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Let $f(z) = \frac{ e^z - 1}{\sin z }$. Im trying to find $Res(f,z_0=0) $

I know that $(z-0)f(z)= \frac{ z(e^z-1)}{\sin z }$. And the limit as $z \to 0$ is of the form $0/0$ so applying lhopitals rule, we get that the limit is actually $0$. Can we conclude that the residue is then $0$ ?

In my notes, it says that the limit $\lim_{z \to z_0} (z-z_0) f(z) $ must be nonzero and then it equals the residue. Is there other ways to find the residue?

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2 Answers 2

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$z=0$ is a pole of order $0$ (it mean it a false singularity). Then $res(f,0)=0$. Indeed $$f(z)=\frac{z}{\sin(z)}\frac{e^z-1}{z}\underset{z\to 0}{\longrightarrow }1$$ and thus $f$ is holomorphe on $0$ and thus the residu is $0$.

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  • $\begingroup$ I thought for this work $p(z_0)$ cannot be $0$ ? $\endgroup$
    – user139708
    Nov 12, 2015 at 8:26
  • $\begingroup$ But, then, $z_0$ is not a simple pole, right? $\endgroup$
    – user139708
    Nov 12, 2015 at 8:37
  • $\begingroup$ But, the theorem in my books states: $f$ analytic on a region $A$ and $z_0$ isolated singularity. Then $z_0$ is a simple pole $\iff$ $\lim_{z \to z_0} (z-z_0)f(z) $ exists and is unequal to zero. This limit is the residue of $f$ at $z_0$. $\endgroup$
    – user139708
    Nov 12, 2015 at 8:39
  • $\begingroup$ Sorry to induced you in error, but in fact you can erase this singularity (it's a false singularity). So forget all what I did before. I edited my answer. $\endgroup$
    – Surb
    Nov 12, 2015 at 8:43
  • $\begingroup$ Quoting Wikipedia: "A few authors allow the order of a pole to be zero, in which case a pole of order zero is either a regular point or a removable singularity. However, it is more usual to require the order of a pole to be positive." $\endgroup$ Nov 12, 2015 at 8:44
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Just to answer your question "Are there other ways to find the residue?"

Its been a little while since I did complex calculus but I'm pretty sure if you find the Laurent Series then find the coefficient $a_{-1}$ then that is your residue.

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