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My teacher said that $2\pi$ the period of $e^{iz}$. I tried to show this but I am not sure how to continue

$$e^{iz+2\pi}=e^{ix+2\pi-y}=e^{2\pi-y}(\cos x+i\sin x)$$

But I am not sure how this equals

$$e^{-y}(\cos x+i\sin x)=e^{-y+i x}=e^{iz}$$

Is $2\pi$ the period of $e^{iz}$?

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    The argument of $e^{i z}$ is $z$, so it's $z$ that's shifted by $2\pi$ (not $iz$). So you should be considering $e^{i(z+2\pi)}$ not $e^{i z+2\pi}$. – 2017-01-17
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    Should n't it be $e^{i 2\pi}$ instead of $e^{2 \pi}$?? – 2017-01-17
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    This is a function on z so you have to calculate $\exp{i (z+2\pi)}$ – 2017-01-17
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    @B.Goddard, If $f(z) = e^{iz}$ then $e^{iz + 2\pi i} = e^{i(z+2\pi)} = f(z+2\pi)$, so the period is $2\pi$, not $2\pi i$. It's $e^z$ that has period $2\pi i$. – 2017-01-17
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    @B.Goddard mmmm..... The period of $e^{iz}$ as a function of real z is $2\pi$. But the period of $e^z$ as a function of complex $z$ (or $e^{iz}$ as a function purely imaginary $iz$) is $2\pi i$. – 2017-01-17
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    I agree it is poorly worded. $z$ is a usually a complex number but the expression $e^{iz}$ implies $z$ a real number in particular the imaginary part of a complex number. If $f(x) = e^{ix}$ then $f$ has a period of $2\pi$. But if $f(z = x + yi) = e^z = e^{x + iy} = e^x{\cos y + i \sin y}$ then $f(z= x + yi) = f(z + 2\pi i= x + yi + 2\pi i = x + (y + 2\pi) i)$ so $f(z) = e^{z}$ has a period of $2\pi i$. Of course $f(z) = e^{iz} = e^y(-\sin x + i\cos x)$ has a period of $2\pi$. – 2017-01-17
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    @fleablood True. Stand by for erasing. – 2017-01-17

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You are misunderstanding the teacher. If the function is $f(z)=e^{iz}$ then saying that the period of $f(z)$ is $2\pi$ means that $f(z+2\pi)=f(z)$, i.e. that $$e^{i(z+2\pi)}=e^{iz}$$ Notice the parentheses in the exponent on the left-hand side, which are missing from the equation in your question.

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    To be precise though, $f(z+2 \pi ) = f(z)$ is a necessary but not sufficient condition to say that $2 \pi$ is the period of $f$; you also need to show that there is no other positive real $a$ such that $f(z+a)=f(z)$ (for all z) and $a<2 \pi$. – 2017-01-17
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    @Pythagoricus A good point, but somewhat tangential, given that the OP's confusion really was about where the parentheses should go. – 2017-01-17
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Let $f(z) = e^{iz}$

The $f(z + 2\pi) = e^{i(z + 2\pi)} = e^{iz + 2\pi i}= e^{iz}e^{2\pi i}$

$e^{2\pi i} = \cos 2\pi + i \sin 2\pi = 1$

So $f(z + 2\pi) = e^{iz} = f(z)$.

So the period of $e^{iz}$ as a function of $z$ is $2\pi$.

But the period of $e^z$ (notice $z$ is not mulitplied by $i$; $z$ is a variable complex number) as a function of $z$ is $2\pi i$