Algebraic operation on Complex Numbers - Practice Questions & MCQ

1. Addition of Two Complex Numbers
$z_1=a+i b$ and $z_2=c+i d$ be any two complex numbers. Then, the sum $z_1+z_2$ is defined as

$
z_1+z_2=(a+i b)+(c+i d)=(a+c)+i(b+d)
$
For example, $z_1=(3-4 i)$ and $z_2=(2+5 i)$, then $z_1+z_2$ is

$
(3-4 i)+(2+5 i)=(3+2)+(-4+5) i=5+i
$

2. Difference of Two Complex Numbers
$z_1=a+i b$ and $z_2=c+i d$ be any two complex numbers. Then, the difference $z_1-z_2$ is defined as

$
z_1-z_2=(a+i b)-(c+i d)=(a-c)+i(b-d)
$
For example, $z_1=(-5+7 i)$ and $z_2=(-11+2 i)$, then $z_1-z_2$ is

$
\begin{aligned}
& (-5+7 i)-(-11+2 i)=-5+7 i+11-2 i \\
& =-5+11+7 i-2 i \\
& =(-5+11)+(7-2) i \\
& =6+5 i
\end{aligned}
$

3. Multiplication of Two Complex Numbers
$\mathrm{z}_1=\mathrm{a}+\mathrm{ib}$ and $\mathrm{z}_2=\mathrm{c}+\mathrm{id}$ be any two complex numbers. Then, the multiplication $\mathrm{z}_1 \cdot \mathrm{z}_2$ is defined as

$
\begin{aligned}
& z_1 \cdot z_2=(a+i b) \cdot(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, $z_1=(4+3 i)$ and $z_2=(2-5 i)$, then $z_1 \cdot z_2$ is

$\begin{aligned} & (4+3 i)(2-5 i)=4(2)-4(5 i)+3 i(2)-(3 i)(5 i) \\ & =8-20 i+6 i-15\left(i^2\right) \\ & =(8+15)+(-20+6) i \\ & =23-14 i\end{aligned}$

4. Division of Two Complex Numbers

z₁ = a + ib and z₂ = c + id (and z₂ is non-zero) be any two complex numbers. Then, the division $\frac{\mathrm{z}_1}{\mathrm{z}_2}$   is defined as 

$\begin{aligned} & \frac{\mathrm{z}_1}{\mathrm{z}_2}=\frac{\mathrm{a}+\mathrm{ib}}{\mathrm{c}+\mathrm{id}} \cdot \frac{\mathrm{c}-\mathrm{id}}{\mathrm{c}-\mathrm{id}} \\ & \text { [multiplying numerator and denominator by } \mathrm{c}-\mathrm{id} \text { where one of } \mathrm{c} \text { and } \mathrm{d} \text { is non }-\mathrm{zero}] \\ & \frac{\mathrm{z}_1}{\mathrm{z}_2}=\frac{\mathrm{ac}-\mathrm{iad}+\mathrm{ibc}-\mathrm{i}^2 \mathrm{bd}}{\mathrm{c}^2-(\mathrm{id})^2}=\frac{\mathrm{ac}+\mathrm{i}(\mathrm{bc}-\mathrm{ad})+\mathrm{bd}}{\mathrm{c}^2-\mathrm{i}^2 \mathrm{~d}^2} \\ & \frac{\mathrm{z}_1}{\mathrm{z}_2}=\frac{\mathrm{ac}+\mathrm{bd}+\mathrm{i}(\mathrm{bc}-\mathrm{ad})}{\mathrm{c}^2+\mathrm{d}^2} \\ & \frac{\mathrm{z}_1}{\mathbf{z}_{\mathbf{2}}}=\frac{\mathbf{a c}+\mathbf{b d}}{\mathbf{c}^{\mathbf{2}}+\mathbf{d}^{\mathbf{2}}+\mathbf{i} \frac{\mathbf{b c}-\mathbf{a d}}{\mathbf{c}^{\mathbf{2}}+\mathbf{d}^{\mathbf{2}}}}\end{aligned}$

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 $z_2$.
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 $z_1, z_2$ and $z_3,\left(z_1+z_2\right)+z_3=z_1+$ $\left(z_2+z_3\right)$
4. Additive identity: if the sum of a complex number $z_1$ with another complex number $z_2$ is $z_1$, then $z_2$ is called the additive identity. We have $z+0=z=0+z$, so 0 is the additive identity.
5. Additive inverse: To every complex number $z=a+i b$, we have the complex number -$\mathrm{a}+\mathrm{i}(-\mathrm{b})$ (denoted as -z ), called the additive inverse or negative of $z$. i.e. $z+(-z)=0(-$ $z$ is additive inverse).