1. ## Almost Cauchy-Goursat theorem

Let $C$ be the positive oriented frontier of the semi disk $0\leq r \leq 1, 0\leq \theta \leq \pi$ and let $f(z)$ be a continuous function defined in the semi disk, given by $f(0)=0$ and $f(z)=z^{\frac{1}{2}}= \sqrt r e^{\frac{i\theta}{2}}$, $(r>0, -\frac{\pi}{2}< \theta < \frac{3\pi}{2})$.
Prove that $\int _C f(z)dz=0$.
Is the C-G theorem applicable?
----------------------------------------------
My attempt : I notice I can't apply C-G theorem because $f(z)$ is not analytic in $z=0$ which is a point in $C$.

I've absolutely no idea how to prove what they ask me.
I'd like an idea...

2. You can avoid the singularity at the origin by making the contour C take a small detour to avoid it. More precisely, replace the straight line section of C (going from –1 to +1) by three segments defined as follows. First, a straight line segment going from –1 to –ε, then a semicircle given by $\varepsilon e^{i\theta}$, with θ going from π to 0, and finally a straight line segment from +ε to +1. The integral round this modified contour will be 0 (by Cauchy's theorem), and as $\varepsilon\to0$ the integral round the modified integral will tend to the original integral round C. The reason for this is that the function f(z) is small when z is close to 0, so the integrals round parts of the contours near the origin will also be small.

3. Originally Posted by Opalg
You can avoid the singularity at the origin by making the contour C take a small detour to avoid it. More precisely, replace the straight line section of C (going from –1 to +1) by three segments defined as follows. First, a straight line segment going from –1 to –ε, then a semicircle given by $\varepsilon e^{i\theta}$, with θ going from π to 0, and finally a straight line segment from +ε to +1. The integral round this modified contour will be 0 (by Cauchy's theorem), and as $\varepsilon\to0$ the integral round the modified integral will tend to the original integral round C. The reason for this is that the function f(z) is small when z is close to 0, so the integrals round parts of the contours near the origin will also be small.
Ah I see what you mean! Thanks for the perfectly clear description.

However I'm having difficulties to formally justify that the original integral is worth $0$.
I know that near the origin the function $f(z)$ is small.

Oh... maybe the integral is worth $0$ because $f(0)=0$ which makes $f$ being continuous in the disk. If $f(0)$ would have been different from $0$ then the integral wouldn't have been worth $0$... or am I mistaking?
And of course the integral over the whole region except the origin is worth 0 as you pointed out, by Cauchy-Goursat.