# Division of Complex Numbers

## Key Questions

By definition if $z = a + b i$ is a complex number, then his conjugate is $\overline{z} = a - b i$

#### Explanation:

A conjugate of a complex number is the other complex number with the same real part and opposite imaginary part

• You can do it by multiplying the numerator and the denominator by the complex conjugate of the denominator.

$\frac{1}{a + b i} = \frac{1}{a + b i} \cdot \frac{a - b i}{a - b i} = \frac{a - b i}{{a}^{2} + {b}^{2}}$

I hope that this was helpful.

• If complex numbers ${z}_{1}$ and ${z}_{2}$ in polar form are

$\left\{\begin{matrix}{z}_{1} = {r}_{1} \left(\cos {\theta}_{1} + i \sin {\theta}_{1}\right) \\ {z}_{2} = {r}_{2} \left(\cos {\theta}_{2} + i \sin {\theta}_{2}\right)\end{matrix}\right.$,

then we can write in exponential form

$\left\{\begin{matrix}{z}_{1} = {r}_{1} {e}^{i {\theta}_{1}} \\ {z}_{2} = {r}_{2} {e}^{i {\theta}_{2}}\end{matrix}\right.$.

So, the quotient ${z}_{1} / {z}_{2}$ can be written as

${z}_{1} / {z}_{2} = \frac{{r}_{1} {e}^{i {\theta}_{1}}}{{r}_{2} {e}^{i {\theta}_{2}}} = {r}_{1} / {r}_{2} {e}^{i \left({\theta}_{1} - {\theta}_{2}\right)}$

$= {r}_{1} / {r}_{2} \left[\cos \left({\theta}_{1} - {\theta}_{2}\right) + i \sin \left({\theta}_{1} - {\theta}_{2}\right)\right]$

I hope that this was helpful.

• Let ${z}_{1} = {a}_{1} + {b}_{1} i$ and ${z}_{2} = {a}_{2} + {b}_{2} i$. We want to find

$q = {z}_{1} / {z}_{2} = \frac{{a}_{1} + {b}_{1} i}{{a}_{2} + {b}_{2} i}$

Generally, we wish to write this in the form

$q = A + B i$

Where $A$ and $B$ are real numbers. To do this, we must amplify the quotient by the conjugate of the denominator:

$q = {z}_{1} / {z}_{2} \cdot \frac{\overline{{z}_{2}}}{\overline{{z}_{2}}} = \frac{{a}_{1} + {b}_{1} i}{{a}_{2} + {b}_{2} i} \cdot \frac{{a}_{2} - {b}_{2} i}{{a}_{2} - {b}_{2} i} = \frac{\left({a}_{1} {a}_{2} + {b}_{1} {b}_{2}\right) + \left({b}_{1} {a}_{2} - {b}_{2} {a}_{1}\right) i}{{a}_{2}^{2} + {b}_{2}^{2}}$

$q = \frac{{a}_{1} {a}_{2} + {b}_{1} {b}_{2}}{{a}_{2}^{2} + {b}_{2}^{2}} + \frac{{b}_{1} {a}_{2} - {b}_{2} {a}_{1}}{{a}_{2}^{2} + {b}_{2}^{2}} i$