Evaluate the integral $\int_{\gamma} \frac{dz}{z+\frac{1}{2}-\frac{i}{3}}$

Question

Let $\gamma(t)=i+e^{it}$, $t \in [0,2\pi]$. Then

Evaluate the integral: $$ \int_{\gamma}\frac{dz}{z+\frac{1}{2}-\frac{i}{3}}$$

Now $\gamma(t)$ is a closed curve . Since $\left(\frac{-1}{2},\frac{1}{3}\right) \not \in B(i,1)$, there must be a branch of logarithm in the given domain and hence that would make the integral zero.

But I am unable to find such a branch.

Is there any other way of doing it??

Answer 1

The point $z_0 := -(1/2) + (1/3)i$ lies inside $\gamma$ since $$|z_0 - i|^2 = (-1/2)^2 + (-2/3)^2 = 1/4 + 4/9 = 25/36 < 1.$$

So by the Cauchy integral formula, the value of the integral is $2\pi i$.