Maximum and Minimum Moduli

**tarheelborn** · November 15th 2011, 07:33 AM

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?

**FernandoRevilla** · November 15th 2011, 08:15 AM

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 $x\geq 0$ .

**Opalg** · November 15th 2011, 11:12 AM

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.

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