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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?

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    $\begingroup$ 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 $2\pi$. $\endgroup$ Aug 18, 2019 at 15:25
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    $\begingroup$ 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. $\endgroup$
    – Burt
    Aug 18, 2019 at 15:28

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$$e^{ix}=i=e^{\frac{iπ}{2}}$$ $$\Longleftrightarrow e^{ix-\frac{iπ}{2}}=1$$ $$\Longleftrightarrow e^{i(x-\frac{π}{2})}=1$$ $$\Longleftrightarrow x-\frac{π}{2}=2nπ$$ $$\Longleftrightarrow x=2nπ+\frac{π}{2}$$

Hope it helps:)

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You got the period wrong. The period of $sin(x)$ and $cos(x)$ is $2\pi$.

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  • $\begingroup$ And I multiplied instead of adding, right? $\endgroup$
    – Burt
    Aug 18, 2019 at 15:28
  • $\begingroup$ Yes, periods are added, not multiplied. $\endgroup$
    – H Huang
    Aug 18, 2019 at 15:30
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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 $k \in \mathbb{Z}$.

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