In class the following situation was presented: suppose $f:\Omega\to \mathbb{C}$ is holomporphic. Let $z_0\in\Omega$ and suppose $f'(z_0)=0$ but $f''(z_0)\not=0$.Now consider the circle $C$ of radius $r$ around $z_0$. The professor then said:
Then, as $z$ moves on $C$, the image $f(z)$ goes around $f(z_0)$ twice since the winding number is the number of solutions to $f(z)=f(z_0)$, which is two in this case.
Can anyone help me see these two facts?
- How is it so that "the winding number is the number of solutions to $f(z)=f(z_0)$"?
- Why is it 2 in this case?
It can be further generalized that if the first non-zero derivative of $f$ is the $n$'th derivative, then the image of $C$ will go arnound $f(z_0)$ a total of $n$ times.
The argument principles gives: $$\frac{1}{2\pi i}\int_{g(C)}\frac{1}{u}\,\mathrm{d}u=\frac{1}{2\pi i}\int_C\frac{g'(z)}{g(z)}\,\mathrm{d}z=Z$$For $Z$ the number of zeroes of $g$ and $g:z\mapsto f(z)-f(z_0)$. $Z$ is the number of solutions to $f(z)=f(z_0)$ enclosed in $C$, and the left hand side of that equation is the winding number of $g(C)$ around zero which is the same as that of $f(C)$ around $f(z_0)$.
We want to show that, if $n$ is the first integer at which $f^{(n)}(z_0)=0$, then: $$\frac{1}{2\pi i}\int_{f(C)}\frac{1}{u-f(z_0)}\,\mathrm{d}u=n!$$I believe it should be $n!$, not $n$. With $n=2$, this is the same.
Picking $r$ suitably small we can expand in analytic series: $$\begin{align}\frac{1}{2\pi i}\int_C\frac{f'(z)}{f(z)-f(z_0)}\,\mathrm{d}z&=\frac{1}{2\pi i}\int_C f'(z)\left(\sum_{k\ge n}a_k(z-z_0)^k\right)^{-1}\,\mathrm{d}z\\&=\frac{1}{2\pi i}\sum_{k\ge -n}b_k\int_Cf'(z)(z-z_0)^k\,\mathrm{d}z\\&=\frac{1}{a_n}(f')^{(n-1)}(z_0)+\sum_{k=1-n}^{-1}b_k(f')^{(-k-1)}(z_0)+0\\&=n!+0\\&=n!\end{align}$$
Using Cauchy's theorem and the Cauchy derivative formulae.