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**tarheelborn** Find the maximum and minimum moduli of $\displaystyle P(z)=z^2-z-1$ in the closed disk $\displaystyle {z:|z|\leq 1}$.

Now, by the maximum modulus theorem, the maximum of $\displaystyle |P(z)|$ is on $\displaystyle |z|=1$. Let $\displaystyle z=e^{i\theta}$, $\displaystyle 0\leq \theta \leq 2\pi$. Then $\displaystyle |P(z)|^2=(z^2-z-1)(\overline{z}^2-\overline{z}-1)$ and then I get a little confused. I do lose the square root part of the modulus at this point, right?