The Algebra of complex numbers involves mathematical operations such as addition, subtraction, multiplication, and division. 

Addition of two complex numbers

Let \(z_1=a+i b\) and \(z_2=c+i d\) be any two complex numbers.
Then, the sum \(z_1+z_2\) is defined as follows:
\(z_1+z_2=(a+c)+i(b+d) \text {, which is again a complex number. }\)

For example, \((2+i 3)+(-6+i 5)=(2-6)+i(3+5)=-4+i 8\) (add the real part and imaginary part separately)

The addition of complex numbers satisfy the following properties:

1. Closure law: The sum of two complex numbers is a complex number, i.e., \(z_1+z_2\) is a complex number for all complex numbers \(z_1\) and \(z_2\). That is \(z_1+z_2\) = complex number.
2. Commutative law: For any two complex numbers \(z_1\) and \(z_2\), \(z_1+z_2=z_2+z_1\)(order does not matter). For Example, \(\begin{aligned}&(2+i 3)+(4+i 5)=(4+i 5)+(2+i 3) \end{aligned}\)
3. Associative law: For any three complex numbers \(z_1, z_2, z_3\), \(\left(z_1+z_2\right)+z_3=z_1+\left(z_2+z_3\right)\). For example,
   \(\begin{aligned}& {[(2+i 3)+(4+i 5)]+(1+i)=(2+i 3)+[(4+i 5)+(1+i)]} \end{aligned}\)
4. Additive identity: There exists the complex number \(0+i 0\) (denoted as 0), called the additive identity or the zero complex number, such that, for every complex number \(z, z+0=z\).
5. Additive inverse: To every complex number \(z=a+i b\), we have the complex number \(-a+i(-b)\) (denoted as \(-z\) ), called the additive inverse or negative of \(z\). We observe that \(z+(-z)=0\) (the additive identity).

Difference of two complex numbers

Given any two complex numbers \(z_1\) and \(z_2\), the difference \(z_1-z_2\) is defined as follows:
\(z_1-z_2=z_1+\left(-z_2\right) \text {. }\) Let’s assume

\(
\begin{aligned}
z_1 &=a+i b \\
z_2 &=c+i d \\
z_1-z_2 &=(a-c)+i(b-d). \\
\end{aligned}\)

For example: \((6+i3)-(2+i) =(6-2)+i(3-1) =4+2 i\)

Multiplication of two complex numbers

\(z_1=a+i b\) and \(z_2=c+i d\) be any two complex numbers. Then, the product \(z_1 z_2\) is defined as follows:

\(
\begin{aligned}
z_1 z_2 &=(a+i b)(c+i d) \\
&=a c+i a d+i b c+i^2 b d \\
&=a c+i(a d+b c)-b d \\
&=(a c-b d)+i(a d+b c) .
\end{aligned}
\)

For example: \((3+i 5)(2+i 6)=(3 \times 2-5 \times 6)+i(3 \times 6+5 \times 2)=-24+i 28\)

The multiplication of complex numbers possesses the following properties:

1. Closure law: The product of two complex numbers is a complex number, the product \(z_1 z_2\) is a complex number for all complex numbers \(z_1\) and \(z_2\).
2. Commutative law: For any two complex numbers \(z_1\) and \(z_2\),\(z_1 z_2=z_2 z_1\)
3. Associative law: For any three complex numbers \(z_1, z_2, z_3\), \(\left(z_1 z_2\right) z_3=z_1\left(z_2 z_3\right)\)
4. Multiplicative identity: There exists the complex number \(1+i 0\) (denoted as 1), called the multiplicative identity such that \(z .1=z\), for every complex number \(z\)
5. Multiplicative inverse: For every non-zero complex number \(z=a+i b\) or \(a+b i(a \neq 0, \mathrm{~b} \neq 0)\), we have the complex number \(\frac{a}{a^2+b^2}+i \frac{-b}{a^2+b^2}\) (denoted by \(\frac{1}{z}\) or \(\left.z^{-1}\right)\), called the multiplicative inverse of \(z\) such that \(z \cdot \frac{1}{z}=1\) (the multiplicative identity).
6. Distributive law: For any three complex numbers \(z_1, z_2, z_3\),
   (a) \(z_1\left(z_2+z_3\right)=z_1 z_2+z_1 z_3\)
   (b) \(\left(z_1+z_2\right) z_3=z_1 z_3+z_2 z_3\)

Division of two complex numbers

Let’s assume the two complex numbers given are \(z_1\) and \(z_2\), where \(z_2 \neq 0\), the quotient \(\frac{z_1}{z_2}\) is defined by
\(\frac{z_1}{z_2}=z_1 \frac{1}{z_2}\)

For example:
\(
\begin{aligned}
z_1 &=2+i 3 \\
z_2 &=2+i 2 \\
\frac{z_1}{z_2} &=z_1 \times \frac{1}{z_2} \\
&=(2+i 3) \times \frac{1}{2+i 2} \\
&=(2+i 3)\left[\frac{1}{2+i 2} \times \frac{2-i 2}{2-i 2}\right] \\
&=(2+i 3) \times \frac{2-i 2}{2^{2}-i^{2} 2^{2}} \\
&=\frac{(2+i 3)(2-i 2)}{4+4} \\
&=\frac{4-i 4+i 6-i^2 6}{8} \\
&=\frac{4+2 i+6}{8} \\
&=\frac{10+2 i}{8} \\
&=\frac{5+i}{4}
\end{aligned}
\)

Power of \(i\) (Integral Powers Of Iota)

We know that
\(i^3=i^2 i=(-1) i=-i, \quad i^4=\left(i^2\right)^2=(-1)^2=1\)
\(i^5=\left(i^2\right)^2 i=(-1)^2 i=i, \quad i^6=\left(i^2\right)^3=(-1)^3=-1\), etc.
Also, we have \(\quad i^{-1}=\frac{1}{i} \times \frac{i}{i}=\frac{i}{-1}=-i, \quad i^{-2}=\frac{1}{i^2}=\frac{1}{-1}=-1\),
\(i^{-3}=\frac{1}{i^3}=\frac{1}{-i} \times \frac{i}{i}=\frac{i}{1}=i, \quad i^{-4}=\frac{1}{i^4}=\frac{1}{1}=1\)
In general, for any integer \(k, i^{4 k}=1, i^{4 k+1}=i, i^{4 k+2}=-1, i^{4 k+3}=-i\)

Example 1: If \(i=\sqrt{-1}\) and \(n \in N\), then \(i^n+i^{n+1}+i^{n+2}+i^{n+3}\) is equal to
(a) 1
(b) \(i\)
(c) \(i^n\)
(d) 0

Solution: (d) We have,
\(
i^n+i^{n+1}+i^{n+2}+i^{n+3}=i^n\left(1+i+i^2+i^3\right)=i^n(1+i-1-i)=0
\)

Example 2: If \(i=\sqrt{-1}\), then \(\left\{i^n+i^{-n}: n \in Z\right\}\) is equal to
(a) \(\{0,2\}\)
(b) \(\{0,-2\}\)
(c) \(\{0,-2,2\}\)
(d) \(\{0,-2 i\}\)

Solution: (c) We have the following cases:
CASE I: When \(n=4 m, m \in Z\)
In this case, we have
\(
\begin{aligned}
& i^n=i^{4 m}=\left(i^4\right)^m=1 \text { and } i^{-n}=\frac{1}{i^n}=\frac{1}{1}=1 \\
\therefore & i^n+i^{-n}=2
\end{aligned}
\)
CASE \(\amalg\): When \(n=4 m+1, m \in Z\)
In this case, we have
\(
\begin{aligned}
&i^n=i^{4 m+1}=\left(i^4\right)^m i^1=i \text { and, } i^{-n}=\frac{1}{i^n}=\frac{1}{i}=-i\\
&\therefore \quad i^n+i^{-n}=i-i=0
\end{aligned}
\)‘
CASE III: When \(n=4 m+2, m \in Z\)
In this case, we have
\(
\begin{array}{ll}
& i^n=i^{4 m+2}=\left(i^4\right)^m i^2=i^2=-1 \\
\therefore \quad & i^n+i^{-n}=(-1)+(-1)=-2
\end{array}
\)
CASEIV: When \(n=4 m+3, m \in Z\)
In this case, we have
\(
\begin{aligned}
& i^n=i^{4 m+3}=\left(i^4\right)^m i^3=i^3=-i \text { and }, i^{-n}=\frac{1}{i^n}=\frac{1}{i^3}=i \\
\therefore \quad & i^n+i^{-n}=(-i)+(i)=0
\end{aligned}
\)
Hence, \(S=\{-2,0,2\}\)

Example 3: The value of \(\sum_{n=1}^{13}\left(i^n+i^{n+1}\right)\), where \(i=\sqrt{-1}\) equals [IIT 1998]
(a) \(i\)
(b) \(i-1\)
(c) \(-i\)
(d) 0

Solution: (b) We have,
\(
\begin{aligned}
& \sum_{n=1}^{13}\left(i^n+i^{n+1}\right) \\
& =\sum_{n=1}^{13} i^n+\sum_{n=1}^{13} i^{n+1} \\
& =i\left(\frac{1-i^{13}}{1-i}\right)+i^2\left(\frac{1-i^{13}}{1-i}\right)=i \frac{(1-i)}{1-i}-\left(\frac{1-i}{1-i}\right)=i-1
\end{aligned}
\)

Example 4: If \(n\) is an odd integer, then \((1+i)^{6 n}+(1-i)^{6 n}\) is equal to
(a) 0
(b) 2
(c) -2
(d) none of these.

Solution: (a) We have,
\(
\begin{aligned}
& (1+i)^{6 n}+(1-i)^{6 n}=\left\{(1+i)^2\right\}^{3 n}+\left\{(1-i)^2\right\}^{3 n} \\
& =(2 i)^{3 n}+(-2 i)^{3 n}=2^{3 n}\left\{i^{3 n}+(-i)^{3 n}\right\}
\end{aligned}
\)
\(
=2^{3 n}\left|i^{3 n}-i^{3 n}\right|=0 \quad[\because n \text { is odd }]
\)

Example 5: If \(m, n, p, q\) are consecutive integers then the value of \(i^m+i^n+i^p+i^q\) is
(a) 1
(b) 4
(c) 0
(d) none of these.

Solution: (c) Since \(m, n, p, q\) are consecutive integers.
\(
\begin{aligned}
\therefore \quad n=m+1, p=m+2, & q=m+3 \\
\Rightarrow \quad i^m+i^n+i^p+i^q & =i^m+i^{m+1}+i^{m+2}+i^{m+3} \\
& =i^m\left(1+i+i^2+i^3\right) \\
& =i^m(1+i-1-i)=0
\end{aligned}
\)

Example 6: \(i^2+i^4+i^6+\ldots .(2 n+1)\) terms \(=?\) [CEE (Delhi) 2000]
(a) -1
(b) 1
(c) \(-i\)
(d) \(i\)

Solution: (a) We have,
\(i^2+i^4+i^6+\ldots(2 n+1)\) terms
\(
=i^2\left\{\frac{1-\left(i^2\right)^{2 n+1}}{1-i^2}\right\}=i^2\left(\frac{1+1}{1+1}\right)=-1
\)

The square roots of a negative real number (Imaginary Quantities)

Note that \(i^2=-1\) and \((-i)^2=i^2=-1\)
Therefore, the square roots of \(-1\) are \(i,-i\). However, by the symbol \(\sqrt{-1}\), we would mean \(i\) only.

Now, we can see that \(i\) and \(-i\) both are the solutions of the equation \(x^2+1=0\) or \(x^2=-1\).
Similarly
\(
\begin{aligned}
&(\sqrt{3} i)^2=(\sqrt{3})^2 i^2=3(-1)=-3 \\
&(-\sqrt{3} i)^2=(-\sqrt{3})^2 i^2=-3
\end{aligned}
\)
Therefore, the square roots of \(-3\) are \(\sqrt{3} i\) and \(-\sqrt{3} i\).
Again, the symbol \(\sqrt{-3}\) is meant to represent \(\sqrt{3} i\) only, i.e., \(\sqrt{-3}=\sqrt{3} i\).
Generally, if \(a\) is a positive real number, \(\sqrt{-a}=\sqrt{a} \sqrt{-1}=\sqrt{a} i\),
We already know that \(\sqrt{a} \times \sqrt{b}=\sqrt{a b}\) for all positive real number \(a\) and \(b\). This result also holds true when either \(a>0, b<0\) or \(a<0, b>0\). What if \(a<0, b<0\)?
Let us examine.
Note that
\(i^2=\sqrt{-1} \sqrt{-1}=\sqrt{(-1)(-1)}\) (by assuming \(\sqrt{a} \times \sqrt{b}=\sqrt{a b}\) for all real numbers) \(=\sqrt{1}=1\), which is a contradiction to the fact that \(i^2=-1\).
Therefore, \(\sqrt{a} \times \sqrt{b} \neq \sqrt{a b}\) if both \(a\) and \(b\) are negative real numbers.
Further, if any of \(a\) and \(b\) is zero, then, clearly, \(\sqrt{a} \times \sqrt{b}=\sqrt{a b}=0\).

Identities

We prove the following identity \(\left(z_1+z_2\right)^2=z_1^2+z_2^2+2 z_1 z_2\), for all complex numbers \(z_1\) and \(z_2\).
Proof We have,
\(
\begin{array}{ll}
\left(z_1+z_2\right)^2=\left(z_1+z_2\right)\left(z_1+z_2\right) & \\
=\left(z_1+z_2\right) z_1+\left(z_1+z_2\right) z_2 & \text { (Distributive law) } \\
=z_1^2+z_2 z_1+z_1 z_2+z_2^2 & \text { (Distributive law) } \\
=z_1^2+z_1 z_2+z_1 z_2+z_2^2 & \text { (Commutative law of multiplication) } \\
=z_1^2+2 z_1 z_2+z_2^2 &
\end{array}
\)
Similarly, we can prove the following identities:
(i) \(\left(z_1-z_2\right)^2=z_1^2-2 z_1 z_2+z_2^2\)
(ii) \(\left(z_1+z_2\right)^3=z_1^3+3 z_1^2 z_2+3 z_1 z_2^2+z_2^3\)
(iii) \(\left(z_1-z_2\right)^3=z_1^3-3 z_1^2 z_2+3 z_1 z_2^2-z_2^3\)
(iv) \(z_1^2-z_2^2=\left(z_1+z_2\right)\left(z_1-z_2\right)\)
In fact, many other identities which are true for all real numbers can be proved to be true for all complex numbers.

Example 7: Express \((5-3 i)^3\) in the form \(a+i b\).

Solution: We have, \((5-3 i)^3=5^3-3 \times 5^2 \times(3 i)+3 \times 5(3 i)^2-(3 i)^3\)
\(
=125-225 i-135+27 i=-10-198 i \text {. }
\)

Example 8: Express \((-\sqrt{3}+\sqrt{-2})(2 \sqrt{3}-i)\) in the form of \(a+i b\)

Solution: We have, \((-\sqrt{3}+\sqrt{-2})(2 \sqrt{3}-i)=(-\sqrt{3}+\sqrt{2} i)(2 \sqrt{3}-i)\)
\(
=-6+\sqrt{3} i+2 \sqrt{6} i-\sqrt{2} i^2=(-6+\sqrt{2})+\sqrt{3}(1+2 \sqrt{2}) i
\)

Theorem: If \(a, b\) are positive real numbers, then
\(
\sqrt{-a} \times \sqrt{-b}=-\sqrt{a b}
\)

Proof: We have,
\(
\begin{array}{ll}
& \sqrt{-a}=\sqrt{-1 \times a}=\sqrt{-1} \times \sqrt{a}=i \sqrt{a} \\
\text { and, } \sqrt{-b}=\sqrt{-1 \times b}=\sqrt{-1} \times \sqrt{b} \quad=i \sqrt{b} \\
\therefore & \sqrt{-a} \times \sqrt{-b}=(i \sqrt{a})(i \sqrt{b}) \\
\Rightarrow & \sqrt{-a} \times \sqrt{-b}=i^2(\sqrt{a} \times \sqrt{b})=-1(\sqrt{a b})=-\sqrt{a b}
\end{array}
\)

• Note 1: For any two real numbers \(\sqrt{a} \times \sqrt{b}=\sqrt{a b}\) is true only when at least one of \(a\) and \(b\) is either positive or zero. In other words, \(\sqrt{a} \times \sqrt{b}=\sqrt{a b}\) is not valid if \(a\) and \(b\) both are negative.
• Note 2: For any positive real number \(a\), we have
  \(
  \sqrt{-a}=\sqrt{-1 \times a}=\sqrt{-1} \times \sqrt{a}=i \sqrt{a}
  \)

Example 9: If \(a, b \in R\) such that \(a b>0\), then \(\sqrt{a} \sqrt{b}\) is equal to
(a) \(\sqrt{|a||b|}\)
(b) \(-\sqrt{|a||b|}\)
(c) \(\sqrt{a b}\)
(d) none of these

Solution: (d) We have, \(a b>0 \Rightarrow\) either \(a>0\) and \(b>0\) or, \(a<0\) and \(b<0\).
So, we have the following cases.
Case-I: When \(a>0\) and \(b>0\)
In this case, we have
\(
\begin{aligned}
& \sqrt{a b}=\sqrt{a} \sqrt{b} \\
\Rightarrow & \sqrt{|a||b|}=\sqrt{a} \sqrt{b} \quad \quad[\because|a|=a,|b|=b]
\end{aligned}
\)
Case-II: When \(a<0\) and \(b<0\)
In this case, we have
\(
\begin{array}{ll}
& \sqrt{a}=\sqrt{-1 \times|a|}=i \sqrt{|a|}, \sqrt{b}=\sqrt{-1|b|}=i \sqrt{|b|} \\
\therefore & \sqrt{a} \sqrt{b}=i \sqrt{|a|} \times i \sqrt{|b|} \\
\Rightarrow & \sqrt{a} \sqrt{b}=i^2 \sqrt{|a||b|} \\
\Rightarrow & \sqrt{a} \sqrt{b}=-\sqrt{|a||b|}
\end{array}
\)

Example 10: If \(a, b \in R\) such that \(a b<0\), then \(\sqrt{a} \sqrt{b}\) is equal to
(a) \(i \sqrt{|a| b}\)
(b) \(i \sqrt{a|b|}\)
(c) \(i \sqrt{|a||b|}\)
(d) \(-\sqrt{|a||b|}\)

Solution: (c) We have,
\(
a b>0 \Rightarrow \text { either } a<0 \text { and } b>0 \text { or, } a>0 \text { and } b<0
\)
So, we have the following cases:
CASE I: When \(a<0\) and \(b>0\)
In this case, we have
\(
\begin{array}{ll}
& \sqrt{a}=\sqrt{-1 \times|a|}=i \sqrt{|a|} \text { and } \sqrt{b}=\sqrt{|b|} \\
\therefore \quad & \sqrt{a} \quad \sqrt{b}=i \sqrt{|a|} \sqrt{|b|}=i \sqrt{|a||b|}
\end{array}
\)
CASE II: When \(a>0\) and \(b<0\)
In this case, we have
\(
\begin{array}{ll}
& \sqrt{a}=\sqrt{|a|} \text { and } \sqrt{b}=\sqrt{-1 \times|b|}=i \sqrt{|b|} \\
\therefore & \sqrt{a} \sqrt{b}=i \sqrt{|a|} \times i \sqrt{|b|}=i \sqrt{|a||b|}
\end{array}
\)
Thus, in both the cases, we have
\(
\sqrt{a} \sqrt{b}=i \sqrt{|a||b|}
\)

Equality Of Complex Numbers

Two complex numbers \(z_1=a_1+i b_1\) and \(z_2=a_2+i b_2\) are equal, if \(a_1=a_2\) and \(b_1=b_2\) i.e. \(\operatorname{Re}\left(z_1\right)=\operatorname{Re}\left(z_2\right)\) and \(\operatorname{Im}\left(z_1\right)=\operatorname{Im}\left(z_2\right)\)

Example 11: If \(\sqrt{3}+i=(a+i b)(c+i d)\), then \(\tan ^{-1} \frac{b}{a}+\tan ^{-1} \frac{d}{c}\) has the value [CEE (Delhi) 2007]
(a) \(\frac{\pi}{3}+2 n \pi, n \in Z\)
(b) \(n \pi+\frac{\pi}{6}, n \in Z\)
(c) \(n \pi-\frac{\pi}{3}, n \in Z\)
(d) \(2 n \pi-\frac{\pi}{3}, n \in \mathrm{Z}\)

Solution: (b) We have,
\(
\begin{aligned}
& \sqrt{3}+i=(a+i b)(c+i d) \\
\Rightarrow & \sqrt{3}+i=(a c-b d)+i(a d+b c) \\
\Rightarrow & \sqrt{3}=a c-b d \text { and } 1=a d+b c
\end{aligned}
\)
Now,
\(
\begin{aligned}
& \therefore \quad \tan ^{-1} \frac{b}{a}+\tan ^{-1} \frac{d}{c}=\tan ^{-1}\left(\frac{\frac{b}{a}+\frac{d}{c}}{1-\frac{b}{a} \times \frac{d}{c}}\right) \\
& \Rightarrow \quad \tan ^{-1} \frac{b}{a}+\tan ^{-1} \frac{d}{c}=\tan ^{-1}\left(\frac{a d+b c}{a c-b d}\right)=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right) \\
& \Rightarrow \quad \tan ^{-1} \frac{b}{a}+\tan ^{-1} \frac{d}{c}=n \pi+\frac{\pi}{6}, n \in Z
\end{aligned}
\)