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Problem.src) Compute $$\oint_C \frac{5\sec 2z}{e^{-3z}-1}\,\mathrm{d}z,$$ where $C:|z|=10^{-4}$.

I solved the problem like this, but I don't know if this is the right way

$$\oint_C \frac{5\sec 2z}{e^{-3z}-1}\,\mathrm{d}z = 2πi\cdot\text{Res}\frac{5\sec 2z}{e^{-3z}-1}$$ $$= 2πi\cdot\text{Res}\frac{5}{cos2z\cdot\ e^{-3z}-1}$$ $$= 2πi\cdot\text{Res z=0 f(z)} = 2πi\cdot\frac{-5}{3}\, $$

$$ \text{Res}\frac{5\sec 2z}{e^{-3z}-1} = \frac{5}{-2sin2z\cdot\ (e^{-3z}-1) + cos2z\cdot\ (-3e^{-3z})} (z=0) =\frac{-5}{3}\ $$

enter image description here

pleaseiwanttograduate
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    Instead of just spamming a downvote why don't you provide a comment which will help the op fix the post. – Mr.Gandalf Sauron Dec 01 '23 at 11:30
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    The main issue is that you are supposed to write out your efforts using Mathjax. You are welcome to add pictures for additional context, but the main portion of your post must be formatted using mathjax. Without this, your post will be subject to closure from the site. – Mr.Gandalf Sauron Dec 01 '23 at 11:32
  • That being said, the residue you have calculated is correct although it is unclear from the picture (which is the reason why you shouln't post picture for the main part of the text) how you have calculated it. – Mr.Gandalf Sauron Dec 01 '23 at 11:40
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    @Mr.Gandalf Sauron, I'm so sorry that I didn't know that I shouldn't post picture because yesterday is my first time to come in this site. I will fix that solution in picture to clear solution using Mathjax. and I'm really thank you to tell that rule! – pleaseiwanttograduate Dec 01 '23 at 11:44

1 Answers1

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Here's an easy formula/principle to remember.

Suppose $f(z)=\frac{p(z)}{q(z)}$ where $p$ and $q$ are analytic and are such that $p(z_{0})\neq 0$ and $q(z)$ has a simple $0$ at $z_{0}$. (i.e. $q(z_{0})=0$ but $q'(z_{0})\neq 0$) , then the Residue of $f$ at $z_{0}$ is given by $\displaystyle\frac{p(z_{0})}{q'(z_{0})}$

The point is that you can avoid the tedius Maclaurin expansions and computations with a bit of cleverness. i.e. you don't need to expand all that is there, you only need to focus on the problematic bit. You are essentially doing the same, but you don't need to even bother with the $\sec(2z)$ part.

To see this, just note that in a neighbourhood of $z_{0}$, $q(z)=(z-z_{0})h(z)$ where $h(z)\neq 0$ for all $z$ in that neighbourhood.

So in fact, in that nbd, $\frac{p(z)}{h(z)}$ is an analytic function and hence has a power series of the form $\sum_{k=0}^{\infty}a_{k}(z-z_{0})^{k}$

Hence, in that neighbourhood, $$\frac{p(z)}{q(z)}=\frac{1}{z-z_{0}}\bigg(\sum_{k=0}^{\infty}a_{k}(z-z_{0})^{k}\bigg)$$

Which directly gives you the residue as $\displaystyle a_{0}=\frac{p(z_{0})}{h(z_{0})}$

But what is $h(z_{0})$? As $h(z)=\frac{q(z)}{z-z_{0}}$ it is simply the limit $$\lim_{z\to z_{0}}\frac{q(z)}{z-z_{0}}=q'(z_{0})$$ due to L'Hospital's rule.

So in this case, what you only need to do find the residue is given by $$\frac{\sec(2\cdot 0)}{\bigg(e^{-3z}-1\bigg)'(0)}=\frac{1}{-3}$$

Mr.Gandalf Sauron
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