How to solve $e^{ix} = i$ 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.
 A: $$e^{ix} = i$$
Euler's formula:
$$e^{ix}=\cos(x)+i\sin(x),$$
so:
$$\cos x+i\sin x = 0+1\cdot i$$
compare real and imaginary parts
$$\sin x =1$$ and $$\cos x =0$$
$$x=\dfrac{(4n+1)\pi}{2}\;\;,n \in \mathbb W$$
(W stands for set of whole number $W=${$0,1,2,3,.......,n$}).
A: Well, the geometric meaning of $e^{ix}$ is the weapon here to use.
This is nothing but the point on the complex plane which has length $1$ and angle $x$ measured from the right half of the real axis, in radian. So that $e^{i\pi}=-1$, for example.
A: So, you say that you're okay with Euler's formula:
$$e^{ix}=\cos(x)+i\sin(x),$$ and it should be clear that (for real numbers $a,b,c,d$), we have $$a+bi=c+di\quad\iff\quad a=c\;\text{ and }\;b=d.$$
So, which values of $x$ will satisfy
$$\cos(x)+\sin(x)i=0+1i\quad ?$$
A: A completely different approach:
$$ e^{i\pi} = -1 \implies \sqrt {e^{i\pi}} = \sqrt {-1} \implies e^{\frac{i\pi}{2}} = i $$
A: 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 $x = \frac{\pi}{2}+2n\pi \quad n\in \mathbb{Z}$
A: $$ e^{ix} = i \implies ix =\ln(i) \implies ix= \ln(|i|)-i \left(\frac{\pi}{2}+2k\pi\right) $$
$$ \implies x = \frac{\pi}{2}+2k\pi,\quad k\in \mathbb{Z}. $$
A: Hint:
$$e^{xi}=i=e^{\frac\pi2i+2k\pi i}\implies x=\frac\pi2+2k\pi\;,\;\;k\in\Bbb Z$$
