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 December 2nd, 2009, 08:17 PM #1 Newbie   Joined: Jan 2009 Posts: 21 Thanks: 0 residue Find the residue of the following. (a) $\text{Res}(f, z_0=0)$, where $f(z)=\frac{e^z}{\sin(2z^2)}$ (b) $\text{Res}(f, z_0=0)$, where $f(z)= \frac{\cos(\frac{1}{z})}{e^z}$ We are just beginning residues right now, and I do not know how to proceed. Should I use the Laurent expansion to find the residue for both parts? I am not very familiar with residues yet. Thanks very much.
 December 3rd, 2009, 10:06 AM #2 Newbie   Joined: Jan 2009 Posts: 21 Thanks: 0 Re: residue I was wondering if you can check my work. For (a) $\text{Res}(f, z_0=0)$, where $f(z)=\frac{e^z}{\sin(2z^2)}$ we have $\frac{1+z+z^2/2!+z^3/3!+\cdots}{2z^2 -(2z^2)^3/3!+(2z^2)^5/5!- \cdots}$ $= \frac{1}{2z^2}+\frac{1}{2z}+1/4+z/12$ So the residue is $1/2$. For (b) $\text{Res}(f, z_0=0)$, where $f(z)= \frac{\cos(\frac{1}{z})}{e^z}$ we have $\cos(\frac{1}{z}) \cdot (1-z+z^2/2-z^3/6+z^4/24+\cdots).$ I am stuck. I don't know how to get the residue. I know that $z_0$ is an essential singularity. I need help on part (b).

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