Solve $(1+i)^z = i$ where $z$ is a complex number.

Question

How can I solve this equation?

$$(1+i)^z = i$$

The only way I can start is to use $e^{z\log(1+i)}$.

Answer 1

Note that $1+i=\sqrt{2}\left(\frac{1+i}{\sqrt{2}}\right)=\sqrt{2}e^{\frac{i\pi}{4}}$. Now let $z=x+iy$. Then \begin{align*} (1+i)^z & =\left(\sqrt{2}e^{\frac{i\pi}{4}}\right)^{x+iy}\\ &=\left(\sqrt{2}e^{\frac{i\pi}{4}}\right)^{x}\cdot \left(\sqrt{2}e^{\frac{i\pi}{4}}\right)^{iy}\\ &=\left(2^{x/2}e^{\frac{ix\pi}{4}}\right)\cdot \left(2^{iy/2}e^{\frac{-y\pi}{4}}\right)\\ &=\left(2^{x/2}e^{\frac{-y\pi}{4}}\right)\cdot \left(e^{iy (\ln 2)/2}e^{\frac{ix\pi}{4}}\right)\\ &=\left(2^{x/2}e^{\frac{-y\pi}{4}}\right)\cdot \left(e^{i\left[\frac{2y\ln 2+x\pi}{4}\right]}\right). \end{align*} In order to solve $(1+i)^z=i$, we do the following: first we equate the magnitude of both sides to get \begin{align*} |(1+i)^z|&=|i|\\ 2^{x/2}e^{\frac{-y\pi}{4}}&=1\\ 2^{x/2}&=e^{\frac{y\pi}{4}}\\ \color{red}{x \ln 4} & \color{red}{=y\pi}. \end{align*} Now we equate the angles to get \begin{align*} \frac{2y\ln 2+x\pi}{4}&=\frac{(4n+1)\pi}{2}\\ \color{blue}{2y\ln2+\pi x}&=\color{blue}{2(2n+1)\pi}. \end{align*}

Now solve for $x$ and $y$ from the two equations to get $z=\frac{i(4n+1)\pi}{2\log(1+i)}$.