1. ## closed curve, trace

Show that if $\gamma$ is a closed curve in the complex plane, with trace $\Gamma$, and if $z_0 \not \in \Gamma$, then $\int_{\gamma} \frac{1}{(z-z_0)^n}dz=0$, $n \geq 2$.

Hint:

For $n \geq 2, (z-z_0)^{-n}$ has primitive $\frac{(z-z_0)^{-n+1}}{1-n}$ on $D$.

Attempt:

So the hint says that the primitive (the antiderivative) is $\frac{(z-z_0)^{-n+1}}{1-n}$. However, I do not see how this will help with showing that $\int_{\gamma} \frac{1}{(z-z_0)^n}dz=0$, $n \geq 2$. Thanks in advance.

2. Originally Posted by Erdos32212
Show that if $\gamma$ is a closed curve in the complex plane, with trace $\Gamma$, and if $z_0 \not \in \Gamma$, then $\int_{\gamma} \frac{1}{(z-z_0)^n}dz=0$, $n \geq 2$.

Hint:

For $n \geq 2, (z-z_0)^{-n}$ has primitive $\frac{(z-z_0)^{-n+1}}{1-n}$ on $D$.

Attempt:

So the hint says that the primitive (the antiderivative) is $\frac{(z-z_0)^{-n+1}}{1-n}$. However, I do not see how this will help with showing that $\int_{\gamma} \frac{1}{(z-z_0)^n}dz=0$, $n \geq 2$. Thanks in advance.
This is a special case of a form of the Fundamental Theorem of Calculus applied to the complex plane:

If $f: D \rightarrow \mathbb{C}$ is continous and such that there exists $g: D \rightarrow \mathbb{C}$ with $g'=f$ on $D$ then for any curve $\gamma: [0,1] \rightarrow D$ we have that $\int_{ \gamma } f= g(\gamma (1))-g(\gamma (0))$. To prove this (assuming $\gamma$ is $C^1$ or piecewise so) we just note that $\int_{ \gamma } f= \int_{0}^{1} f(\gamma (t)) \gamma ' (t) dt = \int_{0}^{1} g'(\gamma (t)) \gamma ' (t)dt$ $= \int_{0}^{1} (g \circ \gamma )' (t)dt = g(\gamma (1))-g(\gamma (0))$

PS. I'm naming $\gamma$ and it's trace $$\Gamma$$ as the same.