NCERT Exemplar MCQs

Hint

(a)

\(
\begin{aligned}
\mid a z_1 &-\left.b z_2\right|^2+\left|b z_1+a z_2\right|^2 \\
&=\left|a z_1\right|^2+\left|b z_2\right|^2-2 \operatorname{Re}\left(a z_1 \times b \bar{z}_2\right)+\left|b z_1\right|^2+\left|a z_2\right|^2+2 \operatorname{Re}\left(b z_1 \times a \bar{z}_2\right) \\
\quad &=\left(a^2+b^2\right)\left|z_1\right|^2+\left(a^2+b^2\right)\left|z_2\right|^2=\left(a^2+b^2\right)\left(\left|z_1\right|^2+\left|z_2\right|^2\right)
\end{aligned}
\)

Hint

(b)

\(
\sqrt{-25} \times \sqrt{-9}=i \sqrt{25} \times i \sqrt{9}=i^2(5 \times 3)=-15
\)

Hint

(d)

\(
\frac{(1-i)^3}{1-i^3}=\frac{(1-i)^3}{(1-i)\left(1+i+i^2\right)}=\frac{(1-i)^2}{i}=\frac{1+i^2-2 i}{i}=\frac{-2 i}{i}=-2
\)

Hint

(c)

\(i+i^2+i^3+\ldots\) upto 1000 terms
\(=\left(i+i^2+i^3+i^4\right)+\left(i^5+i^6+i^7+i^8\right)+\ldots 250\) brackets
\(=0+0+0 \ldots+0 \quad\left[\because i^n+i^{n+1}+i^{n+2}+i^{n+3}=0\right.\), where \(\left.n \in N\right]\)

Hint

(a)

\(
\text { Multiplicative inverse of } 1+i=\frac{1}{1+i}=\frac{1-i}{1-i^2}=\frac{1}{2}(1-i)
\)

Hint

(a)

Let \(z_1=x_1+i y_1\) and \(z_2=x_2+i y_2\) \(z_1+z_2=\left(x_1+x_2\right)+i\left(y_1+y_2\right)\), which is real \(\Rightarrow y_1+y_2=0 \Rightarrow y_1=-y_2\)
\(\because z_2=x_1-i y_1 \quad\) [Assuming \(x_1=x_2\) ] \(\Rightarrow \quad z_2=\bar{z}_1\)

Hint

(b)

\(
\arg (z)+\arg (\bar{z})=\theta+(-\theta)=0
\)

Hint

(d) Given that, \(|z+4| \leq 3\)
For the greatest value of \(|z+1|\),
\(
|z+1|=|z+4-3|
\)
So, greatest value of \(|z+1|\) is 6 .
We know that the least value of the modulus of a complex number is zero. So, the least value of \(|z+1|\) is zero.

Hint

(a) We have \(\left|\frac{z-2}{z+2}\right|=\frac{\pi}{6}\)
\(
\Rightarrow \quad \frac{|x+i y-2|}{|x+i y+2|}=\frac{\pi}{6} \Rightarrow \frac{|x-2+i y|}{|x+2+i y|}=\frac{\pi}{6}
\)
\(
\begin{aligned}
&\Rightarrow \quad 6 \mid x-2+i y]=\pi|x+2+i y| \Rightarrow 36|x-2+i y|^2=\pi^2|x+2+i y|^2 \\
&\Rightarrow \quad 36\left[x^2-4 x+4+y^2\right]=\pi^2\left[x^2+4 x+4+y^2\right]
\end{aligned}
\)
\(\Rightarrow \quad\left(36-\pi^2\right) x^2+\left(36-\pi^2\right) y^2-\left(144+4 \pi^2\right) x+144-4 \pi^2=0\), which is a circle.

Hint

(a) Let \(z=|z|(\cos \theta+i \sin \theta)\)
Where \(\theta=\arg (z)\)
Given that \(|z|=4\) and \(\arg (z)=\frac{5 \pi}{6}\)
\(
\begin{aligned}
\Rightarrow \quad z &=4\left[\cos \frac{5 \pi}{6}+i \sin \frac{5 \pi}{6}\right] \\
&=4\left[-\frac{\sqrt{3}}{2}+i \frac{1}{2}\right]=-2 \sqrt{3}+2 i
\end{aligned}
\)

Hint

(b) False
We can compare two complex numbers when they are purely real. Otherwise, the comparison of complex numbers is not possible or has no meaning.

Hint

Ans: False

Let \(z=x+y i\)
\(z . i=(x+y i) i=x i-y\)
Which rotates at angle of \(180^{\circ}\)

Hint

(a) True
\(
|z|+|z-1|
\)
We know that \(\left|z_1\right|+\left|z_2\right| \geq\left|z_1-z_1\right|\)
\(
\Rightarrow \quad|z|+|z-1| \geq|z-(z-1)| \quad \Rightarrow|z|+|z-1| \geq 1
\)
So, minimum value of \(|z|+|z-1|\) is 1 .

Hint

(a) True
We have, \(|z-1|=|z-\mathrm{i}|\)
Putting \(z=x+i y\), we get
\(
\begin{aligned}
&\Rightarrow \quad|x-1+i y|=|x-i(1-y)| \\
&\Rightarrow \quad(x-1)^2+y^2=x^2+(1-y)^2 \Rightarrow x^2-2 x+1+y^2=x^2+1+y^2-2 y \\
&\Rightarrow \quad-2 x+1=1-2 y \quad \Rightarrow-2 x+2 y=0 \quad \Rightarrow x-y=0
\end{aligned}
\)
Now, equation of a line through the points \((1,0)\) and \((0,1)\) is:
\(
y-0=\frac{1-0}{0-1}(x-1)
\)
or \(\quad x+y=1\)
This line is perpendicular to the line \(x-y=0\).

Hint

(b) True

\(
\begin{aligned}
&z=x+i y \\
&\operatorname{Re} (z)=0 \quad x=0 \\
&z=i y \\
&z^2=i^2 y^2 \\
&=-y^2 \\
&\operatorname{Im}\left(z^2\right)=0
\end{aligned}
\)

Hint

(a) True

Given that: \(|z-4|<|z-2|\)
Let \(z=x+y i\)
\(
\begin{aligned}
&\Rightarrow|x+y i-4|<|x+y i-2| \\
&\Rightarrow|(x-4)+y i|<|(x-2)+y i| \\
&\Rightarrow \sqrt{(x-4)^2+y^2}<\sqrt{(x-2)^2+y^2} \\
&\Rightarrow(x-4)^2+y^2<(x-2)^2+y^2 \\
&\Rightarrow(x-4)^2<(x-2)^2 \\
&\Rightarrow x^2+16-8 x<x^2+4-4 x \\
&\Rightarrow-8 x+4 x<-16+4 \\
&\Rightarrow-4 x<-12 \\
&\Rightarrow x>3
\end{aligned}
\)

Hint

(a) False
Explanation:
Given \(\left|z_1+z_2\right|=\left|z_1\right|+\left|z_2\right|\)
\(
\begin{aligned}
&\Rightarrow\left|z_1+z_2\right|^2=\left|z_1\right|^2+\left|z_2\right|^2+2\left|z_1\right|\left|z_2\right| \\
&\Rightarrow\left|z_1\right|^2+\left|z_2\right|^2+2 \operatorname{Re}\left(z_1 \bar{z}_2\right)=\left|z_1\right|^2+\left|z_2\right|^2+2\left|z_1\right|\left|z_2\right| \\
&\Rightarrow 2 \operatorname{Re}\left(z_1 \bar{z}_2\right)=2\left|z_1\right|\left|z_2\right| \\
&\Rightarrow \cos \left(\theta_1-\theta_2\right)=1 \\
&\Rightarrow \theta_1-\theta_2=0 \\
&\therefore \arg \left(z_1\right)-\arg \left(z_2\right)=0
\end{aligned}
\)

Hint

(a) False

\(
\text { Since, every real number is a complex number. So, } 2 \text { is a complex number. Hence, it is false. }
\)

Hint

(a) Given that, \(z=i+\sqrt{3}\).
So, \(|z|=|i+\sqrt{3}|=\sqrt{1^2+(\sqrt{3})^2}=2\)
Also, \(z\) lies in first quadrant.
\(
\Rightarrow \arg (z)=\tan ^{-1} \frac{1}{\sqrt{3}}=\frac{\pi}{6}
\)
So, the polar from of \(z\) is \(2\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)\).
(b) We have, \(z=-1+\sqrt{-3}=-1+i \sqrt{3}\)
Here \(z\) lies in second quadrant.
\(
\Rightarrow \quad \arg (z)=\operatorname{amp}(z)=\pi-\tan ^{-1}\left|\frac{\sqrt{3}}{-1}\right|=\pi-\tan ^{-1} \sqrt{3}=\pi-\frac{\pi}{3}=\frac{2 \pi}{3}
\)
(c) Given that, \(|z+2|=|z-2|\)
\(\Rightarrow \quad|x+2+i y|=|x-2+i y|\)
\(\Rightarrow \quad(x+2)^2+y^2=(x-2)^2+y^2 \quad \Rightarrow x^2+4 x+4=x^2-4 x+4\)
\(\Rightarrow \quad 8 x=0\)
\(\therefore x=0\)
It is a straight line which is a perpendicular bisector of segment joining the points \((-2,0)\) and \((2,0)\).
1) We have \(|z+2 i|=|z-2 i|\)
Putting \(z=x+i y\), we get
\(
\begin{aligned}
&\Rightarrow|x+i(y+2)|^2=|x+i(y-2)|^2 \quad \Rightarrow x^2+(y+2)^2=x^2+(y-2)^2 \\
&\Rightarrow 4 y=0 \quad \Rightarrow y=0
\end{aligned}
\)
It is a straight line, which is a perpendicular bisector of segment joining \((0,-2)\) and \((0,2)\)
(e) Given that, \(|z+4 i| \geq 3\)
\(
\begin{array}{ll}
\Rightarrow \quad|x+i y+4 i| \geq 3 & \Rightarrow|x+i(y+4)| \geq 3 \\
\Rightarrow \quad \sqrt{x^2+(y+4)^2} \geq 3 & \Rightarrow x^2+y^2+8 y+16 \geq 9 \\
\Rightarrow \quad x^2+y^2+8 y+7 \geq 0 &
\end{array}
\)
This represents the region on or outside the circle having centre \((0,-4)\) and radius 3 .
(f) Given that, \(|z+4| \leq 3\)
\(\Rightarrow|x+i y+4| \leq 3 \quad \Rightarrow|x+4+i y| \leq 3\)
\(\Rightarrow \quad \sqrt{(x+4)^2+y^2} \leq 3 \quad \Rightarrow(x+4)^2+y^2 \leq 9\)
\(\Rightarrow \quad x^2+8 x+16+y^2 \leq 9 \quad \Rightarrow x^2+8 x+y^2+7 \leq 0\)
This represents the region on inside circle having centre \((-4,0)\) and radius 3.
(g) \(z=\frac{1+2 i}{1-i}=\frac{(1+2 i)(1+i)}{(1+i)(1+i)}=\frac{1+2 i+i+2 i^2}{1-i^2}\)
\(
=\frac{1-2+3 i}{1+1}=\frac{-1}{2}+\frac{3 i}{2}
\)
Hence, \(\bar{z}\) lies in the third quadrant.
(h) Given that, \(z=1-i\)
\(
\Rightarrow \quad \frac{1}{z}=\frac{1}{1-i}=\frac{1+i}{(1-i)(1+i)}=\frac{1+i}{1-i^2}=\frac{1}{2}(1+i)
\)
Thus, reciprocal of \(z\) lies in first quadrant.

Hint

(a) We have \(z=\frac{2-i}{(1-2 i)^2}\)
\(
\begin{array}{ll}
\Rightarrow \quad z & =\frac{2-i}{1+4 i^2-4 i}=\frac{2-i}{1-4-4 i}=\frac{2-i}{-3-4 i} \\
& =\frac{(2-i)}{-(3+4 i)}=-\left[\frac{(2-i)(3-4 i)}{(3+4 i)(3-4 i)}\right] \\
& =-\left(\frac{6-8 i-3 i+4 i^2}{9+16}\right)=-\frac{(-11 i+2)}{25} \\
& =\frac{-1}{25}(2-11 i)=\frac{1}{25}(-2+11 i) \\
\therefore \quad \bar{z} & =\frac{1}{25}(-2-11 i)=\frac{-2}{25}-\frac{11}{25} i
\end{array}
\)

Hint

Ans: not necessary

If \(\left|Z_1\right|=\left|Z_2\right|\) then \(z_1\) and \(z_2\) are at the same distance from origin.
But if \(\arg \left(z_1\right) \neq \arg \left(z_2\right)\), then \(z_1\) and \(z_2\) are different.
So, if \(\left|z_1\right|=\left|z_2\right|\), then it is not necessary that \(z_1=z_2\).
Consider \(Z_1=3+4 i\) and \(Z_2=4+3 i\)

Hint

(a)

\(
\begin{aligned}
&\left(a^2+1\right)^2 / 2 a-i=x+i y \\
&\Rightarrow \quad\left|\frac{\left.(a^2+1\right)^2}{2 a-i}\right|=|x+i y| \\
&\Rightarrow \quad \frac{\left|\left(a^2+1\right)^2\right|}{|2 a-i|}=|x+i y| \Rightarrow \frac{\left(a^2+1\right)^2}{\sqrt{(2 a)^2+(-1)^2}}=\sqrt{x^2+y^2} \\
&\Rightarrow \quad x^2+y^2=\frac{\left(a^2+1\right)^4}{4 a^2+1}
\end{aligned}
\)

Hint

(b) Let \(z=|z|(\cos \theta+i \sin \theta)\), where \(\theta=\arg (z)\).
Given that, \(|z|=4\) and \(\arg (z)=\frac{5 \pi}{6}\).
\(
\begin{array}{rlr}
\Rightarrow \quad z & =4\left[\cos \frac{5 \pi}{6}+i \sin \frac{5 \pi}{6}\right] \quad \text { ( } z \text { lies in II quadrant) } \\
& =4\left[-\frac{\sqrt{3}}{2}+i \frac{1}{2}\right]=-2 \sqrt{3}+2 i &
\end{array}
\)

Hint

(b)

\(
\begin{aligned}
&\left|(1+i) \frac{(2+i)}{(3+i)} \times \frac{3-i}{3-i}\right|=\left|(1+i) \cdot \frac{6-2 i+3 i-i^2}{9-i^2}\right| \\
&=\left|\frac{(1+i)(7+i)}{9+1}\right| \\
&=\left|\frac{7+i+7 i+i^2}{9+1}\right| \\
&=\left|\frac{7+i+7 i+i^2}{10}\right| \\
&=\left|\frac{7+8 i-1}{10}\right| \\
&=\left|\frac{6+8 i}{10}\right| \\
&=\left|\frac{3}{5}+\frac{4}{5} i\right| \\
&=\sqrt{\left(\frac{3}{5}\right)^2+\left(\frac{4}{5}\right)^2}=1
\end{aligned}
\)

Hint

(c)

\(
\begin{gathered}
(1+i \sqrt{3})^2=1+i^2 \cdot 3+2 \sqrt{3} i \\
=1-3+2 \sqrt{3} i=-2+2 \sqrt{3} i \\
\tan \alpha=\left|\frac{2 \sqrt{3}}{-2}\right| \\
\tan \alpha=|-\sqrt{3}|=\sqrt{3} \\
\tan \alpha=\tan \frac{\pi}{3} \\
\alpha=\frac{\pi}{3} 
\end{gathered}
\)
Now \(\operatorname{Re}(z)<0\) and image(z)>0 \(\arg (z)=\pi-\alpha=\pi-\frac{\pi}{3}=\frac{2 \pi}{3}\) Hence, the principal \(\arg =\frac{2 \pi}{3}\)

Hint

(b) We have \(|z-5 i / z+5 i|\)
\(
\begin{array}{ll}
\Rightarrow & |z-5 i|=|z+5 i| \quad \Rightarrow|x+i y-5 i|=|x+i y+5 i| \\
\Rightarrow & |x+i(y-5)|^2=|x+i(y+5)|^2 \quad \Rightarrow x^2+(y-5)^2=x^2+(y+5)^2 \\
\Rightarrow & 20 y=0 \quad \Rightarrow y=0
\end{array}
\)
So, \(z\) lies on the \(x\)-axis (real axis).

Hint

(d)
let \(z=\sin x+i \cos 2 x \quad \bar{z}=\sin x-i \cos 2 x\)
But we are given that \(\bar{z}=\cos x-i \sin 2 x\)
\(\sin x-\mathrm{i} \cos 2 x=\cos x-i \sin 2 x\) Comparing the real and imaginary part, we get
\(\sin \mathrm{x}=\cos \mathrm{x} \quad\) and \(\quad \cos 2 \mathrm{x}=\sin 2 \mathrm{x}\)
\(\tan x=1\) and \(\tan 2 x=1\)
Now, \(\tan 2 x=1\)
\(\Rightarrow \frac{2 \tan x}{1-\tan ^2 x}=1\), which is not satisfied by \(\tan x=1\)
Hence, no value of \(x\) is possible.

Hint

(c)

\(z=\frac{1-i \sin \alpha}{1+2 i \sin \alpha}\)
\(
\begin{aligned}
&=\frac{(1-i \sin \alpha)(1-2 i \sin \alpha)}{(1+2 i \sin \alpha)(1-2 i \sin \alpha)}=\frac{1-i \sin \alpha-2 i \sin \alpha+2 i^2 \sin ^2 \alpha}{1-4 i^2 \sin ^2 \alpha} \\
&=\frac{1-3 i \sin \alpha-2 \sin ^2 \alpha}{1+4 \sin ^2 \alpha}=\frac{1-2 \sin ^2 \alpha}{1+4 \sin ^2 \alpha}-\frac{3 i \sin \alpha}{1+4 \sin ^2 \alpha}
\end{aligned}
\)
It is given that \(z\) is a purely real.
\(
\begin{aligned}
&\Rightarrow \quad \frac{-3 \sin \alpha}{1+4 \sin ^2 \alpha}=0 \quad \Rightarrow-3 \sin \alpha=0 \quad \Rightarrow \sin \alpha=0 \\
&\Rightarrow \quad \alpha=n \pi, n \in I
\end{aligned}
\)

Hint

(b) Given that: \(z=x+i y\)
If z lies in the third quadrant.
So \(x<0\) and \(y<0\).
\(
\begin{aligned}
&\bar{z}=\mathrm{x}-\mathrm{i} y \\
&\frac{\bar{z}}{z}=\frac{x-i y}{x+i y} \\
&=\frac{x-i y}{x+i y} \times \frac{x-i y}{x-i y} \\
&=\frac{x^2+i^2 y^2-2 x y i}{x^2-i^2 y^2} \\
&=\frac{x^2-y^2-2 x y i}{x^2+y^2} \\
&=\frac{x^2-y^2}{x^2+y^2}-\frac{2 x y}{x^2+y^2} i
\end{aligned}
\)
When z lies in third quadrant then \(\frac{\bar{z}}{z}\) will also be lie in third quadrant \(\therefore \frac{x^2-y^2}{x^2+y^2}<0\) and \(\frac{-2 x y}{x^2+y^2}<0\)
\(\Rightarrow \mathrm{x}^2-\mathrm{y}^2<0\) and \(2 \mathrm{xy}>0\)
\(\Rightarrow \mathrm{x}^2<\mathrm{y}^2\) and \(\mathrm{xy}>0\)
So \(x<y<0\).

Hint

(a) Let \(z=x+i y\) So, \((z+3)(\bar{z}+3)=(x+i y+3)(x-i y+3)\)
\(
\begin{aligned}
&=[(x+3)+i y][(x+3)-i y] \\
&=(x+3)^2-y^2 i^2=(x+3)^2+y^2 \\
&=|x+3+i y|^2=|z+3|^2
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (b) : Given }\left(\frac{1+i}{1-i}\right)^x=1 \\
& \Rightarrow\left(\frac{2 i}{2}\right)^x=1 \\
& \Rightarrow i^x=1 \\
& \Rightarrow i^x=(i)^{4 n} \\
& \Rightarrow x=4 n, n \in I^{+}
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
&\text { Given that } \frac{3-4 i x}{3+4 i x}=\alpha-i \beta \\
&=\frac{3-4 i x}{3+4 i x} \times \frac{3-4 i x}{3-4 i x}=\alpha-i \beta \\
&\frac{9-12 i x-12 i x+16 i^2 x^2}{9-16 i^2 x^2}=\alpha-i \beta \\
&=\frac{9-16 x^2}{9+16 x^2}-i\frac{24 x}{9+16 x^2}=\alpha-i \beta \ldots \ldots \text { (i) } \\
&\frac{9-16 x^2}{9+16 x^2}+i\frac{24 x}{9+16 x^2}=\alpha+i \beta \ldots \ldots \ldots \text { (ii) }
\end{aligned}
\)
Multiplying eqn (i )and (ii )we get
\(
\begin{aligned}
&\left(\frac{9-16 x^2}{9+16 x^2}\right)^2+\left(\frac{24 x}{9+16 x^2}\right)^2=\alpha^2+\beta^2 \\
&\frac{81+256 x^4-288 x^2+576 x^2}{\left(9+16 x^2\right)^2}=\alpha^2+\beta^2 \\
&\frac{81+256 x^4+288 x^2}{\left(9+16 x^2\right)^2}=\alpha^2+\beta^2 \\
&\frac{\left(9+16 x^2\right)^2}{\left(9+16 x^2\right)^2}=\alpha^2+\beta^2=1
\end{aligned}
\)

Hint

(a) Let, \(z_1=r_1 e^{i \alpha}\) and \(z_2=r_2 e^{i \beta}\)
\(\left|z_1\right|=r_1\) and \(\left|z_2\right|=r_2\)
Option A:
\(z_1 z_2=r_1 \Gamma_2 e^{i(\alpha+\beta)}\)
\(\left|z_1 z_2\right|=r_1 r_2=\left|z_1\right|\left|z_2\right|\)
Option A correct
Option B:
\(\arg \left(z_1 z_2\right)=\alpha+\beta\)
\(=\arg \left(z_1\right)+\arg \left(z_2\right)\)
Option B not correct
Let, \(z_1=a+i b\) and \(z_2=c+i d\)
Option C
\(z_1+z_2=(a+c)+i(b+d)\)
\(\left|\mathrm{z}_1+\mathrm{z}_2\right|=\sqrt{(\mathrm{a}+\mathrm{c})^2+(\mathrm{b}+\mathrm{d})^2}\)
\(\left|\mathrm{z}_1\right|=\sqrt{\mathrm{a}^2+\mathrm{b}^2}\) and \(\left|\mathrm{z}_2\right|=\sqrt{\mathrm{c}^2+\mathrm{d}^2}\)
We cannot say anything about option c and option d

Hint

(b) Given that: \(z=2-i\)
If z rotated through an angle of \(\frac{\pi}{2}\) about the origin in the clockwise direction. Then the new position \(=z \cdot e^{-\left(\frac{\pi}{2}\right)}\)
\(
\begin{aligned}
&=(2-i) e^{-\left(\frac{\pi}{2}\right)} \\
&=(2-i)\left[\cos \left(\frac{-\pi}{2}\right)+i \sin \left(\frac{-\pi}{2}\right)\right] \\
&=(2-\mathrm{i})(0-\mathrm{i}) \\
&=-1-2 \mathrm{i}
\end{aligned}
\)

Hint

(d) \(x+y i\) is a non-real complex number if \(y \neq 0\) If \(x, y \in R\)

Hint

(d) Given that: \(a+i b=c+i d\)
\(
\begin{aligned}
&\Rightarrow|\mathrm{a}+\mathrm{ib}|=|c+i \mathrm{~d}| \\
&\Rightarrow \sqrt{a^2+b^2}=\sqrt{c^2+d^2}
\end{aligned}
\)
Squaring both sides, we get \(a^2+b^2=c^2+d^2\)

Hint

(b) Let \(z=x+i y\)
\(
\begin{aligned}
&\left|\frac{i+z}{i-z}\right|=1 \\
&\Rightarrow|i+z|=|i-z| \\
&\Rightarrow|x+y i+i|=|i-x-i y| \\
&\Rightarrow|x+i(y+1)|=|-x+i(1-y)| \\
&\Rightarrow \sqrt{x^2+(y+1)^2}=\sqrt{x^2+(1-y)^2} \\
&\Rightarrow x^2+(y+1)^2=x^2+(1-y)^2 \\
&\Rightarrow y+1=1-y \\
&\Rightarrow 2 y=0 \\
&\Rightarrow y=0
\end{aligned}
\)

It lies on the x-axis.

Hint

(c) Let \(z_1=r_1\left(\cos \theta_1+i \sin \theta_1\right)\) and \(z_2=r_2\left(\cos \theta_2+i \sin \theta_2\right)\)
Since \(\left|z_1+z_2\right|=\left|z_1\right|+\left|z_2\right|\)
\(
\begin{aligned}
&\left|\mathrm{z}_1+\mathrm{z}_2\right|=\mathrm{r}_1 \cos \theta_1+\mathrm{i} \mathrm{r}_1 \sin \theta_1+\mathrm{r}_2 \cos \theta_2+\mathrm{i} r_2 \sin \theta_2 \\
&\left|\mathrm{z}_1+\mathrm{z}_2\right|=\sqrt{r_1^2 \cos ^2 \theta_{+} r_2^2 \cos ^2 \theta_2+2 r_1 r_2 \cos \theta_1 \cos \theta_2+r_1^2 \sin ^2 \theta_1+r_2^2 \sin ^2 \theta_2+2 r_1 r_2 \sin \theta_1 \sin \theta_2} \\
&=\sqrt{r_1^2+r_2^2+2 r_1 r_2 \cos \left(\theta_1-\theta_2\right)} \\
&\text { But }\left|\mathrm{z}_1+\mathrm{z}_2\right|=\left|\mathrm{z}_1\right|+\left|\mathrm{z}_2\right| \\
&\text { So } \sqrt{r_1^2+r_2^2+2 r_1 r_2 \cos \left(\theta_1-\theta_2\right)}=r_1+r_2 \\
&\text { Squaring both sides, we get } \\
&r_1^2+r_2^2+2 r_1 r_2 \cos \left(\theta_1-\theta_2\right)=r_1^2+r_2^2+2 r_1 r_2 \\
&\Rightarrow 2 r_1 r_2-2 r_1 r_2 \cos \left(\theta_1-\theta_2\right)=0 \\
&\Rightarrow 1-\cos \left(\theta_1-\theta_2\right)=0 \\
&\Rightarrow \cos \left(\theta_1-\theta_2\right)=1 \\
&\Rightarrow \theta_1-\theta_2=0 \\
&\Rightarrow \theta_1=\theta_2 \\
&\text { So, arg }\left(z_1\right)=\arg \left(\mathrm{z}_2\right)
\end{aligned}
\)

Hint

(b) Let \(z=x+y i\)
\(
\begin{aligned}
&|z|=|x+y i| \text { and }|z|^2=|x+y i|^2 \\
&\Rightarrow|z|^2=x^2+y^2 \\
&\text { Now } z^2=x^2+y^2 i^2+2 x y i \\
&z^2=x^2-y^2+2 x y i \\
&|z^2|=\sqrt{\left(x^2-y^2\right)^2+(2x y)^2} \\
&=\sqrt{x^4+y^4-2 x^2 y^2+4 x^2 y^2} \\
&=\sqrt{x^4+y^4+2 x^2 y^2} \\
&=\sqrt{\left(x^2+y^2\right)^2} \\
&\text { So }|z|^2=x^2 + y^2=|z^2| \\
&\text { So }|z|^2=\left|z^2\right|
\end{aligned}
\)

Hint

(c) We have, \(\frac{1+i \cos \theta}{1-2 i \cos \theta}=\frac{1+i \cos \theta}{1-2 i \cos \theta} \times \frac{1+2 i \cos \theta}{1+2 i \cos \theta}\) \(=\frac{1+3 i \cos \theta+2 i^2 \cos ^2 \theta}{1^2-(2 i \cos \theta)^2}=\frac{1-2 \cos ^2 \theta}{1+4 \cos ^2 \theta}+i\left(\frac{3 \cos \theta}{1+4 \cos ^2 \theta}\right)\), Since \(\frac{1+i \cos \theta}{1-2 i \cos \theta}\) is a real number.
\(
\begin{aligned}
&\therefore \frac{3 \cos \theta}{1+4 \cos ^2 \theta}=0 \\
&\Rightarrow \cos \theta=0 \\
&\Rightarrow \cos \theta=\cos \frac{\pi}{2} \\
&\therefore \text { General solution is } \theta=2 n \pi \pm \frac{\pi}{2}
\end{aligned}
\)

Hint

(c) Let \(z=x+0\) i and \(x<0\)
\(
\Rightarrow|z|=\sqrt{(-1)^2+0^2}=1
\)
Since the point \((-x, 0)\) lies on the negative side of the real axis, \(\therefore\) Principal argument \((z)=\pi\)

Hint

(a)

\(
\begin{aligned}
&f(z)=\frac{7-z}{1-z^2} \\
&=\frac{7-(1+2 i)}{1-(1+2 i)^2} \\
&=\frac{7-1-2 i}{1-\left(1^2+2^2 i^2+4 i\right)} \\
&=\frac{6-2 i}{1-1+4-4 i} \\
&=\frac{6-2 i}{4-4 i} \\
&=\frac{6-2 i}{4-4 i} \times \frac{4+4 i}{4+4 i} \\
&=\frac{24+24 i-8 i-8 i^2}{4^2-4^2 i^2} \\
&=\frac{24+16 i+8}{16+16} \\
&=\frac{32+16 i}{32} \\
&=1+\frac{1}{2} i
\end{aligned}
\)
Since \(
\begin{aligned}
&z=1+2 i \\
&\therefore|z|=\sqrt{(1)^2+(2)^2} \\
&=\sqrt{1+4} \\
&=\sqrt{5} \\
&\therefore|f(z)|=\sqrt{(1)^2+\left(\frac{1}{2}\right)^2} \\
&=\sqrt{1+\frac{1}{4}} \\
&=\frac{\sqrt{5}}{2} \\
&=\frac{|z|}{2}
\end{aligned}
\)