$\frac{1}{2 \pi i} \int_{|z|=4} z \frac{f'(z)}{f(z)}dz$

Question

Analytic in $A=\{|z|<5\}$ function $f(z)$ has first order zero at $z_{0}=1+i$ and nowhere else in A. Calculate $$I=\frac{1}{2 \pi i} \int_{|z|=4} z \frac{f'(z)}{f(z)}dz.$$ Prove the answer.

What I've tried so far:

Let $f(z)=g(z)(z-(1+i)),$ $g(z)\ne0$ $\forall z \in A$.

$f'(z)=g'(z)(z-(1+i))+g(z)$

$$I=\frac{1}{2 \pi i} \int_{|z|=4} z \frac{f'(z)}{f(z)}dz=\frac{1}{2 \pi i} \int_{|z|=4} z \frac{g'(z)(z-(1+i))+g(z)}{g(z)(z-(1+i))}dz=\frac{1}{2 \pi i} \int_{|z|=4} z \frac{g'(z)}{g(z)}dz+\frac{1}{2 \pi i} \int_{|z|=4} \frac{z}{z-(1+i)}dz=X+Y$$ To calculate Y it's easy: $\frac{1}{2 \pi i} \int_{|z|=4} \frac{z}{z-(1+i)}dz=1+i$

But I've got some difficulties with X:

$$\frac{1}{2 \pi i} \int_{|z|=4} z \frac{g'(z)}{g(z)}dz=\frac{1}{2 \pi i} \int_0^{2 \pi} 4e^{it} \frac{g'(4e^{it})}{g(4e^{it})}4ie^{it}dt=\frac{1}{2 \pi i} \int_0^{2 \pi} \frac{4e^{it}}{g(4e^{it})}d(g(4e^{it}))$$

I assume that $X=0$ when I take $g(z) \ne 0$ like $g(z)=e^z$ or $g(z)=const$. But how can I prove it?

Can anyone help me?

Answer 1

Why not do this with residue theorem. $zf'(z)$ is holomorphic as $f(z)$ is holomorphic.

So, $F(z) = \frac{zf'(z)}{f(z)} $ has pole of order one only at $z=1+i$ as $f(1+i)=0$

We can claim that $f'(1+i) \not=0$ as according to question, the root is of order one

So, $I = \lim_{z \to1+i} \frac {(z-1-i)zf'(z)}{f(z)} = \lim_{z \to1+i} \frac {zf'(z)}{\left (\frac{f(z) - f(1+i)}{z-(1+i)}\right)}=\lim_{z \to1+i} \frac {zf'(z)}{f'(z)}= 1+i $

Answer 2

Since $g(z)$ has no zero in $A$, the function $z \frac{g'(z)}{g(z)}dz$ is analytic on and everywhere inside $|z| =4$. Therefore, $$X=\frac{1}{2 \pi i} \int_{|z|=4} z \frac{g'(z)}{g(z)}dz=0$$