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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. – Davide Giraudo Oct 23 '11 at 18:46
@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 formula – Freeman Oct 23 '11 at 18:48
You can apply it to $z\mapsto e^{2z}$ at $z_0=-1$, since $-1$ is in the disk of center $0$ and radius $1$. – Davide Giraudo Oct 23 '11 at 18:50
@DavideGiraudo: Ah right.. I was being a fool! what do you think about $\oint_C \ e^{z}(cos(z))^{-1} \ \mathrm dz$ on $C=\{z:z\in \mathbb{C} \}$? – Freeman Oct 23 '11 at 22:19
In my experience "$\oint$" means a line integral, so in "$\oint_C$" we need $C$ to be a curve. – GEdgar Oct 23 '11 at 23:05
up vote 4 down vote accepted

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?) – GEdgar Oct 23 '11 at 20:02
Awesome! How would you use with method to compute $\oint_C \ e^{z}(cos(z))^{-1} \ \mathrm dz$ where $C=\{z:z\in \mathbb{C} \}$ – Freeman Oct 23 '11 at 22:15
@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. – Jesse Madnick Oct 24 '11 at 1:19
@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 $-4\pi i\sinh \frac{\pi}{2}$. – Sasha Oct 24 '11 at 5:02
Thank you greatly, this is far more clear now! – Freeman Oct 24 '11 at 10:03

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