# Effie's question via email about Complex Numbers

#### Prove It

##### Well-known member
MHB Math Helper
If \displaystyle \begin{align*} z = -2 + 2\,\mathbf{i} \end{align*} what is \displaystyle \begin{align*} \textrm{arg}\,\left( z^5 \right) \end{align*}?
First let's write this number in its polar form.

\displaystyle \begin{align*} \left| z \right| &= \sqrt{\left( -2 \right) ^2 + 2^2} \\ &= \sqrt{4 + 4} \\ &= \sqrt{8} \\ &= 2\,\sqrt{2} \end{align*}

and as the number is in Quadrant 2

\displaystyle \begin{align*} \textrm{arg}\,\left( z \right) &= \pi - \arctan{ \left| \frac{2}{-2} \right| } \\ &= \pi - \arctan{ \left( 1 \right) } \\ &= \pi - \frac{\pi}{4} \\ &= \frac{3\,\pi}{4} \end{align*}

thus we can say

\displaystyle \begin{align*} z &= -2 + 2\,\mathrm{i} \\ &= 2\,\sqrt{2}\,\mathrm{e}^{ \frac{3\,\pi}{4}\,\mathrm{i} } \\ z^5 &= \left( 2\,\sqrt{2}\,\mathrm{e}^{\frac{3\,\pi}{4}\,\mathrm{i}} \right) ^5 \\ &= 128\,\sqrt{2}\,\mathrm{e}^{ \frac{15\,\pi}{4}\,\mathrm{i} } \end{align*}

so that means \displaystyle \begin{align*} z^5 \end{align*} makes an angle of \displaystyle \begin{align*} \frac{15\,\pi}{4} \end{align*} with the positive real axis, but as we define \displaystyle \begin{align*} \textrm{arg}\,\left( Z \right) \in \left( -\pi , \pi \right] \end{align*}, that means we keep adding or subtracting integer multiples of \displaystyle \begin{align*} 2\,\pi \end{align*} until we have an angle in this region.

Thus \displaystyle \begin{align*} \textrm{arg}\,\left( z^5 \right) = -\frac{\pi}{4} \end{align*}.