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I'm trying to find the residues of the functions at the points:

1) $f(z) = \dfrac{e^z}{z^2(1-2z)^2}$ at $z = \dfrac{1}{2}$.

So, we have a simple pole and can I take the derivative of the denominator? But why does this seem to still fail?

2) $f(z) = \dfrac{e^2z}{6 \cosh(z) - 10}$ at $z = \log 3$.

I am not sure why we have a pole at $z = \log 3$; it seems like from the Taylor expansion of $\cosh(z)$, nothing happens here.

3) $f(z) = (z^4 + z^2+1)\sin\left(\dfrac{1}{z}\right)$ at $z = 0$.

I think of the residue at infinity where I say: $\frac{1}{z^2}f(\frac{1}{z})$, but not sure if this would apply.

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For (1) it looks more like a double pole, since the factor $(1-2z)$ which is sero at $z=1/2$ appears squared in the denominator. For (2) have you tried to set the bottom to zero and solve? For (3) maybe you can use that $sin(u)$ is $u-u^3/6+...$ somehow. –  coffeemath Oct 22 '12 at 10:01
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1 Answer

Hints:
(1) Why is $z=\frac12$ a simple pole for $f(z)$? Remember that $z=\alpha$ is a pole of order $m$ for $f(z)$ if and only if $\lim_{z\to\alpha}(z-\alpha)^mf(z)$ exists, finite and non-zero. In this case, $${\rm{Res}}(f(z),\alpha)=\lim_{z\to\alpha}\frac{((z-\alpha)^mf(z))^{(m-1)}}{(m-1)!}$$ (2) Remember that $\cosh(z)=\frac{e^z+e^{-z}}{2}$. From here you can find all the singularities of $f(z)$. Also, if a function is analytic at (a small neighborhood of) a point, then the residue at the point is $0$.
(3) Do you know how to find the Laurent series of $\sin\left(\frac1z\right)$ around $z=0$? If so, you can find easily the Laurent series of $f(z)$ around $z=0$. Now find the residue by the definition.

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