Question

what is the integral along c of ln(1-z)dz where c is the boundary of a parallelogram with vertices +i,-i,+(1+i),-(1+i)

Answer 1

Lets call the loop given by the parallelogram's vertices \(\zeta_0\), then \(\zeta_0\) can be continuously deformed into a circle \(\zeta_1\) with center at z=1 and radius 1. We can now parametrize \(\zeta_2\) as:
\(z(t)=1+e^{it}\), \(0\le t\le 2\pi\). (Read about deformation of contours)
Now, we can apply Deformation Invariance theorem:
\[\large \int\limits_{\zeta_0} f(z)dz=\int\limits_{\zeta_1}f(z)dz\]

Answer 2

The problem I have here (You probably need to check this) is that \(f(z)=\ln(1-z)\) is not analytic at \(z=1\).

Answer 3

I calculated quickly and got \(-2i\pi\) as the value of the integral.