Parameterizations of a square in the complex plane

Question

Verify Cauchy's Theorem for the function $3z^2 + iz -4$, where $C$ is the square with vertices at $1+i,\ 1-i,\ -1+i,\ -1-i$

I'm not asking for help with Cauchy's formula, simply with the parameterization of complex lines.

I can see from looking at these points drawn that the segment from $[-1-i, 1-i]$ can be written as $z(t)=t-i$, but what I don't understand is how to use the general case of $$z(t) = (1-t)z_0+tz_1$$

If I plug in $z_0=(-1-i)$ and $z_1=(1-i)$, I clearly don't arrive at $z(t)=t-i$

What are $z_0\ and\ z_1$?

Answer 1

Just divide the square to 4 parts - one part for each segment - and parametrize them independently in such a way that all-in-all $t$ runs from $0$ to $1$.

Answer 2

The curve $\gamma(t) = (1-t) z_0 + t z_1$ is a straight line joining $z_0$ ($t=0$) and $z_1$ ($t=1$).

You can represent the path around the four points by the four segments

$\gamma_1(t) = (1-t) (1+i) + t (1-i)$, $\gamma_2(t) = (1-t) (1-i) + t (-1+i)$, etc.

Then compute $\int_{\gamma_1} f(z) fz + \int_{\gamma_2} f(z) fz + \cdots$.