Root of Unity of any complex power

Question

The problem is finding a complex number z, such that $$z^{a+bi} = z$$ I know for integer powers of z the Roots of Unity are $$z^{n} = e^{\frac{2πik}{n-1}}$$ I basically want to solve, given any complex number, when taken to the power of another complex number is the first complex number. All I know so far is that given: $$z^{a+bi} = z$$ all solutions to z lie on the polar function:$$r = e^{\frac{bθ}{a-1}}$$ and of course also $$0, a≠0$$ and if a = 1 and b ≠ 0, all infinite solutions are real and positive. Forgive my lack of terminology as I have not studied much complex analysis. I feel like the answer may be obvious but I'm not sure.Photo for reference

Answer 1

Let's look at the solutions to $z^w = 1$, where $w$ is some complex number.

Generally, we define $z^w = e^{w \log z}$, where $\log$ is the multi-valued complex logarithm function. So now we have

$$e^{w \log z} = 1 = e^{2 k i \pi}$$

for $k \in \mathbb{Z}$. That gives us

$$\begin{eqnarray} w \log z & = & 2 k i \pi \\ \log z & = & \frac{2 k i \pi}{w} \\ & = & \frac{2 k i \pi \bar{w}}{|w|^2} \end{eqnarray}$$

Then, if we write $z = r e^{i \theta}$ where $r \in (0, \infty)$ and $\theta \in (-\pi, \pi]$, we have $\log z = \ln r + i(\theta + 2m \pi)$, where $m \in \mathbb{Z}$, giving

$$\begin{eqnarray} \log z & = & \frac{2 k i \pi \bar{w}}{|w|^2} \\ \ln r + i \theta + 2 m i \pi & = & \frac{2 k i \pi \bar{w}}{|w|^2} \\ & = & \frac{2 k \pi}{|w|^2} \left(\mathrm{Im}(w) + i \mathrm{Re}(w)\right) \end{eqnarray}$$

So it looks like our final solution set is $z = r e^{i \theta}$ where $r = \exp\left(\frac{2k\pi \mathrm{Im}(w)}{|w|^2}\right)$ and $\theta = \frac{2 k \pi \mathrm{Re}(w)}{|w|^2} + 2 m \pi$ for $k, m \in \mathbb{Z}$. I will leave it to you to determine (a) whether every value of $m$ and $k$ actually gives a valid solution, and (b) whether this collapses to the known solution sets for specific values of $w$, e.g. integers and rational numbers.