P(z)=$a_0+a_1z+a_2z^2+…+a_{n-1}z^{n-1}+z^n$ is a complex polynomial such that |P(z)|$\le1$ on the closed unit disc $\mathbb D$. Prove that P(z)=$z^n$ on the entire complex plane $\mathbb C$.

By applying the formula for the derivatives at 0, I am able to prove that |$a_k| \le 1$ for all k=1,2,...(n-1). But how do I prove that all these coefficients are zero. Please help.

  • $\begingroup$ Thank you very much. I got the solution from the link. $\endgroup$ – Lawrence Mano Oct 25 '20 at 21:01

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