Winding number of a closed path

Question

I have to calculate the winding number of the curve |z|=3 around the point $\alpha=i$. After
parameterizing, the integral is as follows $\frac{1}{2\pi i}\int_0^{2\pi}\frac{3ie^{it}}{3e^{it}-i}dt$

After calculating the integral I get $\left[ ln(3e^{it}-i) \right]_0^{2\pi}$, but I get that it's equal to zero. I know the result of the integral should be $2\pi i$, but I don't know how to get there. I'm missing something but i don't know what. Could you help me with this? Thank you in advance.

Answer 1

Hint : Note that if $\gamma : [0,2\pi] \longmapsto \mathbb{C}$ where $\gamma(t) = Re^{it}, R > 0$ we have that :

If $|z| < R \Rightarrow I(\gamma,z) = 1$,

If $|z| > R \Rightarrow I(\gamma,z) = 0$

Where $I(\gamma,z)$ denotes the winding number of the curve around $z$.

Can you take it from here ?

Answer 2

You should have $1/(2\pi i)\oint_\gamma1/(z-i)\rm dz=1$, by the Cauchy Integral Formula with $f(z)=1$. This is as expected.