# Find all solutions to the equation $e^z = i$

I know that there is an equation for finding the nth roots of a complex number, which easily done once you have the modulus and argument of the complex number in question. There would be n roots. But how do I know how many solutions there are to an exponential raised to a complex number?

Thanks for any help out there!

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Use $e^{r\theta}=r\cdot cos (\theta) + r\cdot sin(\theta)i$ – you Mar 14 '12 at 0:21
@you: I am not sure that is quite correct – Henry Mar 14 '12 at 0:26
oops! It's too late for me to edit it now. I mean "$re^{i\theta}$" – you Mar 14 '12 at 4:14

$\displaystyle e^z=e^{i\frac{\pi}2+2k\pi i}$ so that $\ \displaystyle z=i\frac{\pi}2+2k\pi i\$ with $k\in \mathbb{Z}$
For real $a$ and $b$ you have $$\exp(a+bi) = e^a \cos(b/2\pi) + e^a \sin(b/2\pi) i$$ so now set the right hand side equal to $i$ and solve the real and imaginary parts.
You get $e^a \cos(b/2\pi)=0$ but $e^a$ is positive so $b$ can only be of the form $(n+1/2)\pi$ for integer $n$, implying $\sin(b/2\pi)=\pm 1$, and you want $e^a \sin(b/2\pi)=1$ but $e^a$ is still positive so you must have $\sin(b/2\pi) = 1$ and so $e^a =1$, implying $b=(2n+1/2)\pi$ for integer $n$ and $a=0$, so the solution is $a+bi$ which is $$\left(2n+\frac{1}{2}\right)\pi i.$$