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Let $p(z)$ and $q(z)$ are two polynomial of same degree and zeroes of $p(z)$ and $q(z)$ are inside open unit disc, $|p(z)|=|q(z)|$ on unit circle then show that $p(z)=\lambda q(z)$ where $|\lambda|=1$.

Please just give a hint not the whole solution. Thank you.

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1 Answer 1

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Let $d$ the common degree of $P$ and $Q$. Define $P_1(z):=z^dP(1/z)$ and $P_2(z):=z^dQ(1/z)$. These polynomial doesn't vanish in the unit disk. We apply maximum modulus principle to $P_1/P_2$ and $P_2/P_1$. This gives that $P_1/P_2$ is constant (otherwise we get a contradiction because $P_1/P_2$ would be a holomorphic function with a constant modulus on a connected open set, hence would be itself constant).

Note that the fact that $P$ and $Q$ have the same degree is necessary. Indeed, $P(z)=z$ and $Q(z)=z^2$ have their root in the open unit disk, the same modulus on the unit sphere but of course are not equal up to a constant.

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well, $z^d P(1/z)=a_d+a_{d-1}z+\dots+z^da_0$, could you tell me why this does not vanish on unit disk? as $a_d\neq 0$? –  El Angel Exterminador Jun 24 '12 at 9:15
    
Yes, and because the root of $P$ and $Q$ are in the open unit disk. –  Davide Giraudo Jun 24 '12 at 9:17
    
well, could you tell me what will happen maximum modulas on these two fraction ? I am not getting clearly, will these two attain maximum inside open disk?> –  El Angel Exterminador Jun 24 '12 at 9:33
    
If $P_1/P_2$ is constant, there is nothing to do. Otherwise, $P_1/P_2$ is a holomorphic function on the unit disk of modulus $1$. It actually imply that $P_1/P_2$ is constant. –  Davide Giraudo Jun 24 '12 at 10:00
    
thank you :)... –  El Angel Exterminador Jun 24 '12 at 10:14

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