complex exponential equation

Question

I am trying to solve the following exponential equation: $z^{1+i} = 4$ where the argument of $z$ is between $-\pi$ and $\pi$. Here is what I have gotten so far: If $z = a + bi$ then the magnitude of $z$ is $2$ and $arctan(\frac{b}{a}) = -2$, therefore the two solutions of $z$ are in the 4th and 2nd quadrant respectively. Can anyone confirm this? If this is right then I know how to proceed with the rest of this problem. Thanks

Answer 1

Actually there are infinitely many solutions.

Let $z = re^{i\theta}$, with $r,\theta \in \mathbb{R}$. Then: \begin{align} z^{i+1} &= 4\\ \left( re^{i\theta} \right)^{i+1} &= 4\\ r^{i+1}e^{(i^2+i)\theta} &= 4\\ re^{-\theta} e^{i(\theta + \ln r)} &= 4\\ \end{align} So \begin{align} \theta + \ln r &\equiv 0\ (\operatorname{mod}\ 2\pi)\\ re^{-\theta} &= 4.\\ \end{align} The later of these implies \begin{align} \ln r - \theta &= 2\ln 2\\ 2\theta &\equiv -2\ln 2\ (\operatorname{mod}\ 2\pi)\\ \theta &\equiv -\ln 2\ (\operatorname{mod}\ \pi)\\ \end{align}

Let $\theta = k\pi - \ln 2$ with $k \in \mathbb{Z}$. Then \begin{align} \theta &= k\pi -\ln 2\\ \ln r &= \ln 2 + k \pi\\ r &= e^{ \ln 2 + k\pi }\\ r &= 2 e^{k\pi}\\ z &= 2e^{k\pi + i\left(k \pi -\ln(2) \right)}\\ z &= 2^{1-i} e^{k\pi \left( 1+i \right)}\\ \end{align} One can verify the solution holds $\forall k \in \mathbb{Z}$: \begin{align} z^{1+i} &= 2^{ (1-i)(1+i) }e^{ k\pi(1+i)^2 }\\ &= 2^{ 1-i^2 }e^{ 2k\pi i }\\ &= 4. \end{align}

One can tell that the solutions $z$ are distinct for distinct $k$ because the moduli $|z|$ are distinct.