Algebraic operation on Complex Numbers - Practice Questions & MCQ

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

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

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

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

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

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

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

$\begin{array}{rl}& z_1\cdot z_2=(a+ib)\cdot (c+id)\\ & =ac+iad+ibc+i^{2}bd\\ & =ac+i(ad+bc)-bd\\ & =(ac-bd)+i(ad+bc)\end{array}$
For example, $z_1=(4+3i)$ and $z_2=(2-5i)$, then $z_1\cdot z_2$ is

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

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 

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

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 $z_3,(z_1+z_2)+z_3=z_1+$ $(z_2+z_3)$
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 $z=a+ib$, 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).