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