# Maximum and Minimum Moduli

• November 15th 2011, 07:33 AM
tarheelborn
Maximum and Minimum Moduli
Find the maximum and minimum moduli of $P(z)=z^2-z-1$ in the closed disk ${z:|z|\leq 1}$.

Now, by the maximum modulus theorem, the maximum of $|P(z)|$ is on $|z|=1$. Let $z=e^{i\theta}$, $0\leq \theta \leq 2\pi$. Then $|P(z)|^2=(z^2-z-1)(\overline{z}^2-\overline{z}-1)$ and then I get a little confused. I do lose the square root part of the modulus at this point, right?
• November 15th 2011, 08:15 AM
FernandoRevilla
Re: Maximum and Minimum Moduli
Quote:

Originally Posted by tarheelborn
I do lose the square root part of the modulus at this point, right?

No problem, if $h\geq 0$ then $x_0$ is a point of maximum (minimum) of $h$ iff $x_0$ is a point of maximum (minimum) of $h^2$ because the function $g(x)=x^2$ is strictly increasing for $x\geq 0$ .
• November 15th 2011, 11:12 AM
Opalg
Re: Maximum and Minimum Moduli
Quote:

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

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

If you write $e^{i\theta}$ as $\cos\theta+i\sin\theta$ then $P(e^{i\theta}) = \cos(2\theta) -\cos\theta-1 + i(\sin(2\theta)-\sin\theta).$ Therefore $|P(e^{i\theta})|^2 = \bigl(\cos(2\theta) -\cos\theta-1\bigr)^2 + \bigl(\sin(2\theta)-\sin\theta\bigr)^2.$ If you express everything in terms of $\cos\theta$ then this expression can be greatly simplified to tell you when $|P(z)|$ is maximal.