TITLE: How to solve $e^{ix} = i$
QUESTION [12 upvotes]: I am taking an on-line course and the following homework problem was posed: $$e^{ix} = i$$ I have no idea how to solve this problem. I have never dealt with solving equations that have imaginary parts. What are the steps to solving such equations? I am familiar with Taylor series and the Euler formula if that is any help. 
Thanks.

REPLY [3 votes]: The key here is to think in terms of radial co-ordinates rather than cartesian.  Whenever you see a complex exponential it's best to think of the geometric interpretation first, if possible.
The radial representation of a complex number $z = a + bi$ is $Re^{i\theta}$
where $R = |z|$ and $\theta=arctan(b/a)$.  
So for the number i we ask ourselves what is R, and what is $\theta?$
We look where the point i is on the complex plane - it's on the y axis 1 unit away from the origin. So R is 1 and $\theta$ is $\frac{\pi}{2}$.  (Note that you can add any multiple of $2\pi$ to $\theta$ and things stay the same.)  
So plug these R, $\theta$ values into the radial form of z to get the radial representation for i which is  $1.e^{i\frac{\pi}{2}} = e^{i\frac{\pi}{2}}$
So $e^{ix} = i = e^{i\frac{\pi}{2}}$ and a naive answer is $x = \frac{\pi}{2}$.
But remember we could add any multiple of $2{\pi}$ to our angle for i so we need to add $2n{\pi}$ for the general answer.  
Hence $x = \frac{\pi}{2}+2n\pi \quad n\in \mathbb{Z}$