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How would you evaluate $\oint_C \ e^{2z}(z+1)^{-1} \, \mathrm dz$ where $C=\{z\in \mathbb{C}: |z|=2 \}$?

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    Using Cauchy's integral formula.2011-10-23
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    @DavideGiraudo: but I thought as $e^{2z}(z+1)^{-1}$ is not holomorphic at for instance $z=-1$ then I couldn't use Cauchy's integral formula2011-10-23
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    You can apply it to $z\mapsto e^{2z}$ at $z_0=-1$, since $-1$ is in the disk of center $0$ and radius $1$.2011-10-23
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    @DavideGiraudo: Ah right.. I was being a fool! what do you think about $\oint_C \ e^{z}(cos(z))^{-1} \ \mathrm dz$ on $C=\{z:z\in \mathbb{C} \}$?2011-10-23
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    In my experience "$\oint$" means a line integral, so in "$\oint_C$" we need $C$ to be a curve.2011-10-23

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Recall the Cauchy's integral formula. Use it with $f(z) = \mathrm{e}^{2 z}$ and $a = -1$ and integration contour $\gamma = C$. This gives

$$ \int_C \frac{\mathrm{e}^{2z}}{z+1} \mathrm{d} z = 2 \pi i \mathrm{e}^{2 a} = 2 \pi i \mathrm{e}^{-2} $$

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    Alternate way to say the same thing: the residue theorem. Your integrand has a pole at $-1$ with residue ... (compute) ... so the integral is $2\pi i$ times ... (what?)2011-10-23
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    Awesome! How would you use with method to compute $\oint_C \ e^{z}(cos(z))^{-1} \ \mathrm dz$ where $C=\{z:z\in \mathbb{C} \}$2011-10-23
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    @LHS: That would certainly be more difficult. My first thought would be to use a Laurent series expansion for $\sec(z) = (\cos(z))^{-1}$, and then try to use Cauchy's Integral Formula on each term.2011-10-24
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    @LHS This is really a separate question, but can be done by Cauchy's formula. The denominator $\cos(z)$ has two zeros enclosed by $C$, $z= \pm \frac{\pi}{2}$. Contour $C$ can be bent to reduce to two small circles enclosing each pole. When applying Cauchy's formula for $z=\pi/2$, we should use $f_+(z) = \mathrm{e}^z \frac{z-\pi/2}{\cos(z)}$ which has limit $\lim_{z\to \frac{\pi}{2}} f_+(z) = -\mathrm{e}^{\pi/2}$, and about $z=-\pi/2$, $f_-(z) = \mathrm{e}^z \frac{z+\pi/2}{\cos(z)}$, with $\lim_{z\to \frac{\pi}{2}} f_-(z) = \mathrm{e}^{-\pi/2}$. So the answer is $-4\pi i\sinh \frac{\pi}{2}$.2011-10-24
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    Thank you greatly, this is far more clear now!2011-10-24