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I'm just want to be sure if the function $f(z)=e^{-iz}, z\in \mathbb C$, has no complex or real zeros??

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4 Answers 4

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

In order for $f(z)=0$ you need

$$e^y\cos(x)=0 \,,$$ and $$e^y\sin(x)=0 \,.$$

You can easily see why that is not possible.

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That is correct.

Since both $e^z$ and $e^{-z}$ are entire, they have no poles. Since they are reciprocals of each other, it follows that they have no zeros.

Hope that helps,

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That's all what I need... thanks all. –  Annoli Apr 9 '11 at 18:20
    
How do you know they are both entire? (I'd think by the time figured that out you'd already know there are no zeroes) –  Mitch Apr 9 '11 at 18:42
    
But "entire" does not include that they have no zeros (only that they have no poles) –  Gottfried Helms Apr 9 '11 at 19:33
1  
This is basically the standard way to see that $e^z$ is never zero, but no reference to analyticity is necessary. Since $e^z\cdot e^{-z}=e^{z-z}=e^0=1$ for all $z$ (and $e^z$ is a complex number), $e^z$ is never $0$. –  Jonas Meyer Apr 10 '11 at 7:42
    
@Gottfried: Of course being entire does not mean it has no zeros. But these two functions are entire and reciprocals of each other.... so it does. –  Eric Naslund Apr 10 '11 at 17:07

$e^{-iz}=e^{-ia+b}=e^{-ia}e^{b}$ so if it equals $0$ then $e^{-ia}$ must be zero, since we know that $e^{b}$ is never zero when $b$ is real (graph it on your calculator if you don't want to prove it!). So we have that $\frac{1}{e^{ia}}=0$ which is impossible.

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$\exp(z) \cdot \exp(-z) = \exp(z - z) = \exp(0) = 1$, so $\exp(z) = 0 \implies \frac{1}{\exp(-z)} = 0$

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