How can I convert $(a + bi)^{c+di}$ to polar form?

Question

Specifically, I need to convert $(-8i)^{2+πi/3}$ to polar form. I understand I need to use Euler's formula, $e^{\theta i} =\cos \theta + i \sin \theta$, but I'm not sure about the full process.

Thanks.

Answer 1

For any $z,w \in \mathbb{C}$, with $z = a + bi$ and $w = c+di$, with $a,b,c,d \in \mathbb{R}$.

Then:

$$\begin{split} z^w &= \mathrm{e}^{\log z^w} \\ &= \mathrm{e}^{w \log z} \\ &= \mathrm{e}^{w (\log |z| + i\arg z)} \\ &= \mathrm{e}^{(c + di) (\log |z| + i\arg z)} \\ &= \mathrm{e}^{(c \log |z| - d \arg z+ i(c\arg z + d \log |z|))} \\ &= \mathrm{e}^{(c \log |z| - d \arg z)} \mathrm{e}^{i(c\arg z + d \log |z|)} \\ \end{split} $$

In this last line, the magnitude is $\mathrm{e}^{(c \log |z| - d \arg z)}$ and the angle is $(c\arg z + d \log |z|)$

For your specific case: $z = -8i$ and $w = 2 + \pi/3 i$, so $|z| = 8$, $\arg z = -\pi/2 + 2\pi k \, (\text{with} \, k \in \mathbb{Z})$, $c = 2$, and $d = \pi/3$.

So (taking $k = 0$ for a single example):

$$\begin{split} z^w &= \mathrm{e}^{(c \log |z| - d \arg z)} \mathrm{e}^{i(c\arg z + d \log |z|)} \\ &= \mathrm{e}^{(2 \times \log 8 - \pi/3 \times -\pi/2)} \mathrm{e}^{i(2 \times -\pi/2 + \pi/3 \times \log 8)} \\ &= \mathrm{e}^{(2 \log 8 + \pi^2/6)} \mathrm{e}^{i(-\pi+ \pi/3 \log 8)} \\ &\approx 331.56277234 \mathrm{e}^{-0.964006563 i} \\ \end{split}$$

This is only the principal log though, there are an infinite number of possible solutions.