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

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

Example 1:
\(\begin{aligned} &(6+i 3)-(2+i) =&(6-2)+i(3-1) &=4+2 i\end{aligned}\)

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:

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

Example 2:

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

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
\(\frac{z_1}{z_2}=z_1 \frac{1}{z_2}\)

Example 3:

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

Power of \(i\)

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 \(k, i^{4 k}=1, i^{4 k+1}=i, i^{4 k+2}=-1, i^{4 k+3}=-i\)

The square roots of a negative real number

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 4: 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 5: Express \((-\sqrt{3}+\sqrt{-2})(2 \sqrt{3}-i)\) in the form of \(a+i b\)

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