Solve $i^b = b^i$ for $b$ where $i=\sqrt{-1}$

Question

Solve the equation $i^b = b^i$ for $b$, where $i = \sqrt{-1}$.

This problem appeared on a test. There may or may not be a specific answer for this problem in which case we would have null set.

Answer 1

Let $b=x+iy,\,x,y\in\mathbb{R}$. Then$$\begin{align}i^b = b^i&\implies b\,\text{Log}\,i=i\,\text{Log}\,b\\&\implies b\,\text{Log}\,e^{i\pi/2}=i\,\text{Log}\,b\\&\implies\frac\pi2b=\text{Log}\,b\\&\implies\frac\pi2x+i\frac\pi2y=\frac12\ln(x^2+y^2)+i\,\text{Arg}(x+iy)\end{align}$$ Equating real and imaginary parts we have $$\pi x=\ln(x^2+y^2)$$$$\pi y=2\,\text{Arg}(x+iy)$$ In particular, we can write the first equation as $$y=\sqrt{e^{\pi x}-x^2}.$$

Examples

Let $x=0$. Then $$y=\sqrt{e^0-0}=\pm1$$ so $$b=\pm i.$$ Let $y=0$. Then $$0=2\,\text{Arg}(x)\implies \text{Arg}(x)=0\implies x\in\mathbb{R}$$ but it also has to satisfy $$\pi x=\ln(x^2+0)\implies \ln x=\frac\pi2x$$ which has no real solutions since $x > \ln x$ for all $x > 0$ and $\frac\pi2>1$. Hence $y\neq0$.