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
4
$\begingroup$
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,
-
0Smart and concise! +1 – 2014-10-31
1
$\exp(z) \cdot \exp(-z) = \exp(z - z) = \exp(0) = 1$, so $\exp(z) = 0 \implies \frac{1}{\exp(-z)} = 0$
0
$e^{-iz}=e^{-ia+b}=e^{-ia}e^{b}$ 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.