1. ## Number of solutions

Let $\displaystyle \displaystyle h(z)=\prod_{i=1}^{n}\frac{z-a_i}{1-\bar{a_i} z}$, where $\displaystyle |a_i|<1, i=0,1,2,...,n$. Show that the equation $\displaystyle h(z)=a_0$ has exactly $\displaystyle n$ solutions inside $\displaystyle |z|=1$.

I am trying to use Rouche's theorem here. My initial thought is that the equation above is equivalent to $\displaystyle \displaystyle \prod_{i=1}^{n}(z-a_i)-a_0\prod_{i=1}^{n}(1-\bar{a_i} z)=0$. So we pick $\displaystyle f(z)=\displaystyle \prod_{i=1}^{n}(z-a_i)$ and $\displaystyle \displaystyle g(z)=-a_0\prod_{i=1}^{n}(1-\bar{a_i} z)$. But I am having trouble showing that $\displaystyle |f(z)|>|g(z)|, |z|=1$.

Any help will be appreciated.
SK

2. Originally Posted by skyking
Let $\displaystyle \displaystyle h(z)=\prod_{i=1}^{n}\frac{z-a_i}{1-\bar{a_i} z}$, where $\displaystyle |a_i|<1, i=0,1,2,...,n$. Show that the equation $\displaystyle h(z)=a_0$ has exactly $\displaystyle n$ solutions inside $\displaystyle |z|=1$.

I am trying to use Rouche's theorem here. My initial thought is that the equation above is equivalent to $\displaystyle \displaystyle \prod_{i=1}^{n}(z-a_i)-a_0\prod_{i=1}^{n}(1-\bar{a_i} z)=0$. So we pick $\displaystyle f(z)=\displaystyle \prod_{i=1}^{n}(z-a_i)$ and $\displaystyle \displaystyle g(z)=-a_0\prod_{i=1}^{n}(1-\bar{a_i} z)$. But I am having trouble showing that $\displaystyle |f(z)|>|g(z)|, |z|=1$.

Any help will be appreciated.
SK

Put $\displaystyle z=x+iy\,,\,a_k=\alpha_k+i\beta_k$ , and show that $\displaystyle |z-a_k|^2= |1-\overline{a_k}z|^2$

for $\displaystyle |a_k|<1\,,\,|z|=1$

Tonio

3. Originally Posted by tonio
Put $\displaystyle z=x+iy\,,\,a_k=\alpha_k+i\beta_k$ , and show that $\displaystyle |z-a_k|^2= |1-\overline{a_k}z|^2$

for $\displaystyle |a_k|<1\,,\,|z|=1$

Tonio
Tonio

Thank you, I was able to show it with your help.

Thanks again

SK