Number of solutions

**skyking** · April 10th 2011, 06:51 PM

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

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

Any help will be appreciated.
SK

**tonio** · April 10th 2011, 07:36 PM

Originally Posted by skyking

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

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

Any help will be appreciated.
SK

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

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

Tonio

**skyking** · April 11th 2011, 05:07 AM

Originally Posted by tonio

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

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

Tonio

Tonio

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

Thanks again

SK

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