Find the singularities and residues that correspond. Show details

Question

Find all the singularities in the finite plane and the corresponding residues. Show the details.

$$f(z) = \frac{1}{1-e^z}$$

So I'm trying to find the singularities. So the singularities are located when the denominator = 0 and that occurs when $e^z = 1$ which occurs when $z = 2 \pi in$ when $n \in \mathbb{R}$

Now what? How can I find the residues?

so the residue formula that I know is:

$$\operatorname{Res}_{z = z_0}f(z) = \lim_{z \to z_0} (z-z_0) f(z)$$

That's what I'm trying to use. But how do I determine this when $z_0$ isn't fixed?

score 1 · Answer 1 · answered Nov 30 '20 at 21:56

1

For each $n\in\Bbb Z$, you have\begin{align}\operatorname{res}_{z=2\pi in}\frac1{1-e^z}&=\lim_{z\to2\pi in}\frac{z-2\pi in}{1-e^z}\\&=-\frac1{\lim_{z\to2\pi in}\frac{e^z-e^{2\pi in}}{z-2\pi in}}\\&=-\frac1{\exp'(2\pi in)}\\&=-\frac1{e^{2\pi in}}\\&=-1.\end{align}

answered Nov 30 '20 at 21:56

José Carlos Santos

427,504

How did you get from the 1st to the 2nd line? – Jwan622 Nov 30 '20 at 22:00
$$\lim_{z\to a}f(z)=\frac1{\lim_{z\to a}\frac1{f(z)}}$$ – José Carlos Santos Nov 30 '20 at 22:02
where did the $e^z - e^{2 \pi n}$ come from? – Jwan622 Nov 30 '20 at 22:15
Anyway you can show me using thes residue formula above? I don't know how you got from 2nd to third or 1st to 2nd – Jwan622 Nov 30 '20 at 22:16
There is no $e^z-e^{2\pi n}$ in my answer. – José Carlos Santos Nov 30 '20 at 22:18
sorry in line 2, there's a $e^z - e^{2 \pi i n}$, where did that come from – Jwan622 Nov 30 '20 at 22:21
From the fact that $e^{2\pi in}=1$. Therefore$$\frac{e^z-1}{z-2\pi in}=\frac{e^z-e^{2\pi in}}{z-2\pi in}.$$ – José Carlos Santos Nov 30 '20 at 22:21
ohhhh. If you're interested I also found this: https://www.slader.com/discussion/question/find-all-the-singularities-in-the-finite-plane-and-the-6/ – Jwan622 Nov 30 '20 at 22:33

Find the singularities and residues that correspond. Show details

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