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Use Euler's formula $\sin \pi z =\frac{ e^{i\pi z} - e^{-i\pi z}}{2i}$ to find the residue of $\frac{1}{\sin \pi z}$. Show that the complex zeros of $\sin \pi z$ are exactly at the integers, and they are each of order $1$. Calculate the residue of $\frac{1}{\sin \pi z}$.

For $\sin \pi z$ , equating it to zero we get using Euler's formula $e^{2i\pi z}=1$ So $2i\pi z= 2\pi in$ so $z=n$ for $n$ integer. Hence complex zeros are at exactly integers. How can I show they are of order $1$? Because there are countably many zeros, i.e., integers? Also I need to find residue.

But then I will get residue for each zero so how should I proceed?

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3 Answers

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The function has simple poles at the integers. If $f$ has a simple pole at $z_0$, its Laurent expansion is $$ f(z)=\frac{c_{-1}}{z-z_0}+c_0+c_1(z-z_0)+\dotsb $$ so it's clear that $$ c_{-1}=\lim_{z\to z_0}(z-z_0)f(z) $$ Conversely, if this limit exists (finite) and is not zero, then $f$ has a simple pole at $z_0$.

If $n$ is an integer, we have $$ \lim_{z\to n}\frac{z-n}{\sin(\pi z)}= \lim_{w\to0}\frac{w}{\sin(\pi w+n\pi)} $$ using $z-n=w$. Distinguish between $n$ odd and even, remembering a basic trig formula.

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$\sin \pi z =\dfrac{ e^{i\pi z} - e^{-i\pi z}}{2i}=0$ imply $e^{2i\pi z}=1$ then $2i\pi z=2k\pi i$ or $z=k$.

also the residues in $z=k$ are $$\lim_{z\to k}(z-k)\dfrac{1}{\sin \pi z}=\dfrac{(-1)^k}{\pi}$$ with l'hopital rule.

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Hint Use the formula: If $f(z_0)\neq 0$ and $g$ has a simple pole at $z_0$ then $$Res(\frac{f}{g},z_0)= =\frac{f(z_0)}{g'(z_0)}$$

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