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Let $C(0, 1)$ be the unit circle centered at the origin with radius $1$. Then I need to evaluate the following complex contour integral: $$ \int_{C(0,1)} \frac{z e^z }{\tan^2 z}dz$$ I know the function $f(z) = \frac{z e^z}{\tan^2 z}$ has singularities at $z = n\pi$ with $n = 0, 1, \ldots $ But only only the first singularity lies inside the contour. So should I compute the residue $\text{Res}(f(z), z=0)$? Would this be a pole of order 2?

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    $\begingroup$ "Yes" and "no". $\endgroup$ – Martin R Nov 23 '15 at 15:45
  • $\begingroup$ How do I compute the residue then? I cannot use the formula $g(z_0) / h'(z_0)$, since in this case $g(z_0) = 0$. $\endgroup$ – Kamil Nov 23 '15 at 16:01
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A hint: $$ \frac{z e^z}{\tan^2 z} = \frac{z e^z}{\sin^2 z}\cos^2 z =\\ \frac{z(1+z+z^2/2+\dots)}{(z - z^3/6+\dots)^2}(1-z^2/2+\dots)^2=\frac{1}{z}+\dots. $$ From above $\text{Res}(f(z), z=0) = 1$.

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  • $\begingroup$ How did you deduce that this can be written as $1 / z + \ldots $ ? Did you plug in $z = 0$ ? $\endgroup$ – Kamil Nov 23 '15 at 16:40
  • $\begingroup$ @Kamil No, I just multiply all the staff in mind. Also note that $\frac{1}{(z - z^3/6+\dots)^2} = \frac{1}{z^2}(1+z^2/3+\dots)$. $\endgroup$ – Nikita Evseev Nov 23 '15 at 16:51
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$0$ is a pole of order $1$. So by residual theorem $$ \int_{C(0,1)} \frac{z e^z}{\tan^2 z}dz=2\pi iRes(f(z),0)=\lim_{z\to0}\frac{z^2 e^z}{\tan^2 z}=\lim_{z\to0}\frac{z^2 e^z \cos^2 z}{\sin^2 z}=1 $$ Note that $$ \lim_{z\to0}\frac{z^2 }{\sin^2 z}=1 $$

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