what is $c^{a+\mathrm i b}$ for $c \in \mathbb{R}$?

Question

How can be $c^{a+\mathrm i b}$ for $c,a,b \in \mathbb{R}$ rewritten in the form $e+ \mathrm i d$ for $d,e \in \mathbb{R}$ (i.e. as a $\mathbb{R}$-linear combination of $1, \mathrm i$)?

Answer 1

We know that $e^{ix} = \cos x + i\sin x$, and $e^{\ln c} = c$.

For $c > 0$ $$c^{a+ib} = c^a\cdot c^{ib} = c^ae^{ib\ln c} = c^a(\cos (b\ln c) + i\sin(b\ln c))$$

For $c < 0$, we can write $c = |c|e^{i(2n+1)\pi}$ where $n\in \mathbb{N}$, (since $e^{i\pi} = e^{3\pi i} = e^{5\pi i} = \cdots = -1$), so $\ln c = \ln|c| + i(2n+1)\pi$, after which we can use the previous equation.

Answer 2

Assuming that, in your posing of the question, the "$c$" in the expression you wish to rewrite is not the same as the "$c$" in the general form you seek.

Note $c^{a+ib} = c^a\cdot e^{ib(\log c)}$. It is multivalued in general.

Answer 3

Use Euler's formula:

$$ e ^ {ix} = \cos x + i\sin x $$

with

$$ x ^ y = e ^ {y \ ln x} $$