Division of Complex Numbers

Dividing Complex Numbers of the form a + bi

Tip: This isn't the place to ask a question because the teacher can't reply.

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.

• This key question hasn't been answered yet.

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