Almost Cauchy-Goursat theorem

Printable View

September 16th 2009, 10:49 AM
arbolis

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...
September 16th 2009, 11:33 AM
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.
September 16th 2009, 11:56 AM
arbolis

Quote:

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.