A simple question about complex integration

Question

I've been asked if there exists a pair of curves $\gamma_1$ and $\gamma_2$ contained in an open set $G\subset \mathbb{C}$ connecting $z_1=-2i$ and $z_2=2i$ such that $$\int_{\gamma_1}\dfrac{1}{z}dz \neq \int_{\gamma_2}\dfrac{1}{z}dz$$ I thought that if $0\in G$, then the statement is true, but wasn't able to come up with a solution without this hypothesis.

Answer 1

I thought that if $0\in G$, then the statement is true

You would not let run the curves through 0, anyway, hence 0 need not be in $G$. There is no need for $G$ to be simply connected, hence you might take a ring, for example like $G=\{z\in\mathbb C\,\big/\,0.5 < |z| < 2\}$ and integrate along half-circles $|z|=1$ around 0.

Or you could use a half-circle for $\gamma_1$ and $\gamma2=\gamma_1+\text{one-full-turn-around-0}$ so that the integrals will differ by he integral over that one full turn.

Answer 2

Being asked this question, it really seem like you should have already covered what the winding number of a curve is.

The two expressions you list compute the winding numbers of $\gamma_1$ and $\gamma_2$. Therefore, if you can think of two with different winding behaviors around $0$, then you will have different integrals.