Complex analysis review 4

The zeros of analytic functions

Use Liouville theorem, we can prove the fundamental theorem of algebra.

Theorem 1

If $p(z)$ is a polynomial, then there is at least one $z_0$ such that $p(z_0)=0$.

Let

$$f(z) = \frac{1}{p(z)}$$
then if $p$ has no roots, $f$ is analytic on $C$. Implying that $f$ is a constant, so as $p$.

Theorem 2

Suppose that $f(z)$ is analytic on $U\subset C$, then except for being the constant function 0, there is no accumulation point of the set $\{z\in U| f(z) = 0\}$.

If it has a accumulation point, i.e., suppose that $z_1,\cdots,z_n,\cdots$ , are zeros of $f$, and there is an accumulation point $z_0$. Without loss of generality, we assume that $z_0=0$. Then since $f$ is analytic on $U$, there is a Taylor expansion of $f$ at point 0

$$f(z) = a_0+a_1z+\cdots$$
Then

$$\lim_{n\rightarrow \infty} f(z_n) = f(\lim_{n\rightarrow \infty}z_n) = f(0) = a_0 = 0.$$
Next we can consider $f(z)/z$, for the same reason we have $a_1=0$. The process continues.

Now we can conclude that if $E$ is a subset of $U$ and $E$ has an accumulation point, $h_1 = h_2$ on $E$. Then $h_1=h_2$ on $U$. So for some triangular equations, if they hold in $\mathbb{R}$, then they also hold in $\mathbb{C}$.

The argument principle

Theorem 3

If f(z) is an analytic function inside and on some closed contour $\gamma$, and f has no zeros on $\gamma$ as well as finitely many zeros inside the contour. Then the number $k$ of zeros insider the contour $\gamma$ is

$$k = \frac{1}{2\pi i} \int_\gamma \frac{f'(z)}{f(z)}dz.$$
Denote the zeros and their multiple number as $z_1,\cdots,z_n$, with respect to $k_1,\cdots,k_n$. They by Cauchy integral formula,
$$\frac{1}{2\pi i} \int_\gamma \frac{f'(z)}{f(z)} dz= \sum_{i=1}^n\frac{1}{2\pi i}\int_{\gamma_i}\frac{f'(z)}{f(z)}dz.$$
Inside of each $\gamma_i$, we can write
$$f(z) = (z-z_i)^{k_i}h_i(z).$$
Where $h_i$ is nonzero, and then
$$\frac{f'(z)}{f(z)} = \frac{k_i}{z-z_i}+\frac{h_i'(z)}{h_i(z)}$$
which give the desired result.

Theorem 4 (Hurwitz)

Suppose that $\{f_j\}$ is a sequence of analytic function on $U\subset C$, for any compact subset of $U$, it converges uniformly to a function $f$. If all of the function in the sequence are not identity to zero, then $f$ is either equals to constant function 0 or never equals to zero.

Choose any closed contour $\gamma$, then for $z$ insider the contour

$$f_j (z)= \frac{1}{2\pi i}\int_\gamma \frac{f_j(\xi)}{\xi- z}d\xi$$
By hypothesis, let $j\rightarrow \infty$, then

$$f(z)= \frac{1}{2\pi i}\int_\gamma \frac{f(\xi)}{\xi- z}d\xi$$
So $f(z)$ is an analytic function and by the same reason $f_j'$ converges to $f'$ on any compact subset of $U$. If $f$ is not constant function 0 then we choose $\gamma$ not pass through its zeros, which is guaranteed by the discreteness of zeros of analytic functions. And note that
$$\frac{1}{2\pi i}\int_\gamma \frac{f_j'(z)}{f_j(z)}dz =0.$$
We concluded that
$$\frac{1}{2\pi i}\int_\gamma \frac{f'(z)}{f(z)}dz =0.$$

Rouche’s Theorem

Theorem 4

Suppose that $f(z),g(z)$ are analytic on $U\subset \mathbb{C}$, $\gamma$ is a contour that is rectifable, on $\gamma$ there holds

$$|f(z)-g(z)|< |f(z)|,$$
then $f,g$ have the same number zeros insider the contour $\gamma$.

Let $N_1,N_2$ denote the number of zeros of $f,g$ respectively. Then

$$N_2-N_1 = \frac{1}{2\pi i}\int_\gamma \frac{F'(z)}{F(z)}dz.$$
Where $F(z) = \frac{g(z)}{f(z)}$. And from the hypothese, we know that
$$|F(z)-1|< 1.$$
Which shows that $N_2-N_1=0$.

Let

$$p(z) = a_nz^n+\cdots+a_0,\quad a_n\neq 0.$$
Choose $g(z) = a_nz^n$, then $g$ has $n$ roots. When $R$ is large enough, on contour $|z|=R$, we have
$$|p(z)-g(z)|<|g(z)| = |a_n|R^n.$$
By Rouche’s theorem, $p$ has exactly $n$ roots.

Theorem 5

Suppose that $f$ is analytic on $U\subset \mathbb{C}$, $w_0 = f(z_0),z_0\in U$. If $z_0$ has multiple number $m$, then for $\rho>0$ small enough, there is $\delta>0$, such that for any $A\in D(w_0,\delta)$, the number of zeros of $f(z)-A$ on $D(z_0,\rho)$ is exactly $m$.

There is some $\rho>0$ small enough such that $f(z)-f(z_0)$ has no zeros except for $z_0$ on $\bar{D}(z_0,\rho)\subset U$. But in the circle $|z-z_0|=\rho$, there holds

$$|f(z)-f(z_0)|\geq \delta \quad (\delta>0).$$
Then for $A\in D(w_0,\delta)$,
$$|A-w_0|< |f(z)-f(z_0)|.$$
That is when $|z-z_0|=\rho$,
$$|(f(z)-f(z_0))-(f(z)-A)|<|f(z)-f(z_0)|.$$
This shows that $f(z)-A$ has the same number of zeros in $D(z_0,\rho)$ which is .