Complex Integration

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1

Let be the region of the complex plane enclosed by $x = \pm 2 \text{ and } y = \pm 2$

Compute $\int_C \frac{\cos z}{z(z^2+8)} dz$

$\frac{\pi i }{4}$

CORRECT

$\frac{\pi i }{2}$

0

$\frac{\pi i}{8}$

0

$\pi i$

0

$2 \pi i$

0

Explanation

Recall Cauchy's integral formula, which states

$f(z) = \frac{1}{2\pi i}\int_C\frac{f(s)}{s-z}ds$

In this case, we have $f(s) = \frac{cos(s)}{s^2 + 8}$

Plugging in gives us

$\int_C \frac{\frac{\cos z}{z^2+8}}{z} dz = 2\pi i f(0) = 2\pi i \frac{1}{8} = \frac{\pi i }{4}$

2

Let be the unit circle.

Compute $\int_C z^2 sin\left(\frac{1}{z}\right) dz$

$-\frac{\pi i}{3}$

CORRECT

$-\frac{\pi i}{2}$

0

$-\frac{\pi i}{6}$

0

$-\pi i$

0

0

Explanation

Recall the Taylor expansion of

$\sin(z) = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} + ...$

Use this to write

$z^2\sin\left(\frac{1}{z}\right) = z - \frac{1}{z3!} + \frac{1}{z^35!} - \frac{1}{z^57!} + ...$

The coefficient of $\frac{1}{z}$ is the residue, so by the residue theorem, the value of the integral is

$-\frac{\pi i}{3}$

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