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 April 7th, 2013, 02:30 PM #21 Member   Joined: Jan 2012 Posts: 86 Thanks: 0 Re: solutions to problems in Schaum's Complex Variables The residue theorem for exterior domains states if $f(z)$ is analytic outside (and on) a simple closed contour $C$ expect at isolated points (i.e., poles and isolated essential singularities), then $\int_{C} f(z) \ dz= -2 \pi i a_{-1} - 2 \pi i \sum_{k} \text{Res} [f(z),z_{k}]$ where $z_{k}$ are the poles/isolated essential singularities outside of the contour and $a_{-1}$ is the coefficient of the $\frac{1}{z}$ term in the Laurent expansion of $f(z)$ that converges for $z$ greater than some $R$. The value $a_{-1}$ is referred to as the residue at infinity. So you can restate the theorem to say that $\int_{C} f(z) \ dz= -2 \pi i \text{Res} [f(z),\infty] - 2 \pi i \sum_{k} \text{Res} [f(z),z_{k}]$. The negative signs are there because if we're moving counterclockwise around the contour, then we're moving clockwise relative to the poles/isolated essential singularities outside of the contour. You don't need to actually find the Laurent series since it turns out that calculating the residue at infinity is equivalent to calculating the residue of $- \frac{1}{z^{2}} f \left(\frac{1}{z} \right)$ at the origin. So $\int_{|z|=4} z^{2} \csc \left(\frac{1}{z} \right) \ dz = -2 \pi i \text{Res}[ z^{2} \csc \left(\frac{1}{z} \right),\infty] - 2 \pi i (0) = -2 \pi i \text{Res} [ -\frac{1}{z^{2}} \left(\frac{1}{z} \right)^{2} \csc \left(\frac{1}{\frac{1}{z}} \right), 0] = -\text{Res} [ -\frac{1}{z^{4}} \csc z, 0]$ You don't have to expand $\csc z$ to find that residue, of course. Since $\frac{1}{z^{4}}$ has a pole of order 4 at the origin and $\csc z$ has a simple pole at the origin, $- \frac{1}{z^{4}} \csc z$ has a pole of order $5$ at the origin. And thus $\text{Res} [ -\frac{1}{z^{4}} \csc z, 0]= -\frac{1}{4!} \lim_{z \to 0} \ \frac{d^{4}}{dz^{4}} \ (z-0)^{5} \frac{1}{z^{4}} \csc z$. I'll try to think of an example where using the residue theorem for exterior domains to calculate a real-valued integral is much easier than using the regular residue theorem for interior domains. Or maybe you can think of an example.
 April 8th, 2013, 11:05 AM #22 Senior Member   Joined: May 2011 Posts: 501 Thanks: 6 Re: solutions to problems in Schaum's Complex Variables Thanks G for the tutorial on the matter.

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# solution manual schaum's complex variables

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