# Solving $e^{ix}=i$

I was assigned this problem: $$e^{ix}=i$$ I understand that with Euler's formula, $$e^{ix}=\cos x+i\sin x$$. I then set up the problem as $$i=\cos x +i\sin x$$ This means that $$\cos x = 0$$ and $$\sin x =1$$. This works for $$\frac{\pi}2$$. It has to be multiplied $$n$$, so the answer should be $$\frac{n\pi}2$$.

This is how I did this problem. However, I believe it was marked as incorrect. Did I do something wrong? Is there another possible answer that I am missing? How should I go about solving a problem like this?

• What do you mean by "it has to be multiplied by $n$"? It clearly doesn't work for $x=\pi$. You instead should have $x=\pi/2+2\pi n$ because the trigonometric functions are periodic with period $2\pi$. Aug 18, 2019 at 15:25
• I see - I got the $\frac{\pi}2$ part, but I messed up on the $2\pi n$. That is how you take care of the fact that the function repeats itself over and over again.
– Burt
Aug 18, 2019 at 15:28

$$e^{ix}=i=e^{\frac{iπ}{2}}$$ $$\Longleftrightarrow e^{ix-\frac{iπ}{2}}=1$$ $$\Longleftrightarrow e^{i(x-\frac{π}{2})}=1$$ $$\Longleftrightarrow x-\frac{π}{2}=2nπ$$ $$\Longleftrightarrow x=2nπ+\frac{π}{2}$$
You got the period wrong. The period of $$sin(x)$$ and $$cos(x)$$ is $$2\pi$$.
If $$\cos(x) = 0$$ and $$\sin(x)=1$$, therefore $$x = \frac{\pi}{2} \text{ mod }2\pi$$ Thus $$x= \frac{\pi}{2} + 2k\pi$$ where $$k \in \mathbb{Z}$$.