# Powers of Complex Numbers

## Key Questions

• Given a complex number of form $a + b i$,it can be proved that any power of it will be of the form $c + \mathrm{di}$.
For example, ${\left(a + b i\right)}^{2} = \left({a}^{2} - {b}^{2}\right) + 2 a b i$

Knowing that, its less scary to try and find bigger powers, such as a cubic or fourth.
Whatsoever, any negative power of a complex number will look like this:
${\left(a + b i\right)}^{-} n = \frac{1}{a + b i} ^ n = \frac{1}{c + \mathrm{di}}$
This final form is not acceptable, as it has a division by $i$, but we can use a factoring method to make it better.
$\left(m + n\right) \left(m - n\right) = {m}^{2} - {n}^{2}$ $\to$ $\left(m + n i\right) \cdot \left(m - n i\right) = {m}^{2} + {n}^{2}$

$\frac{1}{c + \mathrm{di}} \cdot \frac{\left(c - \mathrm{di}\right)}{\left(c - \mathrm{di}\right)} = \frac{c - \mathrm{di}}{{c}^{2} + {d}^{2}}$//

• You could use the complex number in rectangular form ($z = a + b i$) and multiply it ${n}^{t h}$ times by itself but this is not very practical in particular if $n > 2$.
What you can do, instead, is to convert your complex number in POLAR form: $z = r \angle \theta$ where $r$ is the modulus and $\theta$ is the argument.
Graphically:

so that now the ${n}^{t h}$ power becomes:

${z}^{n} = {r}^{n} \angle n \cdot \theta$

Let's look at an example:
Suppose you want to evaluate ${z}^{4}$ where $z = 4 + 3 i$
Using this notation you should evaluate: ${\left(4 + 3 i\right)}^{4}$ which is difficult and...well...boring!
But if you change it in polar form you get:

Your number in polar form becomes: z=5 angle 37° and:
z^4=5^4 angle (4*37°)=625 angle 148°

You can now wonder what is the rectangular form of your result.
We get there using the trigonometric form and do some math.
Looking at your ${1}^{s t}$ graph you can see that:
$a = r \cdot \cos \left(\theta\right)$
$b = r \cdot \sin \left(\theta\right)$

$z = a + b i = r \cdot \cos \left(\theta\right) + r \cdot \sin \left(\theta\right) \cdot i$
That gives you:
$z = - 530 + 331 i$

(I rounded a little bit to make it clearer)

• We can find the value of a power of i by using ${i}^{2} = - 1$.

Let us look at some examples.

${i}^{5} = {i}^{2} \cdot {i}^{2} \cdot i = \left(- 1\right) \cdot \left(- 1\right) \cdot i = i$

${i}^{6} = {i}^{2} \cdot {i}^{2} \cdot {i}^{2} = \left(- 1\right) \cdot \left(- 1\right) \cdot \left(- 1\right) = - 1$

I hope that this was helpful.