How to solve $e^{ix}=i$? This is a question related to another posted question:
The answer to the following question "Find all solutions to: $e^{ix}=i$" is as follows: 
"Euler's formula: $e^{ix}=\cos(x)+i\sin(x)$,
so: $ \cos x+i\sin x=0+1⋅i$
compare real and imaginary parts
$\sin(x)=1$
and
$\cos(x)=0$
$x=\frac{(4n+1)π}2$, $n∈$
(W stands for set of whole number W={0,1,2,3,.......,n})."
My question: Where does $x=\frac{(4n+1)π}2$, $n∈$ come from? 
My steps: 


*

*$\cos(x) + i\sin(x) = 0 + i(1)$

*$\cos(x) = i(1 - \sin(x))$

*... 

*how does $x=\frac{(4n+1)π}2$ follow? 
 A: $\sin$ and $\cos$ functions are $2\pi$-periodic which is: $\sin(x+2n\pi)=\sin x$, $\cos(x+2n\pi)=\cos(x)$. So when you find that $x=\pi/2$ is a solution, then also $x_n = \pi/2 + 2n\pi$ is a solution for every $n\in \mathbb Z$ (where $\mathbb Z$ are whole numbers: $0, 1, -1, 2, -2,\dots$)
Notice that
$$
  \frac \pi 2 + 2n\pi = \frac{4n+1}{2}\pi.
$$
A: I always find it easier to use a fixed method, and I thought you might find this explanation easier, so I'm posting it. 
Start by putting everything into exponential form. Now $i = e^{\frac{i\pi}{2}}$. You can derive this from $e^{i\pi} = -1$ and taking square roots on both sides.
Now note that for any $\theta$, $e^{i\theta} = e^{i(\theta + 2k\pi)}, k \in \mathbb{Z}$, and this is because $e^{2k\pi i} = 1$. Essentially, this can be viewed as the periodicity of the exponential form. To compute general solutions or roots, you would be well-advised to include this term so that you don't miss any solutions. 
Hence you can now write $i = e^{i(\frac{\pi}{2} + 2k\pi)} = e^{i\pi\frac{4k+1}{2}}$
Note that the final step is just an algebraic rearrangement of the exponent.
You can now immediately solve the equation by taking logs of both sides, i.e.
$e^{ix} = e^{i\pi\frac{4k+1}{2}}\\ \implies x = \pi\frac{4k+1}{2}$ which is essentially the required form.
A: You said it, you compare real and imaginary parts. $\cos(x)=0$ and $\sin(x)=1$. When is this true?
A: Hint:


*

*$i$ is a point on the unit circle.

*$e^{ix}$ is also a point on the unit circle, lying x radians away from $(1,0)$, in trigonometric or counterclockwise direction.
So, to answer a question with a question, What is the position of $i$ on the unit circle, and How many radians away from $(1,0)$ does it lie, in trigonometric or counterclockwise direction ?
