Integral of 1/f(z) over a unit circle.

Question

My lecturer has given solutions to this problem but I don't understand them.

Below is what she wrote

My issue is with the second part of the proof. I bounded (1/f(z) - 1/(f(z0)(z-z0)) = phi(z)/(f'(z)f(z)). But from this point now I don't understand what she did after this point to achieve the result.

Answer 1

The point is that the residue of $1/f(z)$ at $z=z_0$ is $1/f'(z_0)$.

In more detail, $f$ is nonzero except at $z=z_0$ so the only singularity of $f$ inside the circle is at $z_0$. So the integral is $2\pi i$ times the residue of $1/f(z)$ at $z=z_0$. By the conditions given, $f(z)$ has Taylor series $$f(z)=f'(z_0)(z-z_0)+a_2(z-z_0)^2+\cdots$$ and so $$\frac1{f'(z)}=\frac1{f'(z_0)}\frac1{z-z_0}+b_0+b_1(z-z_0)^2+\cdots $$ so the residue is $1/f'(z_0)$

Answer 2

There are two ideas in the proof.

The first is that if $g$ is analytic on $B(z_0,2) \setminus \{z_0\}$ then using Cauchy's integral theorem we can show that for any $\delta>0$ (and less than $2$), we have $\int_{|z-z_0|=1} g(z) dz = \int_{|z-z_0|=\delta} g(z) dz$. Hence we can choose our circle as small as we want.

The second is that since $f(z) = f'(z_0) (z-z_0)+ \phi(z)(z-z_0)$ (where $\phi(z) \to 0$ as $z \to z_0$), we would like if ${1 \over f(z)} \approx {1 \over f'(z_0) (z-z_0)}$ since we know how to compute the latter.

Note that since $\phi(z) \to 0$ we can find a $\delta>0$ such that for $|z-z_0| \le \delta$ we have $|f(z)| \ge { 1\over 2}|f'(z_0)| |z-z_0|$.

Note that $|{1 \over a} - {1 \over b}| \le {|a-b| \over |ab|}$ for $a,b \neq 0$.

Hence $|{1 \over f(z)} - {1 \over f'(z_0) (z-z_0)}| \le {|\phi(z)(z-z_0) | \over |f(z) f'(z_0)(z-z_0)| } = {|\phi(z) | \over |f(z) f'(z_0)| } \le {2|\phi(z) | \over |f'(z_0)|^2|z-z_0| }$ as long as $0< |z-z_0| \le \delta$.

Now choose $\epsilon>0$ and $\delta' \le \delta$ such that if $0<|z-z_0| \le \delta'$ then $|\phi(z) | < \epsilon$, which gives $|{1 \over f(z)} - {1 \over f'(z_0) (z-z_0)}| < {2 \epsilon \over |f'(z_0)|^2|z-z_0| } $.

Now integrate around the circle $|z-z_0| = \delta'$ to get \begin{eqnarray} |\int_{|z-z_0| = \delta'}({1 \over f(z)} - {1 \over f'(z_0) (z-z_0)}) dz| &\le& \int_{|z-z_0| = \delta'}|{1 \over f(z)} - {1 \over f'(z_0) (z-z_0)}| |dz| \\ &<& \int_{|z-z_0| = \delta'} {2 \epsilon \over |f'(z_0)|^2|z-z_0| } |dz| \\ &=& {4 \pi \epsilon \over |f'(z_0)|^2 } \end{eqnarray} Hence we have $\int_{|z-z_0| = 1}{1 \over f(z)} dz = \int_{|z-z_0| = 1} {1 \over f'(z_0) (z-z_0)}dz = {2 \pi i \over f'(z_0)}$.