Lets say you have an equation with the following form.
$$(a+ib)^{c+id}$$
How do you go about doing it? I am very familiar with De Moivre's Theorem where d is zero and have that all set. I am unfamiliar with d being real not equal to zero.
Any additional information would be very helpful.
On
Let $z = a + b \mathrm{i}$. Then $z = |z| \mathrm{e}^{\mathrm{i} \arg z}$ and $$ z^{c+d \mathrm{i}} = |z|^{c+d \mathrm{i}} \mathrm{e}^{(c+d \mathrm{i})\mathrm{i} \arg z} \text{.}$$
You say you can finish from here.
On
When $\exists\space\text{z}_1\space\wedge\space\exists\space\text{z}_2\in\mathbb{C}$:
$$\text{z}_1^{\text{z}_2}=\left(\left|\text{z}_1\right|e^{\left(\arg\left(\text{z}_1\right)+2\pi\text{k}_1\right)i}\right)^{\text{z}_2}=\left|\text{z}_1\right|^{\text{z}_2}e^{\text{z}_2\left(\arg\left(\text{z}_1\right)+2\pi\text{k}_1\right)i}$$
Where $\left|\text{z}_1\right|=\sqrt{\Re^2\left[\text{z}_1\right]+\Im^2\left[\text{z}_1\right]}$,$\arg\left(\text{z}_1\right)$ is the complex argument of $\text{z}_1$ and $\exists\space\text{k}_1\in\mathbb{Z}$
So:
If $a, ,b, c, d\in\Re$ We first compute $$(a+bi)=r\times e^{i\theta}, r\neq 0$$
$\begin{eqnarray*}(a+bi)^{c+di}&=& (r\times e^{i\theta})^{c+di}\\ &=& (r^c\times e^{ci\theta})\times r^{di}\times e^{i\theta \times di}\\ &=&(r^c e^{-d\theta})\times e^{i(c\theta + d\times \ln{r})} \end{eqnarray*}\\$
Where $(r^c e^{-d\theta})$ is your radios and $e^{i(c\theta + d\times \ln{r})}$ is your unit complex number.