Why some complex integrals are 0

This question asks to give an answer to why the following integrals are $0$:

(Notation: $\gamma(a,r)$ is a circle that has center $a$ and radius $r$.)

$$\int_{\gamma(i,3)}\frac{dz}{(z-2)^3}$$

My attempt: All three zeros of the denominator are in the disc. If they were outside the disc, then this would be zero. But I'm not sure what to do with this next.

$$\int_{\gamma(0,1)}z|z|^4$$

My attempt: The integrand is not holomorphic, so I'm not sure how to proceed.

EDIT: The problem above was answered by a hint in the comments.

$$\int_{\gamma(1,2)}\frac{\sin(z)dz}{z}$$

My attempt: $0\leq|\int_{\gamma(1,2)}\frac{\sin(z)dz}{z}|\leq\int_{\gamma(1,2)}|\frac{\sin(z)}{z}|dz\leq\int_{\gamma(1,2)}|\frac{1}{z}|dz\leq\int_{\gamma(1,2)}\frac{1}{|z|}dz$

I want to say $\int_{\gamma(1,2)}\frac{1}{|z|}dz\leq 0$ but $\frac{1}{|z|}$ isn't holomorphic, so I'm not sure what to do here.

Answer

The first one: $\frac{1}{(z-2)^3}$ has the antiderivative $-\frac{1}{2(z-2)^2}$.

The second one: It is the same as $\int_{\gamma(0,1)}z\cdot 1^4dz$ because $|z|=1$ on the curve.

The third one: $\frac{\sin z}{z}$ has a removable singularity at $z=0$, so the integral can actually be viewed as an integral of a holomorphic function. Alternatively, use Cauchy's integral formula $\frac{1}{2\pi i}\int_{\gamma}\frac{f(z)}{z-a}dz=f(a)$ and apply it to the point $a=0$ and function $f(z)=\sin z$, noting that $\sin 0=0$.