1 \dots(i)\)
\(\Rightarrow a^2+b^2=1\) and \(c^2+d^2=1\).
Also \(\operatorname{Re}\left(z_1 \overline{z_2}\right)=0 \Rightarrow a c+b d=0\)
\(
\Rightarrow \frac{a}{b}=\frac{-d}{c}=\alpha(s a y) \dots(ii)
\)
From (i) and (ii), we get
\(
b^2 \alpha^2+b^2=c^2 \alpha^2+c^2 \Rightarrow b^2=c^2 \text {; }
\)

Similarly, \(\quad a^2=d^2\)
\(\therefore \quad\left|\omega_1\right|=\sqrt{a^2+c^2}=\sqrt{c^2+b^2}=1\)
and \(\left|\omega_2\right|=\sqrt{b^2+d^2}=\sqrt{c^2+d^2}=1\)
Also, \(\operatorname{Re}\left(\omega_1 \bar{\omega}_2\right)=a b+c d=(b \alpha) b+c(-c \alpha)\)
\(
=\alpha\left(b^2-c^2\right)=0
\)

Hint

(c) (P) \(\rightarrow(1): z_k=\cos \frac{2 k \pi}{10}+i \sin \frac{2 k \pi}{10}, k=1\) to 9
\(
\therefore z_k=e^{i \frac{2 k \pi}{10}}
\)
Now \(z_k \cdot z_j=1 \Rightarrow z_j=\frac{1}{z_k}=e^{-i \frac{2 k \pi}{10}}=\overline{z_k}\)
We know if \(z_k\) is \(10^{\text {th }}\) root of unity so will be \(\bar{z}_k\).
\(\therefore \quad\) For every \(z_k\), there exist \(z_i=\bar{z}_k\)
Such that \(z_k \cdot z_j=z_k \bar{z}_k=1\)
Hence the statement is true.
(Q) \(\rightarrow\) (2) \(z_1=z_k \Rightarrow z=\frac{z_k}{z_1}\) for \(z_1 \neq 0\)
\(\therefore\) We can always find a solution of \(z_1 \cdot z=z_k\)
Hence the statement is false.
\((\mathrm{R}) \rightarrow(3):\) We know \(z^{10}-1=(z-1)\left(z-z_1\right) \ldots\left(z-z_9\right)\)
\(
\Rightarrow\left(z-z_1\right)\left(z-z_2\right) \ldots . .\left(z-z_9\right)=\frac{z^{10}-1}{z-1}
\)
\(=1+z+z^2+\ldots z^9\)
For \(z=1\), we get \(\left(1-z_1\right)\left(1-z_2\right) \ldots .\left(1-z_9\right)=10\)
\(
\therefore \frac{\left|1-z_1\right|\left|1-z_2\right| \ldots .\left|1-z_9\right|}{10}=1
\)
\((\mathrm{S}) \rightarrow(4): 1, Z_1, Z_2, \ldots, Z_9\) are 10 th roots of unity. \(\therefore Z^{10}-1=0\)
From equation \(1+Z_1+Z_2+\ldots .+Z_9=0\),
\(\operatorname{Re}(1)+\operatorname{Re}\left(Z_1\right)+\operatorname{Re}\left(Z_2\right)+\ldots .+\operatorname{Re}\left(Z_9\right)=0\)
\(\Rightarrow \operatorname{Re}\left(Z_1\right)+\operatorname{Re}\left(Z_2\right)+\ldots . \operatorname{Re}\left(Z_9\right)=-1\)
\(\Rightarrow \sum_{K=1}^9 \cos \frac{2 k \pi}{10}=-1 \Rightarrow 1-\sum_{K=1}^9 \cos \frac{2 k \pi}{10}=2\)
Hence (c) is the correct option.

Hint

\(
\begin{aligned}
&S_1: x^2+y^2<16 \\
&S_2: \operatorname{Im}\left[\frac{(x-1)+i(y+\sqrt{3})}{1-i \sqrt{3}}\right]>0 \\
&\quad \Rightarrow \sqrt{3}(x-1)+(y+\sqrt{3})>0 \Rightarrow y+\sqrt{3} x>0 \\
&S_3: x>0
\end{aligned}
\)
Then \(S: S_1 \cap S_2 \cap S_3\) is as shown in the figure given below.

(b) Area of shaded region
\(
=\frac{\pi}{4} \times 4^2+\frac{\pi \times 4^2 \times 60^{\circ}}{360^{\circ}}=4 \pi+\frac{8 \pi}{3}=\frac{20 \pi}{3}
\)

Hint

\(
\begin{aligned}
&S_1: x^2+y^2<16 \\
&S_2: \operatorname{Im}\left[\frac{(x-1)+i(y+\sqrt{3})}{1-i \sqrt{3}}\right]>0 \\
&\quad \Rightarrow \sqrt{3}(x-1)+(y+\sqrt{3})>0 \Rightarrow y+\sqrt{3} x>0 \\
&S_3: x>0
\end{aligned}
\)
Then \(S: S_1 \cap S_2 \cap S_3\) is as shown in the figure given below.

(c) \(\min _{z \in s}|1-3 i-z|=\min\) distance between \(z\) and \((1,-3)\)
Clearly (from figure) minimum distance between \(\mathrm{z} \in \mathrm{S}\) and \((1,-3)\) from line \(y+x \sqrt{3}=0\) i.e. \(\left|\frac{\sqrt{3}-3}{\sqrt{3+1}}\right|=\frac{3-\sqrt{3}}{2}\)

Hint

Given : \(A=\{z: \operatorname{Im}(z) \geq 1\}=\{(x, y): y \geq 1\}\)
Clearly \(A\) is the set of all points lying on or above the line \(y=1\) in cartesian plane.
\(
B=\{z:|z-2-i|=3\}=\left\{(x, y):(x-2)^2+(y-1)^2=9\right\}
\)
\(\Rightarrow B\) is the set of all points lying on the boundary of the circle with centre \((2,1)\) and radius 3.
\(C=\{z: \operatorname{Re}[(1-i) z]=\sqrt{2}\}=\{(x, y): x+y=\sqrt{2}\}\)
\(\Rightarrow \mathrm{C}\) is the set of all points lying on the straight line represented by \(x\) \(+y=\sqrt{2}\).
Graphically, the three sets are represented as shown below:

(b) From graph \(A \cap B \cap C\) consists of only one point \(P\) [the common point of the region \(\mathrm{y} \geq 1,(x-2)^2+(y-1)^2=9\) and \(\left.x+y=\sqrt{2}\right] \therefore n(A\) \(\cap B \cap C)=1\)

Hint

Given : \(A=\{z: \operatorname{Im}(z) \geq 1\}=\{(x, y): y \geq 1\}\)
Clearly \(A\) is the set of all points lying on or above the line \(y=1\) in cartesian plane.
\(
B=\{z:|z-2-i|=3\}=\left\{(x, y):(x-2)^2+(y-1)^2=9\right\}
\)
\(\Rightarrow B\) is the set of all points lying on the boundary of the circle with centre \((2,1)\) and radius 3.
\(C=\{z: \operatorname{Re}[(1-i) z]=\sqrt{2}\}=\{(x, y): x+y=\sqrt{2}\}\)
\(\Rightarrow \mathrm{C}\) is the set of all points lying on the straight line represented by \(x\) \(+y=\sqrt{2}\).
Graphically, the three sets are represented as shown below:

(c) Since, \(z\) is a point of \(A \cap B \cap C \Rightarrow z\) represents the point \(P\)
\(
\therefore|z+1-i|^2+|z-5-i|^2
\)
\(
\Rightarrow|z-(-1+i)|^2+|z-(5+i)|^2
\)
\(\Rightarrow \mathrm{PQ}^2+\mathrm{PR}^2=\mathrm{QR}^2=6^2=36\), which lies between 35 and 39
\(\therefore\) (c) is correct option.

Hint

Given : \(A=\{z: \operatorname{Im}(z) \geq 1\}=\{(x, y): y \geq 1\}\)
Clearly \(A\) is the set of all points lying on or above the line \(y=1\) in cartesian plane.
\(
B=\{z:|z-2-i|=3\}=\left\{(x, y):(x-2)^2+(y-1)^2=9\right\}
\)
\(\Rightarrow B\) is the set of all points lying on the boundary of the circle with centre \((2,1)\) and radius 3.
\(C=\{z: \operatorname{Re}[(1-i) z]=\sqrt{2}\}=\{(x, y): x+y=\sqrt{2}\}\)
\(\Rightarrow \mathrm{C}\) is the set of all points lying on the straight line represented by \(x\) \(+y=\sqrt{2}\).
Graphically, the three sets are represented as shown below:

(d) Given : \(|w-2-i|<3\)
\(\Rightarrow\) Distance between \(w\) and \(2+i\) i.e. \(S\) is smaller than 3 .
\(\Rightarrow w\) is a point lying inside the circle with centre \(S\) and radius 3 .
\(\Rightarrow\) Distance between \(z\) (i.e. the point \(P\) ) and \(w\) should be smaller than 6 (the diameter of the circle)
i.e. \(|z-w|<6\)
But we know that \(\|\mathrm{z}|-| \mathrm{w}\|<|\mathrm{z}-\mathrm{w}|\)
\(
\Rightarrow|| \mathrm{z}|-| \mathrm{w} \|<6 \Rightarrow-6<|\mathrm{z}|-|\mathrm{w}|<6
\)
\(
-3<|\mathrm{z}|-|\mathrm{w}|+3<9
\)

Hint

(b) Let \(z=x+i y, \bar{z}=x-i y\)
Now, \(z=1-\bar{z}\)
\(\Rightarrow x+i y=1-(x-i y)\)
\(\Rightarrow 2 x=1 \Rightarrow x=\frac{1}{2}\)
Now, \(|z|=1 \Rightarrow x^2+y^2=1 \Rightarrow y^2=1-x^2\)
\(
\Rightarrow y=\pm \frac{\sqrt{3}}{2}
\)
Now, \(\tan \theta=\frac{y}{x}\) ( \(\theta\) is the argument \()\)
\(=\frac{\sqrt{3}}{2} \div \frac{1}{2} \quad\) (+ve since only principal argument)
\(
\begin{aligned}
&=\sqrt{3} \\
&\Rightarrow \quad \theta=\tan ^{-1} \sqrt{3}=\frac{\pi}{3}
\end{aligned}
\)
Hence, \(z\) is not a real number
So, statement-1 is false and 2 is true.

Hint

\(
\begin{aligned}
&z_1=a+b i \\
&z_2=c+d i
\end{aligned}
\)
By given condition:
\(
a^2+b^2<1<c^2+d^2
\)
To prove: \(\left|1-z_1 \bar{z}_2\right|<\left|z_1-z_2\right|\)
\(
\begin{aligned}
|1-(a+b i)(c-d i)|<|a-c+(b-d) i| \\
|1-a c+b d+(a d-b c) i|<\sqrt{(a-c)^2+(b-d)^2} \\
(1-a c+b d)^2+(a d-b c)^2<(a-c)^2+(b-d)^2 \\
1+\left(a^2+b^2\right)\left(c^2+d^2\right)+4 b d-4 a b c d<a^2+b^2+c^2+d^2 \\
\left(a^2+b^2\right)\left(c^2+d^2\right)<a^2+b^2 \text { From given first half of inequality } \\
1<\mathrm{c}^2+d^2 \text { From second half } \\
b d<a b c d \text { Since its all positive } \\
\Rightarrow \text { LHS }<c^2+d^2+\left(a^2+b^2\right)+0=\text { RHS }
\end{aligned}
\)
Hence proved

Hint

Given \(: z_1=10+6 i\) and \(z_2=4+6 i\)
Also \(\arg \left(\frac{z-z_1}{z-z_2}\right)=\frac{\pi}{4}\)
\(
\begin{aligned}
&\Rightarrow \arg \left(z-z_1\right)-\arg \left(z-z_2\right)=\frac{\pi}{4} \\
&\Rightarrow \quad \arg ((x+i y)-(10+6 i))-\arg ((x+i y)-(4+6 i))=\frac{\pi}{4} \\
&\Rightarrow \quad \arg [(x-10)+i(y-6)]-\arg [(x-4)+i(y-6)]=\frac{\pi}{4} \\
&\Rightarrow \quad \tan ^{-1}\left(\frac{y-6}{x-10}\right)-\tan ^{-1}\left(\frac{y-6}{x-4}\right)=\frac{\pi}{4} \\
&\Rightarrow \quad \tan ^{-1}\left(\frac{\frac{y-6}{x-10}-\frac{y-6}{x-4}}{1+\frac{(y-6)^2}{(x-4)(x-10)}}\right)=\frac{\pi}{4} \\
&\Rightarrow \quad \frac{(x-4)(y-6)-(x-10)(y-6)}{(x-4)(x-10)+(y-6)^2}=\tan \frac{\pi}{4} \\
&\Rightarrow \quad(x-4-x+10)(y-6)=(x-4)(x-10)+(y-6)^2 \\
&\Rightarrow 6 y-36=x^2+y^2-14 x-12 y+40+36 \\
&\Rightarrow \quad x^2+y^2-14 x-18 y+112=0
\end{aligned}
\)
\(
\begin{aligned}
&\Rightarrow\left(x^2-14 x+49\right)+\left(y^2-18 y+81\right)=18 \\
&\Rightarrow(x-7)^2+(y-9)^2=(3 \sqrt{2})^2 \\
&\Rightarrow|(x+i y)-(7+9 i)|=3 \sqrt{2} \Rightarrow|z-(7+9 i)|=3 \sqrt{2}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\frac{(1+i) x-2 i}{3+i}+\frac{(2-3 i) y+i}{3-i}=i \\
&\Rightarrow \quad(4+2 i) x-6 i-2+(9-7 i) y+3 i-1=10 i \\
&\Rightarrow \quad(4 x+9 y-3)+(2 x-7 y-3) i=10 i \\
&\Rightarrow 4 x+9 y-3=0 \text { and } 2 x-7 y-3=10
\end{aligned}
\)
On solving these two equations, we get \(x=3, y=-1\)

Hint

Let \(A=z=x+i y, B=i z=-y+i x\), \(C=z+i z=(x-y)+i(x+y)\)
Now, area of \(\triangle A B C=\frac{1}{2}\left|\begin{array}{ccc}x & y & 1 \\ -y & x & 1 \\ x-y & x+y & 1\end{array}\right|\)
On applying, \(R_2-R_1, R_3-R_1\), we get
\(
\begin{aligned}
&\Delta=\frac{1}{2}\left|\begin{array}{ccc}
x & y & 1 \\
-y-x & x-y & 0 \\
-y & x & 0
\end{array}\right| \\
&=\frac{1}{2}\left|-x y-x^2+x y-y^2\right|=\frac{1}{2}\left|-x^2-y^2\right| \\
&=\frac{1}{2}\left|x^2+y^2\right|=\frac{1}{2}|z|^2
\end{aligned}
\)

Hint

Given : \(x+i y=\sqrt{\frac{a+i b}{c+i d}}\)
\(
\Rightarrow(x+i y)^2=\frac{a+i b}{c+i d} \dots(i)
\)
Taking conjugate on both sides, we get
\(
(x-i y)^2=\frac{a-i b}{c-i d} \dots(ii)
\)
On multiplying (i) and (ii), we get
\(
\left(x^2+y^2\right)^2=\frac{a^2+b^2}{c^2+d^2}
\)

Hint

\(
\begin{aligned}
\frac{1}{1-\cos \theta+2 i \sin \theta} \\
=\frac{1}{2 \sin ^2 \theta / 2+4 i \sin \theta / 2 \cos \theta / 2} \\
=\frac{1}{2 \sin \theta / 2} \left[\frac{\sin \theta / 2-2 i \cos \theta / 2}{(\sin \theta / 2+2 i \cos \theta / 2)(\sin \theta / 2-2 i \cos \theta / 2)}\right] \\
=\frac{1}{2 \sin \theta / 2}\left[\frac{\sin \theta / 2-2 i \cos \theta / 2}{\left(\sin ^2 \theta / 2+4 \cos ^2 \theta / 2\right.}\right] \\
=\frac{1}{2 \sin \theta / 2}\left[\frac{2 \sin \theta / 2-4 i \cos \theta / 2}{1-\cos \theta+4+4 \cos \theta}\right] \\
=\frac{1}{2 \sin \theta / 2}\left[\frac{2 \sin \theta / 2-4 i \cos \theta / 2}{5+3 \cos \theta}\right] \\
=\left(\frac{1}{5+3 \cos \theta}\right)+\left(\frac{-2 \cot \theta / 2}{5+3 \cos \theta}\right) i \\
\text { which is of the form } x+i y .
\end{aligned}
\)

Hint

(c) Let \(z=x+i y\)
\(
\begin{aligned}
&\because z^2=i|z|^2 \\
&\therefore x^2-y^2+2 i x y=i\left(x^2+y^2\right) \\
&\Rightarrow x^2-y^2=0 \text { and } 2 x y=x^2+y^2 \\
&\Rightarrow(x-y)(x+y)=0 \text { and }(x-y)^2=0 \\
&\Rightarrow x=y
\end{aligned}
\)

Hint

\(
\text { (a) Given that, } \alpha=\frac{-1+\sqrt{3} i}{2}=\omega
\)
\(
\begin{aligned}
&\therefore(2+\omega)^4=a+b \omega \Rightarrow\left(4+\omega^2+4 \omega\right)^2=a+b \omega \\
&\Rightarrow\left(\omega^2+4(1+\omega)\right)^2=a+b \omega \\
&\Rightarrow\left(\omega^2-4 \omega^2\right)^2=a+b \omega \quad \quad \quad \quad\left(\because 1+\omega=-\omega^2\right]\\
&\Rightarrow\left(-3 \omega^2\right)^2=a+b \omega \Rightarrow 9 \omega^4=a+b \omega \\
&\Rightarrow 9 \omega=a+b \omega \quad\left(\because \omega^3=1\right)\\
&\text { On comparing, } a=0, b=9 \\
&\Rightarrow a+b=0+9=9 .
\end{aligned}
\)

Hint

(c) \(\left(\frac{1+\cos \frac{5 \pi}{18}+i \sin \frac{5 \pi}{18}}{1+\cos \frac{5 \pi}{18}-i \sin \frac{5 \pi}{18}}\right)^3\)
\(
=\left(\frac{2 \cos ^2 \frac{5 \pi}{36}+i 2 \sin \frac{5 \pi}{36} \cdot \cos \frac{5 \pi}{36}}{2 \cos ^2 \frac{5 \pi}{36}-i 2 \sin \frac{5 \pi}{36} \cdot \cos \frac{5 \pi}{36}}\right)^3
\)
\(
=\left(\frac{\cos \frac{5 \pi}{36}+i \sin \frac{5 \pi}{36}}{\cos \frac{5 \pi}{36}-i \sin \frac{5 \pi}{36}}\right)^3=\left(\cos \frac{5 \pi}{36}+i \sin \frac{5 \pi}{36}\right)^6
\)
\(
=\cos \left(6 \times \frac{5 \pi}{36}\right)+i \sin \left(6 \times \frac{5 \pi}{36}\right)=\cos \frac{5 \pi}{6}+i \sin \frac{5 \pi}{6}
\)
\(
=-\frac{\sqrt{3}}{2}+i \frac{1}{2}=-\frac{1}{2}(\sqrt{3}-i)
\)

Hint

(b) \(3+2 \sqrt{-54}=3+6 \sqrt{6} i\)
Let \(\sqrt{3+6 \sqrt{6} i}=a+i b\)
\(
\begin{aligned}
&\Rightarrow a^2-b^2=3 \text { and } a b=3 \sqrt{6} \\
&\Rightarrow a^2+b^2=\sqrt{\left(a^2-b^2\right)^2+4 a^2 b^2}=15
\end{aligned}
\)
So, \(a=\pm 3\) and \(b=\pm \sqrt{6}\)
\(
\sqrt{3+6 \sqrt{6} i}=\pm(3+\sqrt{6} i)
\)
Similarly, \(\sqrt{3-6 \sqrt{6} i}=\pm(3-\sqrt{6} i)\)
\(\operatorname{lm}(\sqrt{3+6 \sqrt{6} i}-\sqrt{3-6 \sqrt{6} i})=\pm 2 \sqrt{6}\)

Hint

(b) Let \(\alpha=\omega, b=1+\omega^3+\omega^6+\ldots \ldots=101\)
\(a=(1+\omega)\left(1+\omega^2+\omega^4+\ldots \omega \omega^{198}+\omega^{200}\right)\)
\(
\begin{aligned}
&=(1+\omega) \frac{\left(1-\left(\omega^2\right)^{101}\right)}{1-\omega^2}=\frac{(\omega+1)\left(\omega^{202}-1\right)}{\left(\omega^2-1\right)} \\
&\Rightarrow \quad a=\frac{(1+\omega)(1-\omega)}{1-\omega^2}=1
\end{aligned}
\)
Required equation \(=x^2-(101+1) x+(101) \times 1=0\) \(\Rightarrow x^2-102 x+101=0\)

Hint

(d) \(z=x+i y\)
\(
\begin{aligned}
&\left(\frac{z-1}{2 z+i}\right)=\frac{(x-1)+i y}{2(x+i y)+i} \\
&=\frac{(x-1)+i y}{2 x+(2 y+1) i} \times \frac{2 x-(2 y+1) i}{2 x-(2 y+1) i} \\
&\operatorname{Re}\left(\frac{z+1}{2 z+i}\right)=\frac{2 x(x-1)+y(2 y+1)}{(2 x)^2+(2 y+1)^2}=1 \\
&\Rightarrow\left(x+\frac{1}{2}\right)^2+\left(y+\frac{3}{4}\right)^2=\left(\frac{\sqrt{5}}{4}\right)^2
\end{aligned}
\)

Hint

(d) \(\frac{\sqrt{3}}{2}+\frac{i}{2}=-i\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)=-i \omega\)
where \(\omega\) is imaginary cube root of unity.
Now, \(\left(1+i z+z^5+i z^8\right)^9\)
\(=\left(1+\omega-i \omega^2+i \omega^2\right)^9=(1+\omega)^9\) \(=\left(-\omega^2\right)^9=-\omega^{18}=-1\)
\(\left(\therefore 1+\omega+\omega^2=0\right)\)

Hint

\(
\text { (a) }-(6+i)^3=x+i y
\)
\(
\begin{aligned}
&-\left[216+i^3+18 i(6+i)\right]=x+i y \\
&-[216-i+108 i-18]=x+i y \\
&-216+i-108 i+18=x+i y \\
&-198-107 i=x+i y \\
&x=-198, y=-107 \\
&y-x=-107+198=91
\end{aligned}
\)

Hint

(d) \(\mathrm{S}:|\mathrm{z}-2+\mathrm{i}| \geq \sqrt{5}\) represents boundary and outer region of circle with centre \((2,-1)\) and radius \(\sqrt{5}\) units.
\(\mathrm{z}_0 \in \mathrm{S}\), such that \(\sqrt{5}\) is the maximum.
\(\therefore\left|\mathrm{z}_0-1\right|\) is minimum
\(\mathrm{z}_0 \in \mathrm{S}\) with \(\left|\mathrm{z}_0-1\right|\) as minimum will be a point on boundary of circle of region \(\mathrm{S}\) which lies on radius of this circle, which passes through (1, \(0)\).
\(\therefore \quad z_0, 1,2-i\) are collinear, or \(\left(x_0, y_0\right),(1,0),(2,-1)\) are collinear.
\(\therefore \quad\) Using slopes of paralled lines, \(x\) ‘ \(\frac{y_0}{x_0-1}=\frac{-1}{2-1} \Rightarrow y_0=1-x_0\)
Now
\(
\begin{aligned}
&\frac{4-z_0-\bar{z}_0}{z_0-\bar{z}_0+2 i}=\frac{4-\left(z_0+\bar{z}_0\right)}{\left(z_0-\bar{z}_0\right)+2 i} \\
&=\frac{4-2 x_0}{2 i y_0+2 i}=\frac{4-2 x_0}{2 i-2 x_0 i+2 i}
\end{aligned}
\)
\(
\begin{aligned}
==& \frac{4-2 x_0}{2 i y_0+2 i}=\frac{4-2 x_0}{2 i-2 x_0 i+2 i} \\
& \quad \therefore \quad \operatorname{Arg}\left(\frac{4-z_0-\bar{z}_0}{z_0-\bar{z}_0-2 i}\right)=\operatorname{Arg}(-i)=\frac{-\pi}{2}
\end{aligned}
\)

Hint

d) Let \(l=\left(\frac{1+i \sqrt{3}}{1-i \sqrt{3}}\right)\).
\(
\begin{aligned}
&\therefore l=\left(\frac{1+i \sqrt{3}}{1-i \sqrt{3}}\right) \times\left(\frac{1+i \sqrt{3}}{1+i \sqrt{3}}\right) \\
&=\left(\frac{-2+i 2 \sqrt{3}}{4}\right)=\left(\frac{1-i \sqrt{3}}{-2}\right)
\end{aligned}
\)
Also, \(l=\left(\frac{1+i \sqrt{3}}{1-i \sqrt{3}}\right) \times\left(\frac{1-i \sqrt{3}}{1-i \sqrt{3}}\right)\)
\(
=\left(\frac{4}{-2-i 2 \sqrt{3}}\right)=\left(\frac{-2}{1+i \sqrt{3}}\right)
\)
\(\left(\frac{1+i \sqrt{3}}{1-i \sqrt{3}}\right)^3=\left(\frac{1+i \sqrt{3}}{1-i \sqrt{3}}\right) \times\left(\frac{1+i \sqrt{3}}{1-i \sqrt{3}}\right) \times\left(\frac{1+i \sqrt{3}}{1-i \sqrt{3}}\right)\)
\(
=\left(\frac{1+i \sqrt{3}}{1-i \sqrt{3}}\right) \times\left(\frac{-2}{1+i \sqrt{3}}\right) \times\left(\frac{1-i \sqrt{3}}{-2}\right)=1
\)
\(\therefore\) least positive integer \(\mathrm{n}\) is 3 .

Hint

\(
\begin{aligned}
&(b) \left|\frac{z_1-2 z_2}{2-z_1 \bar{z}_2}\right|=1 \\
&\Rightarrow\left|z_1-2 z_2\right|^2=\left|2-z_1 \bar{z}_2\right|^2 \\
&\Rightarrow \quad\left(z_1-2 z_2\right) \overline{\left(z_1-2 z_2\right)}=\left(2-z_1 \bar{z}_2\right)\left(\overline{2-z_1 \bar{z}_2}\right) \\
&\Rightarrow \quad\left(z_1-2 z_2\right)\left(\bar{z}_1-2 \bar{z}_2\right)=\left(2-z_1 \bar{z}_2\right)\left(2-\bar{z}_1 z_2\right) \\
&\Rightarrow \quad\left(z_1 \bar{z}_1\right)-2 z_1 \bar{z}_2-2 \bar{z}_1 z_2+4 z_2 \bar{z}_2 \\
&=4-2 \bar{z}_1 z_2-2 z_1 \bar{z}_2+z_1 \bar{z}_1 z_2 \bar{z}_2
\end{aligned}
\)
\(
\begin{aligned}
&\Rightarrow\left|\mathrm{z}_1\right|^2+4\left|\mathrm{z}_2\right|^2=4+\left|\mathrm{z}_1\right|^2\left|\mathrm{z}_2\right|^2 \\
&\Rightarrow\left|\mathrm{z}_1\right|^2+4\left|\mathrm{z}_2\right|^2-4-\left|\mathrm{z}_1\right|^2\left|\mathrm{z}_2\right|^2=0
\end{aligned}
\)
\(
\left(\left|z_1\right|^2-4\right)\left(1-\left|z_2\right|^2\right)=0
\)
\(
\begin{aligned}
&\quad \because \quad\left|z_2\right| \neq 1 \\
&\therefore \quad\left|z_1\right|^2=4 \\
&\Rightarrow \quad\left|z_1\right|=2
\end{aligned}
\)
\(\Rightarrow \quad\) Point \(z_1\) lies on circle of radius 2.

Hint

(c) Since, \(\alpha\) lies on the circle \(\left(x-x_0\right)^2+\left(y-y_0\right)^2=r^2\)
\(
\therefore \quad\left|\alpha-z_0\right|^2=r^2
\)
\(
\begin{aligned}
&\Rightarrow\left(\alpha-z_0\right)\left(\bar{a}-\bar{z}_0\right)=r^2 \\
&\Rightarrow \alpha \bar{\alpha}-\alpha \bar{z}_0-\bar{\alpha} z_0+z_0 \bar{z}_0=r^2 \\
&\Rightarrow|\alpha|^2+\left|z_0\right|^2-\alpha \bar{z}_0-\bar{\alpha} z_0=r^2
\end{aligned}
\)
Also \(\frac{1}{\bar{\alpha}}\) lies on the circle \(\left(x-x_0\right)^2+\left(y-y_0\right)^2=4 r^2\)
\(
\therefore\left|\frac{1}{\bar{\alpha}}-z_0\right|^2=4 r^2 \Rightarrow \quad\left(\frac{1}{\bar{\alpha}}-z_0\right)\left(\frac{1}{\alpha}-\bar{z}_0\right)=4 r^2
\)
\(
\begin{aligned}
&\Rightarrow \quad \frac{1}{\alpha \bar{\alpha}}-\frac{z_0}{\alpha}-\frac{\bar{z}_0}{\bar{\alpha}}+z_0 \bar{z}_0=4 r^2 \\
&\Rightarrow \quad \frac{1}{|\alpha|^2}-\frac{z_0 \bar{\alpha}}{|\alpha|^2}-\frac{\bar{z}_0 \alpha}{|\alpha|^2}+\left|z_0\right|^2=4 r^2 \\
&\Rightarrow \quad 1+|\alpha|^2\left|z_0\right|^2-z_0 \bar{\alpha}-\bar{z}_0 \alpha=4 r^2|\alpha|^2
\end{aligned}
\)
On subtracting equation (i) from (ii), we get
\(
1-|\alpha|^2+\left|z_0\right|^2\left(|\alpha|^2-1\right)=r^2\left(4|\alpha|^2-1\right)
\)
or \(\quad\left(|\alpha|^2-1\right)\left(\left|z_0\right|^2-1\right)=r^2\left(4|\alpha|^2-1\right)\)
Using \(\left|z_0\right|^2=\frac{r^2+2}{2}\), we get
\(
\begin{gathered}
\quad\left(|\alpha|^2-1\right) \frac{r^2}{2}=r^2\left(4|\alpha|^2-1\right) \\
\Rightarrow \quad|\alpha|^2-1=8|\alpha|^2-2 \Rightarrow|\alpha|=\frac{1}{\sqrt{7}}
\end{gathered}
\)

Hint

\(
\begin{aligned}
&\text { Clearly, this equation does not have real roots if }\\
&\begin{aligned}
& \mathrm{D}<0 \\
& \Rightarrow 1-4(1-\mathrm{a})<0 \\
& \Rightarrow 4 \mathrm{a}<3 \\
& \Rightarrow a<3 / 4
\end{aligned}
\end{aligned}
\)

Hint

(a) Given : \(z=x+i y\), where \(x\) and \(y\) are integer
Also, \(z \bar{z}^3+z^3\bar{z}=350 \Rightarrow|z|^2\left(\bar{z}^2+z^2\right)=350\)
\(
\begin{aligned}
&\Rightarrow \quad\left(x^2+y^2\right)\left(x^2-y^2\right)=175 \\
&\Rightarrow \quad\left(x^2+y^2\right)\left(x^2-y^2\right)=25 \times 7  \dots(i)\\
&\text { or } \quad\left(x^2+y^2\right)\left(x^2-y^2\right)=35 \times 5 \dots(ii)
\end{aligned}
\)
\(\because x\) and \(y\) are integers,
\(\therefore x^2+y^2=25 \quad\) and \(x^2-y^2=7 \quad\) [From eq (i)]
\(\Rightarrow x^2=16\) and \(y^2=9\)
\(\Rightarrow x=\pm 4\) and \(y=\pm 3\)
\(\therefore \quad\) Vertices of rectangle are
\((4,3),(4,-3),(-4,-3),(-4,3)\).
\(\therefore \quad\) Area of rectangle \(=8 \times 6=48\) sq. units
Now from eq. (ii),
\(x^2+y^2=35\) and \(x^2-y^2=5\)
\(\Rightarrow x^2=20\), which is not possible for any integral value of \(x\)

Hint

\(
\begin{aligned}
&\text { (d) } \quad z=\cos \theta+i \sin \theta \\
&\Rightarrow \quad z^{2 m-1}=(\cos \theta+i \sin \theta)^{2 m-1} \\
&=\cos (2 m-1) \theta+i \sin (2 m-1) \theta
\end{aligned}
\)
\(
\begin{aligned}
&\left[\begin{array}{l}
\text { By De Moivre’s theorem : } \\
(\cos \theta+i \sin \theta)^n=\cos n \theta+i \sin n \theta
\end{array}\right]\\
&\therefore \quad \operatorname{Im}\left(z^{2 m-1}\right)=\sin (2 m-1) \theta\\
&\therefore \quad \sum_{m=1}^{15} \operatorname{Im}\left(z^{2 m-1}\right)=\sum_{m=1}^{15} \sin (2 m-1) \theta\\
&=\sin \theta+\sin 3 \theta+\sin 5 \theta+\ldots \ldots+\text { upto } 15 \text { terms }\\
&=\frac{\sin \left[15\left(\frac{2 \theta}{2}\right)\right] \cdot \sin [\theta+14 \times \theta]}{\sin \theta}\\
&\left[\begin{array}{l}
\because \sin \alpha+\sin (\alpha+\beta)+\sin (\alpha+2 \beta)+\ldots n \text { terms } \\
=\frac{\sin (n \beta / 2) \cdot \sin [\alpha+(n-1) \beta / 2]}{\sin (\beta / 2)}
\end{array}\right]\\
&=\frac{\sin 15 \theta \cdot \sin 15 \theta}{\sin \theta}=\frac{\sin 30^{\circ} \cdot \sin 30^{\circ}}{\sin 2^{\circ}}=\frac{1}{4 \sin 2^{\circ}}
\end{aligned}
\)

Hint

(d) The initial position of point is \(Z_0=1+2 i\)
\(
\therefore Z_1=(1+5)+(2+3) i=6+5 i
\)
Now \(Z_1\) is moved through a distance of \(\sqrt{2}\) units in the direction \(\hat{i}+\hat{j}\). (i.e. by \(1+i)\)
\(\therefore\) It becomes \(Z_1^{\prime}=Z_1+(1+i)=7+6 i\)
Now \(\mathrm{OZ}_1\) ‘ is rotated through an angle \(\frac{\pi}{2}\) in anticlockwise direction, therefore \(Z_2=i Z_1{ }^{\prime}=-6+7 i\)

Hint

(d) Given : \(|z|=1\) and \(z \neq \pm 1\)
To find the locus of \(\omega=\frac{z}{1-z^2}\)
Now, \(\omega=\frac{z}{1-z^2}=\frac{z}{z \bar{z}-z^2}\)
\(
\left[\because|z|=1 \Rightarrow|z|^2=z \bar{z}=1\right]
\)
\(=\frac{1}{\bar{z}-z}=\) purely imaginary number
\(
\therefore \omega \text { must lie on } y \text {-axis. }
\)

Hint

\(
\begin{aligned}
&\text { (d) } \overrightarrow{O P}=\overrightarrow{O A}+\overrightarrow{A P} \\
&\Rightarrow \overrightarrow{O P}=\overrightarrow{O A}+\overrightarrow{O B} \\
&\Rightarrow \overrightarrow{O P}=3 e^{i \pi / 4}+4 e^{i(\pi / 2+\pi / 4)} \\
&=3 e^{i \pi / 4}+4 e^{i \pi / 2} \cdot e^{i \pi / 4} \\
&=3 e^{i \pi / 4}+4 i e^{i \pi / 4}=e^{i \pi / 4}(3+4 i) .
\end{aligned}
\)

Hint

(b) Given that \(a, b, c\) are integers not all equal and \(\omega\) is cube root of unity \(\neq 1\), then \(\left|a+b \omega+c \omega^2\right|\)
\(=\left|a+b\left(\frac{-1+i \sqrt{3}}{2}\right)+c\left(\frac{-1-i \sqrt{3}}{2}\right)\right|\)
\(=\left|\left(\frac{2 a-b-c}{2}\right)+i\left(\frac{b \sqrt{3}-c \sqrt{3}}{2}\right)\right|\)
\(=\frac{1}{2} \sqrt{(2 a-b-c)^2+3(b-c)^2}\)
\(=\sqrt{\frac{1}{2}\left[(a-b)^2+(b-c)^2+(c-a)^2\right]}\)
R.H.S. will be minimum when \(a=b=c\), but according to the question, we cannot take \(a=b=c\).
The minimum value is obtained when any two are zero and third is a minimum magnitude integer i.e. 1.

Hint

(a) In the figure, we see that.
\(
A B=A C=A D=2
\)
\(\therefore B C D\) is an arc of a circle with centre at \(A\) and radius 2 . Shaded region is exterior part of this sector \(A B C D A\).
\(\therefore\) For any point represented by \(z\) on arc \(B C D\) we should have \(|z-(-1)|=2\)
and for shaded region, \(|z+1|>2\)
For shaded region, we also have
\(
-\pi / 4<\arg (z+1)<\pi / 4
\)
or \(|\arg (z+1)|<\pi / 4\)
From (i) and (ii), we get (a) is the correct option.

Hint

(b) \(\left(1+\omega^2\right)^n=\left(1+\omega^4\right)^n\)
\(
\Rightarrow(-\omega)^n=(1+\omega)^n=\left(-\omega^2\right)^n \Rightarrow \omega^n=1 \Rightarrow n=3
\)

Hint

(a) Given that \(|z|=1\) and \(\omega=\frac{z-1}{z+1}(z \neq-1)\)
Now we know that \(z \bar{z}=|z|^2\)
\(
\begin{gathered}
\Rightarrow z \bar{z}=1 \quad(\text { for }|z|=1) \\
\therefore \omega=\left(\frac{z-1}{z+1}\right) \times \frac{(\bar{z}+1)}{(\bar{z}+1)}=\frac{z \bar{z}+z-\bar{z}-1}{z \bar{z}+z+\bar{z}+1}=\frac{2 i y}{2+2 x} \\
{[\quad z \bar{z}=1 \text { and taking } z=x+i y \text { so that }} \\
z+\bar{z}=2 x \text { and } z-\bar{z}=2 i y] \\
\quad \Rightarrow \operatorname{Re}(\omega)=0
\end{gathered}
\)

Hint

(b) Applying \(R_1 \rightarrow R_1+R_2+R_3\), we get
\(\left|\begin{array}{ccc}1 & 1 & 0 \\ 1 & -1-\omega & \omega^2 \\ 1 & \omega^2 & \omega^4\end{array}\right|=\left|\begin{array}{ccc}3 & 0 & 0 \\ 1 & -1-\omega^2 & \omega^2 \\ 1 & \omega^2 & \omega^4\end{array}\right|\)
\(
=3[-\omega-1-\omega]=3\left(\omega^2-\omega\right)
\)

Hint

\(
\begin{aligned}
&\text { (c) } \frac{z_1-z_3}{z_2-z_3}=\frac{1-i \sqrt{3}}{2} \\
&\Rightarrow \arg \left(\frac{z_1-z_3}{z_2-z_3}\right)=\arg \left(\frac{1-i \sqrt{3}}{2}\right) \\
&\Rightarrow \arg (\cos (-\pi / 3)+i \sin (-\pi / 3)) \\
&\Rightarrow \text { angle between }\left(z_1-z_3\right) \text { and }\left(z_2-z_3\right) \text { is } 60^{\circ}
\end{aligned}
\)
and \(\left|\frac{z_1-z_3}{z_2-z_3}\right|=\left|\frac{1-i \sqrt{3}}{2}\right|\)
\(
\Rightarrow\left|\frac{z_1-z_3}{z_2-z_3}\right|=1 \Rightarrow\left|z_1-z_3\right|=\left|z_2-z_3\right| \quad \text { (Imp Step) }
\)
\(\Rightarrow\) The \(\Delta\) with vertices \(z_1, z_2\) and \(z_3\) is isosceles with vertical angle
\(60^{\circ}\). Hence rest of the two angles should also be \(60^{\circ}\) each.
\(\Rightarrow\) Required triangle is an equilateral triangle.

Hint

(d) Let \(z=(1)^{1 / n}=(\cos 2 k \pi+i \sin 2 k \pi)^{1 / n}\) \(z=\cos \frac{2 k \pi}{n}+i \sin \frac{2 k \pi}{n}, k=0,1,2, \ldots ., n-1\).
Let \(z_1=\cos \left(\frac{2 k_1 \pi}{n}\right)+i \sin \left(\frac{2 k_1 \pi}{n}\right)\) and \(z_2=\cos \left(\frac{2 k_2 \pi}{n}\right)+i \sin \frac{2 k_2 \pi}{n}\)
be the two values of \(z\). Such that they subtend right angle at origin.
\(
\therefore \frac{2 k_1 \pi}{n}-\frac{2 k_2 \pi}{n}=\pm \frac{\pi}{2} \Rightarrow 4\left(k_1-k_2\right)=\pm n
\)
As \(k_1\) and \(k_2\) are integers and \(k_1 \neq k_2\)
\(
\therefore \quad n=4 k, k \in \mathrm{I}
\)

Hint

(c) \(E=4+5(\omega)^{334}+3(\omega)^{365}=4+5 \omega+3 \omega^2\) \(=1+2 \omega+3\left(1+\omega+\omega^2\right)=1+(-1+i \sqrt{3)}=i \sqrt{3}\)

Hint

\(
\begin{array}{ll}
\text { (b) } & (1+\omega)^7=A+B \omega \\
\Rightarrow & \left(-\omega^2\right)^7=A+B \omega \quad\left(\because 1+\omega+\omega^2=0\right) \\
\Rightarrow & -\omega^{14}=A+B \omega \\
\Rightarrow & -\omega^2=A+B \omega \quad\left(\because \omega^3=1\right) \\
\Rightarrow & 1+\omega=A+B \omega \Rightarrow A=1, B=1
\end{array}
\)

Hint

(b) Let \(A B C\) be the \(\triangle\) whose vertices are represented by complex numbers \(a, b, c\) and \(P Q R\) be the \(\triangle\) with whose vertices are represented by complex numbers \(u, v, w\).

\(
\text { Then } c=(1-r) a+r b
\)
\(
\begin{aligned}
&\Rightarrow c-a=r(b-a) \Rightarrow \frac{c-a}{b-a}=r\\
&\Rightarrow w=(1-r) u+r v \quad \Rightarrow \frac{w-u}{v-u}=r\\
&\text { From (i) and (ii), } \quad\left|\frac{c-a}{b-a}\right|=\left|\frac{w-u}{v-u}\right|\\
&\text { and } \arg \left(\frac{c-a}{b-a}\right)=\arg \left(\frac{w-u}{v-u}\right)\\
&\Rightarrow \frac{A C}{A B}=\frac{P R}{P Q} \text { and } \angle C A B=\angle R P Q\\
&\Rightarrow \quad \triangle A B C \sim \triangle P Q R
\end{aligned}
\)

Hint

(b) If vertices of a parallelogram are \(z_1, z_2, z_3, z_4\) then as diagonals bisect each other
\(
\therefore \frac{z_1+z_3}{2}=\frac{z_2+z_4}{2} \Rightarrow z_1+z_3=z_2+z_4
\)

Hint

\(
\begin{aligned}
&\text { (b) }|\omega|=1 \Rightarrow\left|\frac{1-i z}{z-i}\right|=1\\
&\Rightarrow|1-i z|=|z-i|\\
&\Rightarrow|1-i(x+i y)|=|x+i y-i|\\
&\Rightarrow|(y+1)-i x|=|x+i(y-1)|\\
&\Rightarrow x^2+(y+1)^2=x^2+(y-1)^2\\
&\Rightarrow 4 y=0 \Rightarrow y=0 \Rightarrow z \text { lies on real axis }
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (d) }|z-4|<|z-2| \\
&\Rightarrow|(x-4)+i y|<|(x-2)+i y|
\end{aligned}
\)
\(
\begin{aligned}
&\Rightarrow(x-4)^2+y^2<(x-2)^2+y^2 \\
&\Rightarrow-8 x+16<-4 x+4 \Rightarrow 4 x-12>0 \\
&\Rightarrow x>3 \Rightarrow \operatorname{Re}(z)>3
\end{aligned}
\)

Hint

(b) \(\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)=-i\left(\frac{-1}{2}+\frac{i \sqrt{3}}{2}\right)=i \omega\)
\(
\frac{\sqrt{3}}{2}-\frac{i}{2}=i\left(\frac{-1}{2}-\frac{i \sqrt{3}}{2}\right)=i \omega^2
\)
\(
\begin{aligned}
\therefore \quad & z=(-i \omega)^5+\left(i \omega^2\right)^5=-i \omega^2+i \omega \\
=& i\left(\omega-\omega^2\right)=i(i \sqrt{3})=-\sqrt{3}
\end{aligned}
\)
\(\Rightarrow \operatorname{Re}(z)<0\) and \(\operatorname{Im}(z)=0\)

Hint

(a) Since, \(\mathrm{z}=x+i y\) satisfies the equation \(\left|\frac{z-5 i}{z+5 i}\right|=1\)
\(
\begin{aligned}
&\therefore |x+i y-5 i|=| x+i y+5 i \mid \\
&\Rightarrow|x+(y-5) i|=|x+(y+5) i| \\
&\Rightarrow x^2+(y-5)^2=x^2+(y+5)^2 \\
&\Rightarrow x^2+y^2-10 y+25=x^2+y^2+10 y+25 \\
&\Rightarrow 20 y=0 \Rightarrow y=0
\end{aligned}
\)
‘ \(a\) ‘ is the correct alternative.

Hint

(b) \((x-1)^3+8=0\)
\(
\Rightarrow(x-1)^3=-8=(-2)^3
\)
\(\Rightarrow x-1=-2\) or \(-2 \omega\) or \(-2 \omega^2\)
\(
\Rightarrow \quad x=-1,1-2 \omega, 1-2 \omega^2
\)

Hint

(b) Let \(z=x+i y\)
\(
z^4-|z|^4=41 z^2
\)
\(
\Rightarrow \mathrm{z}^4-(z \bar{z})^2=4 \mathrm{iz}^2 \Rightarrow \mathrm{z}^2\left(\mathrm{z}^2-\bar{z}^2\right)=4 \mathrm{i} z^2
\)
\(
\begin{aligned}
&\Rightarrow z=0 \text { or } z^2-(\bar{z})^2=4 i \\
&\Rightarrow 4 i x y=4 i \Rightarrow x y=1
\end{aligned}
\)
Locus of \(\mathrm{z}\) is a rectangular hyperbola \(x y=1\)
Given that \(\operatorname{Re}\left(\mathrm{z}_1\right)>0\) and \(\operatorname{Re}\left(\mathrm{z}_2\right)<0\)
\(
\begin{aligned}
&\therefore\left|z_1-z_2\right|_{\min }=\sqrt{(1+1)^2+(1+1)^2}=\sqrt{8} \\
&\Rightarrow\left|z_1-z_2\right|_{\text {min }}^2=8
\end{aligned}
\)

Hint

(c) \(a, b, c\) are distinct non-zero integers
Min. value of \(\left|a+b \omega+c \omega^2\right|^2\) is to be found \(\left|a+b \omega+c \omega^2\right|^2\)
\(
=\left|a+b\left(\frac{-1+i \sqrt{3}}{2}\right)+c\left(\frac{-1-i \sqrt{3}}{2}\right)\right|^2
\)
\(
\begin{aligned}
&=\left|\frac{1}{2}(2 a-b-c)+\frac{i \sqrt{3}}{2}(b-c)\right|^2 \\
&=\frac{1}{4}(2 a-b-c)^2+\frac{3}{4}(b-c)^2 \\
&=\frac{1}{4}\left(4 a^2+b^2+c^2-4 a b+2 b c-4 a c+3 b^2+3 c^2-6 b c\right) \\
&=a^2+b^2+c^2-a b-b c-c a \\
&=\frac{1}{2}\left[(a-b)^2+(b-c)^2+(c-a)^2\right]
\end{aligned}
\)
For minimum value, let us consider \(a=3, b=2, c=1\)
\(
\therefore \quad \text { minimum value }=\frac{1}{2}[1+1+4]=\frac{6}{2}=3
\)

Hint

(a)
As the question is wrong. The answer is not unique for different values, we get different answers.
For \(\omega=e^{i 2 \pi / 3}\) we get a genuine answer
\(\frac{|x|^2+|y|^2+|z|^2}{|a|^2+|b|^2+|c|^2}=\frac{x \bar{x}+y \bar{y}+z \bar{z}}{|a|^2+|b|^2+|c|^2}\)
\((a+b+c)(\bar{a}+\bar{b}+\bar{c})+\left(a+b \omega+c \omega^2\right)+\)
\(=\frac{\left(\bar{a}+\bar{b} \omega^2+\bar{c} \omega\right)+\left(a+b \omega^2+c \omega\right)\left(\bar{a}+\bar{b} \omega+\bar{c} \omega^2\right)}{|a|^2+|b|^2+|c|^2}\)
\(=\frac{3\left(|a|^2+|b|^2+|c|^2\right)}{|a|^2+|b|^2+|c|^2}=3\)

Hint

rth term of the given series
\(
\begin{aligned}
&\left.=r[(r+1)-\omega](r+1)-\omega^2\right] \\
&=r\left[(r+1)^2-\left(\omega+\omega^2\right)(r+1)+\omega^3\right] \\
&=r\left[(r+1)^2-(-1)(r+1)+1\right] \\
&=r\left[\left(r^2+3 r+3\right]=r^3+3 r^2+3 r\right. \\
&\therefore \quad \text { Sum of the given series }=\sum_{r=1}^{(n-1)}\left(\mathrm{r}^3+3 r^2+3 r\right) \\
&=\frac{1}{4}(n-1)^2 n^2+3 \cdot \frac{1}{6}(n-1)(n)(2 n-1)+3 \cdot \frac{1}{2}(n-1) n \\
&=(n-1)(n)\left[\frac{1}{4}(n-1) n+\frac{1}{2}(2 n-1)+\frac{3}{2}\right] \\
&=\frac{1}{4}(n-1) n\left[n^2-n+4 n-2+6\right] \\
&=\frac{1}{4}(n-1) n\left[n^2+3 n+4\right]
\end{aligned}
\)

Hint

(a) Let \(z_1, z_2, z_3\) be the vertices \(A, B\) and \(C\) respectively of equilateral \(\angle A B C\), inscribed in a circle \(|z|=2\) with centre \((0,0)\) and radius \(=2\)
\(
\begin{aligned}
&\text { Given } \begin{array}{l}
z_1=1+i \sqrt{3} \\
z_2=e^{\frac{2 \pi i}{3}} z_1 \\
=\left(\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right)(1+i \sqrt{3)} \\
=\frac{-1-3}{2}=-2 \\
\text { and } z_3=e^{4(\pi / 3) i} z_1 \\
=\left(\cos \frac{4 \pi}{3}+i \sin \frac{4 \pi}{3}\right)(1+i \sqrt{3)} \\
=\left(\frac{-1-i \sqrt{3}}{2}\right)(1+i \sqrt{3})=\frac{-1-2 i \sqrt{3}+3}{2}=1-i \sqrt{3}
\end{array}
\end{aligned}
\)

Hint

(a) As \(D\) and \(m\) are represented by complex numbers \((1+i)\) and \((2-i)\) respectively
\(\therefore \quad D \equiv(1,1)\) and \(M \equiv(2,-1)\)
We know that diagonals of rhombus bisect each other at right angles.
\(\therefore \quad A C\) passes through \(M\) and is \(\perp\) to \(B D\)
\(\therefore \quad\) Eq. of \(A C\) in symmetric form can be written as \(\frac{x-2}{2 / \sqrt{5}}=\frac{y+1}{1 / \sqrt{5}}=r\)
Now for pt. A, as
\(
A M=\frac{1}{2} D M=\frac{1}{2} \sqrt{(2-1)^2+(-1-1)^2}=\sqrt{5} / 2
\)
On putting \(r=\pm \sqrt{5} / 2\), we get
\(
\begin{aligned}
&\frac{x-2}{2 / \sqrt{5}}=\frac{y+1}{1 / \sqrt{5}}=\pm \sqrt{5} / 2 \\
&\Rightarrow x=\pm 1+2, y=\pm \frac{1}{2}-1 \\
&\Rightarrow x=3 \quad \text { or } 1, y=\frac{-1}{2} \text { or } \frac{-3}{2}
\end{aligned}
\)
Therefore, point \(A\) is represented by \(3-i / 2\) or \(1-(3 / 2) i\)

Hint

(c) Distance between two points represented by \(z_1\) and \(z_2\) \(=\left|z_1-z_2\right|\)
Since \(z_1=a+i, \quad z_2=1+b i\) and \(z_3=0\) form an equilateral triangle, therefore \(\left|z_1-z_3\right|=\left|z_2-z_3\right|=\left|z_1-z_2\right|\)
\(
\begin{aligned}
& \Rightarrow|a+i|=|1+b i|=|(a-1)+i(1-b)| \\
& \Rightarrow a^2+1=1+b^2=(a-1)^2+(1-b)^2 \\
& \Rightarrow a^2=b^2=a^2+b^2-2 a-2 b+1 \\
& \Rightarrow a=b \\
(\because \quad a&\neq-b \text { because } 0<a, b<1) \\
\text { and } & b^2-2 a-2 b+1=0 \\
& \Rightarrow a^2-2 a-2 b+1=0 \\
& \Rightarrow a^2-2 a-2 a+1=0 \\
& \Rightarrow a^2-4 a+1=0 \\
& \therefore a=\frac{4 \pm 2 \sqrt{3}}{2}=2 \pm \sqrt{3}
\end{aligned}
\)
But \(0<a, b<1\)
\(
\therefore \quad a=2-\sqrt{3} \quad \because b=a \quad \therefore b=2-\sqrt{3}
\)

Hint

(b) \(\left|a z_1-b z_2\right|^2+\left|b z_1+a z_2\right|^2\)
\(=a^2\left|z_1\right|^2+b^2\left|z_2\right|^2-2 a b \operatorname{Re}\left(z_1 \bar{z}_2\right)+b^2\left|z_1\right|^2\)
\(
+a^2\left|z_2\right|^2+2 a b \operatorname{Re}\left(z_1 \bar{z}_2\right)
\)
\(=\left(a^2+b^2\right)\left(\left|z_1\right|^2+\left|z_2\right|^2\right)\)

Hint

(True) \(\because\) Cube roots of unity are \(1, \frac{-1+i \sqrt{3}}{2}, \frac{-1-\sqrt{3}}{2}\)
\(\therefore \quad\) Vertices of triangle are
\(
A(1,0), B\left(\frac{-1}{2}, \frac{\sqrt{3}}{2}\right), C\left(\frac{-1}{2}, \frac{-\sqrt{3}}{2}\right)
\)
\(\Rightarrow A B=B C=C A, \quad \therefore \triangle\) is equilateral.

Hint

(False) If \(z_1, z_2, z_3\) are in A.P. then, \(\frac{z_1+z_3}{2}=z_2\) \(\Rightarrow z_2\) is mid pt. of line joining \(z_1\) and \(z_3\).
\(\Rightarrow z_1, z_2, z_3\) lie on a st. line
\(\therefore \quad\) Given statement is false

Hint

(True)
As \(\left|z_1\right|=\left|z_2\right|=\left|z_3\right|\)
\(\therefore z_1, z_2, z_3\) are equidistant from origin. Hence \(\mathrm{O}\) is the circumcentre of \(\triangle A B C\).
But \(\triangle A B C\) is equilateral and hence circumcentre and centriod of \(\triangle\) \(A B C\) coincide.
\(\therefore \quad\) Centriod of \(\triangle A B C=0\)
\(\Rightarrow \frac{z_1+z_2+z_3}{3}=0\)
\(\Rightarrow z_1+z_2+z_3=0\)
\(\therefore \quad\) Statement is true.

Hint

Let \(Z=x+\) iy
Then \(( x -1)^2+ y ^2=1 \quad \rightarrow(1)\)
& \((\sqrt{2}-1)(2 x)-i(2 i y)=2 \sqrt{2}\)
\(
\Rightarrow(\sqrt{2}-1) x+y=\sqrt{2} \rightarrow(2)
\)
Solving (1) & (2)
we get
Either \(x =1\) or \(x=\frac{1}{2-\sqrt{2}} \rightarrow(3)\)
On solving (3) with (2)
we get
For \(x =1 \Rightarrow y =1 \Rightarrow Z _2=1+ i \&\) for
\(
x=\frac{1}{2-\sqrt{2}} \Rightarrow y=\sqrt{2}-\frac{1}{\sqrt{2}} \Rightarrow Z_1=\left(1+\frac{1}{\sqrt{2}}\right)+\frac{i}{\sqrt{2}}
\)
Now
\(
\begin{aligned}
& \left|\sqrt{2} z_1-z_2\right|^2 \\
& =\left|\left(\frac{1}{\sqrt{2}}+1\right) \sqrt{2}+i-(1+i)\right|^2 \\
& =(\sqrt{2})^2 \\
& =2
\end{aligned}
\)

Hint

Clearly for the shaded region \(z _1\) is the intersection of the circle and the line passing through \(P \left( L _1\right)\) and \(z _2\) is intersection of line \(L _1 \& L _2\)
Circle: \((x+2)^2+(y-3)^2=1\)
\(L _1: x + y -1=0\)
\(L _2: x – y +4=0\)
On solving circle \(\& L _1\) we get
\(
z _1:\left(-2-\frac{1}{\sqrt{2}}, 3+\frac{1}{\sqrt{2}}\right)
\)
On solving \(L _1\) and \(z _2\) is intersection of line \(L _1 \& L _2\) we get \(z _2:\left(\frac{-3}{2}, \frac{5}{2}\right)\)
\(
\begin{aligned}
& \left|z_1\right|^2+2\left|z_2\right|^2=14+5 \sqrt{2}+17 \\
& =31+5 \sqrt{2}
\end{aligned}
\)
So \(\alpha=31\)
\(
\begin{aligned}
& \beta=5 \\
& \alpha+\beta=36
\end{aligned}
\)

Hint

\(|Z| \geq 1\) is a region outside & periphery of circle with centre \((0,0)\) & radius = 1 unit
Minimum value of \(\left|Z-\left(-\frac{3}{2}-2 i\right)\right|\) is minimum distance of p point in \(|Z| \geq 1\) from \(\left(-\frac{3}{2}-2\right)\)
Therefore minimum value of \(\left|Z+\frac{1}{2}(3+4 i)\right|=0\)

Hint

\(
|z-i|=|z+i|=|z-1|
\)

ABC is a triangle. Hence its circum-centre will be the only point whose distance from \(A , B , C\) will be same.
\(
\text { So } n(S)=1
\)

Hint

\(
x^2+x+1=0 \Rightarrow x=\omega, \omega^2=\alpha
\)
Let \(\alpha =\omega\)
Now \((1+\alpha)^7=-\omega^{14}=-\omega^2=1+\omega\)
\(
\begin{aligned}
& A =1, B =1, C =0 \\
& \therefore 5(3 A-2 B-C)=5(3-2-0)=5
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \left|z-z_0\right|^2=4 \\
& \quad \Rightarrow\left(\alpha-z_0\right)\left(\bar{\alpha}-\bar{z}_0\right)=4 \\
& \Rightarrow \alpha \bar{\alpha}-\alpha \bar{z}_0-z_0 \bar{\alpha}+\left|z_0\right|^2=4 \\
& \quad \Rightarrow|\alpha|^2-\alpha \bar{z}_0-z_0 \bar{\alpha}=2 \ldots \ldots \ldots (1)
\end{aligned}
\)
\(
\begin{aligned}
& \left|z-z_0\right|^2=16 \\
& \Rightarrow\left(\frac{1}{\bar{\alpha}}-z_0\right)\left(\frac{1}{\alpha}-\bar{z}_0\right)=16 \\
& \Rightarrow\left(1-\bar{\alpha} z_0\right)\left(1-\alpha \bar{z}_0\right)=16|\alpha|^2 \\
& \Rightarrow 1-\bar{\alpha} z_0-\alpha \bar{z}_0+|\alpha|^2\left|z_0\right|^2=16|\alpha|^2 \\
& \Rightarrow 1-\bar{\alpha} z_0-\alpha \bar{z}_0=14|\alpha|^2 \ldots \ldots \ldots \ldots (2)
\end{aligned}
\)
\(
\begin{aligned}
& \text { From (1) and (2) } \\
& \Rightarrow 5|\alpha|^2=1 \\
& \Rightarrow 100|\alpha|^2=20
\end{aligned}
\)

Hint

\(
\begin{aligned}
& z =\frac{1}{2}-2 i \\
& | z +1|=\alpha z+\beta(1+ i ) \\
& \left|\frac{3}{2}-2 i \right|=\frac{\alpha}{2}-2 \alpha i +\beta+\beta i \\
& \left|\frac{3}{2}-2 i \right|=\left(\frac{\alpha}{2}+\beta\right)+(\beta-2 \alpha) i \\
& \beta=2 \alpha \text { and } \frac{\alpha}{2}+\beta=\sqrt{\frac{9}{4}+4} \\
& \alpha+\beta=3
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \alpha^6+\alpha^4+\beta^4-5 \alpha^2 \\
& =\alpha^4(\alpha-2)+\alpha^4-5 \alpha^2+(\beta-2)^2 \\
& =\alpha^5-\alpha^4-5 \alpha^2+\beta^2-4 \beta+4 \\
& =\alpha^3(\alpha-2)-\alpha^4-5 \alpha^2+\beta-2-4 \beta+4 \\
& =-2 \alpha^3-5 \alpha^2-3 \beta+2 \\
& =-2 \alpha(\alpha-2)-5 \alpha^2-3 \beta+2 \\
& =-7 \alpha^2+4 \alpha-3 \beta+2 \\
& =-7(\alpha-2)+4 \alpha-3 \beta+2 \\
& =-3 \alpha-3 \beta+16=-3(1)+16=13
\end{aligned}
\)

Hint

\(
\begin{aligned}
z & =2-i\left(2 \tan \frac{5 \pi}{8}\right)=x+i y(\text { let }) \\
r & =\sqrt{x^2+y^2} \& \theta=\tan ^{-1} \frac{y}{x} \\
r & =\sqrt{(2)^2+\left(2 \tan \frac{5 \pi}{8}\right)^2} \\
& =\left|2 \sec \frac{5 \pi}{8}\right|=\left|2 \sec \left(\pi-\frac{3 \pi}{8}\right)\right| \\
& =2 \sec \frac{3 \pi}{8} \\
\& \theta & =\tan ^{-1}\left(\frac{-2 \tan \frac{5 \pi}{8}}{2}\right) \\
& =\tan ^{-1}\left(\tan \left(\pi-\frac{5 \pi}{8}\right)\right) \\
& =\frac{3 \pi}{8}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& x^2-\sqrt{6} x+6=0_{\Delta \beta}^\alpha \\
& x=\frac{\sqrt{6} \pm i \sqrt{6}}{2}=\frac{\sqrt{6}}{2}(1 \pm i) \\
& \alpha=\sqrt{3}\left(e^{i \frac{\pi}{4}}\right), \beta=\sqrt{3}\left(e^{-i \frac{\pi}{4}}\right) \\
& \therefore \frac{\alpha^{99}}{\beta}+\alpha^{98}=\alpha^{98}\left(\frac{\alpha}{\beta}+1\right) \\
& =\frac{\alpha^{98}(\alpha+\beta)}{\beta}=3^{49}\left(e^{i 99 \frac{\pi}{4}}\right) \times \sqrt{2} \\
& =3^{49}(-1+ i ) \\
& =3^{ n }( a + ib ) \\
& \therefore n =49, a =-1, b =1 \\
& \therefore n + a + b =49-1+1=49
\end{aligned}
\)

Hint

\(
\begin{aligned}
& z ^2=- i \bar{z} \\
& \left| z ^2\right|=| iz | \\
& \left|z^2\right|=|z| \\
& |z|^2-|z|=0 \\
& |z|(|z|-1)=0 \\
& |z|=0 \text { (not acceptable) }
\end{aligned}
\)
\(
\begin{aligned}
& \therefore|z|=1 \\
& \therefore|z|^2=1
\end{aligned}
\)

Hint

\(
\begin{array}{ll}
z^{1985}+z^{100}+1=0 \& z^3+2 z^2+2 z+1=0 \\
& (z+1)\left(z^2-z+1\right)+2 z(z+1)=0 \\
& (z+1)\left(z^2+z+1\right)=0 \\
\Rightarrow \quad & z=-1, \quad z=w, w^2
\end{array}
\)
Now putting \(z=-1\) not satisfy
Now put \(z = w\)
\(
\begin{aligned}
& \Rightarrow \quad w^{1985}+w^{100}+1 \\
& \Rightarrow \quad w^2+w+1=0
\end{aligned}
\)
Also, \(z = w ^2\)
\(
\begin{array}{ll}
\Rightarrow & w^{3970}+w^{200}+1 \\
\Rightarrow & w+w^2+1=0
\end{array}
\)
Two common root

Hint

\(
\begin{aligned}
& (\sqrt{2})^x=2^x \Rightarrow x=0 \Rightarrow \alpha=1 \\
& z=\frac{\pi}{4}(1+i)^4\left[\frac{\sqrt{\pi}-\pi i-i-\sqrt{\pi}}{\pi+1}+\frac{\sqrt{\pi}-i-\pi i-\sqrt{\pi}}{1+\pi}\right] \\
& =-\frac{\pi i}{2}\left(1+4 i+6 i^2+4 i^3+1\right) \\
& =2 \pi i \\
& \beta=\frac{2 \pi}{\frac{\pi}{2}}=4
\end{aligned}
\)
Distance from \((1,4)\) to \(4 x-3 y=7\)
Will be \(\frac{15}{5}=3\)

Hint

\(
\begin{aligned}
& z_1+z_2=5 \\
& z_1^3+z_2^3=20+15 i \\
& z_1^3+z_2^3=\left(z_1+z_2\right)^3-3 z_1 z_2\left(z_1+z_2\right) \\
& z_1^3+z_2^3=125-3 z_1 \cdot z_2(5) \\
& \Rightarrow 20+15 i=125-15 z_1 z_2 \\
& \Rightarrow 3 z_1 z_2=25-4-3 i \\
& \Rightarrow 3 z_1 z_2=21-3 i \\
& \Rightarrow z_1 \cdot z_2=7-i \\
& \Rightarrow\left(z_1+z_2\right)^2=25 \\
& \Rightarrow z_1^2+z_2^2=25-2(7-i) \\
& \Rightarrow 11+2 i \\
& \left(z_1^2+z_2^2\right)^2=121-4+44 i \\
& \Rightarrow z_1^4+z_2^4+2(7-i)^2=117+44 i \\
& \Rightarrow z_1^4+z_2^4=117+44 i-2(49-1-14 i) \\
& \Rightarrow\left|z_1^4+z_2^4\right|=75
\end{aligned}
\)

Hint

\(
\begin{aligned}
& (1-\sqrt{3} i )^{200}=2^{199}( p + iq ) \\
& 2^{200}\left(\cos \frac{\pi}{3}- i \sin \frac{\pi}{3}\right)^{200}=2^{199}( p + iq ) \\
& 2\left(-\frac{1}{2}- i \frac{\sqrt{3}}{2}\right)= p + iq \\
& p =-1, q =-\sqrt{3} \\
& \alpha= p + q + q ^2=2-\sqrt{3} \\
& \beta= p – q + q ^2=2+\sqrt{3} \\
& \alpha+\beta=4 \\
& \alpha \cdot \beta=1 \\
& \text { equation } x ^2-4 x +1=0
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { Let } \sin \frac{2 \pi}{9}+ i \cos \frac{2 \pi}{9}= z \\
& \left(\frac{1+ z }{1+\bar{z}}\right)^3=\left(\frac{1+ z }{1+\frac{1}{ z }}\right)^3= z ^3 \\
& \Rightarrow\left( i \left(\cos \frac{2 \pi}{9}- i \sin \frac{2 \pi}{9}\right)\right)^3 \\
& =- i \left(\cos \frac{2 \pi}{3}- i \sin \frac{2 \pi}{3}\right)=- i \left(\frac{-1}{2}- i \frac{\sqrt{3}}{2}\right) \\
& \Rightarrow \frac{-1}{2}(\sqrt{3}-i)
\end{aligned}
\)

Hint

\(
\begin{aligned}
& (x -2)^2+( y -3)^2-(x -3)^2-( y -4)^2=2 \\
& \Rightarrow(x-2)^2+(y-3)^2-(x-3)^2-(y-4)^2=1+1 \\
& \Rightarrow-4 x+4+9-6 y-9+6 x-16+8 y=2 \\
& \Rightarrow x + y =7 \quad \Rightarrow 2 x+2 y=14
\end{aligned}
\)

Hint

\(
\begin{aligned}
& (z-2 i)(\bar{z}+2 i)=4(z+i)(\bar{z}-i) \\
& z \bar{z}+4+2 i(z-\bar{z})=4(z \bar{z}+1+i(\bar{z}-z)) \\
& 3 z \bar{z}-6 i(z-\bar{z})=0 \\
& x^2+y^2-2 i(2 i y)=0 \\
& x^2+y^2+4 y=0
\end{aligned}
\)

Hint

\(
\begin{aligned}
& z_1=x_1+i y_1 \\
& z_2=x_2+i y_2 \\
& \operatorname{Re}\left(z_1 z_2\right)=x_1 x_2-y_1 y_2=0 \\
& \operatorname{Re}\left(z_1+z_2\right)=x_1+x_2=0
\end{aligned}
\)
\(x_1 \& x_2\) are of opposite sign \(y _1 \& y _2\) are of opposite sign

Hint

\(
\begin{aligned}
&\alpha=8-14 i\\
&\begin{aligned}
& z=x+i y \\
& a z=(8 x+14 y)+i(-14 x+8 y) \\
& z+\bar{z}=2 x \quad z-\bar{z}=2 i y
\end{aligned}
\end{aligned}
\)
Set A: \(\frac{2 i(-14 x+8 y)}{i(4 x y-112)}=1\)
\(
(x-4)(y+7)=0
\)
\(x =4 \quad\) or \(y=-7\)
Set B: \(x^2+(y+3)^2=16\)
when \(x=4\)
\(
y=-3
\)
when \(y=-7\)
\(
x =0
\)
\(
\therefore A \cap B=\{4-3 i , 0-7 i \}
\)
So, \(\sum_{z \in A \cap B}(\operatorname{Re} z-\operatorname{Im} z)=4-(-3)+(0-(-7))=14\)

Hint

Let \(w=z+\frac{1}{z}=4 e^{i \theta}+\frac{1}{4} e^{-i \theta}\)
\(
\Rightarrow w =\frac{17}{4} \cos \theta+ i \frac{15}{4} \sin \theta
\)
So locus of \(w\) is ellipse \(\frac{x^2}{\left(\frac{17}{4}\right)^2}+\frac{y^2}{\left(\frac{15}{4}\right)^2}=1\)
Locus of \(z\) is circle \(x^2+y^2=16\)