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Evaluate the given integral by using Cauchy residue theorem

$$\int_{|z|=1} \frac{(1-\cos z)}{(e^z-1)\sin z}$$

My attempts : I take $f(z) = \frac{(1-\cos z)}{\sin z}$ as Residue $z=0$ now I got $f(0)=0$ $$\int_{|z|=1} \frac{(1-\cos z)}{(e^z-1)\sin z}= 2\pi if(z) _{\operatorname{res}z=0}=0$$

is its true ?

any hints/solution

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    $\begingroup$ Your result is correct. Recall $1-\cos(z)=2\sin^2(z/2)$ and $\sin(z)=2\sin(z/2)\cos(z/2)$. $\endgroup$ – Mark Viola Feb 20 at 17:50
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The integrand $\frac{1-\cos(z)}{(e^z-1)\sin(z)}=\frac{\tan(z/2)}{e^z-1}$ has a removable discontinuity at $z=0$. Hence, there is no singularity in or on $|z|\le 1$.

Cauchy's Integral Theorem guarantees that

$$\oint_{|z|=1}\frac{1-\cos(z)}{(e^z-1)\sin(z)}\,dz=0$$

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No, that is not correct. The answer is indeed $0$, but in order to prove that the best way is to see that$$\frac{1-\cos(z)}{(e^z-1)\sin(z)}=\frac{\frac{z^2}{2}-\frac{z^4}{24}+\cdots}{z^2+\frac{z^3}{2}-\frac{z^5}{24}+\cdots}=\frac{\frac12-\frac{z^2}{24}+\cdots}{1+\frac z2-\frac{z^3}{24}+\cdots}$$and that therefore $0$ is a removable singularity.

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Note that the singularity at $z= 0$ is removable, since

  • $\lim_{z\to 0}\frac{1-\cos z}{z^2} = \frac{1}{2}$ and
  • $\lim_{z\to 0}\frac{e^z-1}{z} = 1$ and
  • $\lim_{z\to 0}\frac{\sin z}{z} = 1$

Now write

$$\frac{1-\cos z}{(e^z-1)\sin z}=\frac{1-\cos z}{z^2}\frac{z^2}{(e^z-1)\sin z} $$

So, the residue is equal to $0$ as the only singularity within the region bounded by $|z|=1$ is removable.

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  • $\begingroup$ ..@trancelocation i thinks $\lim_{z\to 0}\frac{1-e^z}{z} = 0$ $\endgroup$ – jasmine Feb 20 at 17:59
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    $\begingroup$ @jasmine : No. It is $\frac{d}{dz}\left(e^z\right)$ at $z = 0$. Btw.: I edited my typo from $1-e^z$ to $e^z-1$. $\endgroup$ – trancelocation Feb 20 at 18:01
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    $\begingroup$ @jasmine Using L'Hospital's Rule, we see that $\lim_{z\to0}\frac{e^z-1}{z}=\lim_{z\to0}e^z=1$. Alternatively, $e^z=1+z+O(z^2)$ so that $\frac{e^z-1}{z}=1+O(z)\to 1$ as $z\to0$. $\endgroup$ – Mark Viola Feb 20 at 18:02
  • $\begingroup$ @Tancelocation ya i missed that $\endgroup$ – jasmine Feb 20 at 18:02

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