# Thread: Maximum and Minimum Moduli

1. ## Maximum and Minimum Moduli

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?

2. ## Re: Maximum and Minimum Moduli

Originally Posted by tarheelborn
I do lose the square root part of the modulus at this point, right?
No problem, if $\displaystyle h\geq 0$ then $\displaystyle x_0$ is a point of maximum (minimum) of $\displaystyle h$ iff $\displaystyle x_0$ is a point of maximum (minimum) of $\displaystyle h^2$ because the function $\displaystyle g(x)=x^2$ is strictly increasing for $\displaystyle x\geq 0$ .

3. ## Re: Maximum and Minimum Moduli

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