Compute the integral $\int \frac { \sin(z)}{z^3+16z} dz$ over $ |z+2i|=1$

Question

Compute this integral over $ |z+2i|=1$ : $$\int \frac { \sin(z)}{z^3+16z} dz$$

I am not sure how to start this question.

If I split the denominator into $z(z+4i)(z-4i)$, I feel there is a way that I can use the extended version of the Cauchy Integral formula however I cannot seem to figure out what my n would be, to get the integral $f(z)/(z-z_0)^{n+1}$

Answer 1

Denominator: $z^3+16z=z(z+4i)(z-4i)$

Note that $\sin(z)$ is an entire function.

Observe the singularities of the integrand: $0, -4i, 4i$.

But:

$|0+2i|=2>1$, $|-4i+2i|=2>1$, $|4i+2i|=6>1$.

Therefore, no poles are enclosed.

By Cauchy’s theorem, this integral is zero.

Answer 2

HINT.

Find the singularities.

$$z^3 + 16z = z(z^2 + 16) = z(z-4i)(z+4i)$$

hence by residues theorem...

Notice that you have a circle of radius 1 centered at $-2i$.

Evaluate the poles that lie within your circle and you'll fine the integral is zero.