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Math Help - Maximum and Minimum Moduli

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    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?
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Maximum and Minimum Moduli

    Quote Originally Posted by tarheelborn View Post
    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 .
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    Re: Maximum and Minimum Moduli

    Quote Originally Posted by tarheelborn View Post
    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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