I'm just want to be sure if the function $f(z)=e^{-iz}, z\in \mathbb C$, has no complex or real zeros??
Zeros of exponential
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$\begingroup$
complex-analysis
4 Answers
10
$f(x+iy)=e^{-ix+y}=e^{y}(\cos(x)-i\sin(x))$
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.
8
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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0That's all what I need... thanks all. – 2011-04-09
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0How do you know they are both entire? (I'd think by the time figured that out you'd already know there are no zeroes) – 2011-04-09
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0But "entire" does *not* include that they have no zeros (only that they have no poles) – 2011-04-09
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2This 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 $0$. – 2011-04-10
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0@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. – 2011-04-10
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0@Mitch: I take $e^z$ to be defined by its power series which converges uniformly on all of $\mathbb{C}$ by the Weierstrass M-test. – 2011-04-10
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0@Eric: well, that's correct. But the consequences of the symmetry and reciprocity are just *the* key idea/argument; so I'd always explicate it: "Since e^z is entire so is e^-z. Then they both have no poles. But because they are also their mutual reciprocals they also cannot have a zero" - or something like that. (Just to focus the attention of the OP at the key idea) – 2011-04-11
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0Smart and concise! +1 – 2014-10-31
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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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$e^{-iz}=e^{-ia+b}=e^{-ia}e^{b}$ 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.