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Math Help - Number of solutions

  1. #1
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    Number of solutions

    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
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  2. #2
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    Quote Originally Posted by skyking View Post
    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
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  3. #3
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    Quote Originally Posted by tonio View Post
    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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