Ngô Quốc Anh

August 29, 2009

On a polynomials of degree n, having all its zeros in the unit dis

I found a very useful inequality involving a polynomials of degree n, having all its zeros in the unit disk from the following paper, doi:10.1016/j.jmaa.2009.07.049, published in J. Math. Anal. Appl.

Statement. If P(z) is a polynomial of degree n, having all its zeros in the disk |z| \leq 1, then

\left| {zP'(z)} \right| \geq \displaystyle\frac{n} {2}\left| {P(z)} \right|

for |z|=1.

Proof. Since all the zeros of P(z) lie in $latex|z| \leq 1$. Hence if z_1, z_2,...,z_n are the zeros of P(z), then |z_j| \leq 1 for all j =1,2,...,n. Clearly,

\displaystyle\Re \frac{{{e^{i\theta }}P'\left( {{e^{i\theta }}} \right)}}{{P\left( {{e^{i\theta }}} \right)}} = \sum\limits_{j = 1}^n {\Re \frac{{{e^{i\theta }}}}{{{e^{i\theta }} - {z_j}}}} ,

for every point e^{i\theta}, 0 \leq \theta < 2 \pi which is not a zero of P(z). Note that

\displaystyle\sum\limits_{j = 1}^n {\Re \frac{e^{i\theta }}{e^{i\theta } - z_j}}\geq\sum\limits_{j = 1}^n{\frac{1}{2}}=\frac{n}{2}.

This implies

\displaystyle\left| {\frac{{{e^{i\theta }}P'\left( {{e^{i\theta }}} \right)}}{{P\left( {{e^{i\theta }}} \right)}}} \right| \geq \Re \frac{{{e^{i\theta }}P'\left( {{e^{i\theta }}} \right)}}{{P\left( {{e^{i\theta }}} \right)}} \geq \frac{n}{2},

for every point e^{i \pi}, 0 \leq \theta < 2\pi. Hence

\left| {zP'(z)} \right| \geq \displaystyle \frac{n}{2}\left| {P(z)} \right|

for |z|=1 and this completes the proof.

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