I am studying the Goursat proof of the Cauchy Integral Theorem
It relies on the remark that the integrals $\int dz$ and $\int zdz$, taken along any closed contour are null.
I don't know how to prove that. I can imagine that $\int dz=0$ because when you return from where you went, the sum of all your trips should be null (it is not a very rigorous mathematical proof), but for $\int zdz$ it is much harder for me.
Can you help me to solve that problem?
Edit: we could use Cauchy's integral theorem for that but since it is exactly what we are looking to prove it doesn't make sense...
The fundamental theorem calculus applies to complex differentiable functions in a similar way to how it applies to real differentiable functions. To spell it out, if $f(z)$ is complex differentiable, and if $\gamma$ is a path from $a$ to $b$, then $$ \int_\gamma f'(z) dz = f(b) - f(a).$$
In particular, if $\gamma$ is a closed path, then $\gamma$ starts and finishes at the same place, so the $a$ and $b$ in the equation above are the same point. So if $\gamma$ is a closed path, you have $$ \oint_\gamma f'(z) dz = 0.$$
Now if you apply this result with the differentiable functions $f(z) = z$ and $f(z) = \frac 1 2 z^2$, you will be able to deduce that $\oint_\gamma 1 dz = 0$ and $\oint_\gamma z dz = 0$ along any closed path $\gamma$.