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Math Help - Prove the equation won't be satisfied for any complex number Z?

  1. #1
    Super Member fardeen_gen's Avatar
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    Prove the equation won't be satisfied for any complex number Z?

    If |Z| < \frac{1}{3}, then prove that the equation 1 + 2\cos^2 \theta_{1}Z + 2\cos^2\theta_{2}Z^2 + ... + 2\cos^2\theta_{n}Z^n = 0, will not be satisfied by any complex number z.
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  2. #2
    Super Member Showcase_22's Avatar
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    I haven't solved it yet, but here's what I have so far:

    <br /> <br />
1 + 2\cos^2 \theta_{1}Z + 2\cos^2\theta_{2}Z^2 + ... + 2\cos^2\theta_{n}Z^n = 0<br />

    1+(1+ \cos 2 \theta_1)Z+(1+ \cos 2 \theta_2)Z^2+....+ (1+\cos 2 \theta_n)Z^n=0

    (1+Z+Z^2+.......+Z^n)+(Z \cos 2\theta_1+ Z^2 \cos 2\theta_2+.....+ Z^n \cos 2 \theta_n)=0

    0=(1+Z+Z^2+.......+Z^n)+(Z \cos \theta_2+ Z^2 \cos \theta_2+.....+ Z^n \cos 2 \theta_n) < \left(1+\frac{1}{3}+\frac{1}{9}+......+\frac{1}{3^  n} \right)+ \left( \frac{1}{3} \cos 2\theta_1+\frac{1}{9} \cos 2\theta_2+.....+ \frac{1}{3^n} \cos 2 \theta_n \right)

    \left( 1+\frac{1}{3}+\frac{1}{3^2}+...+ \frac{1}{3^n} \right)= \frac{1-\left( \frac{1}{3} \right)^n}{1-\frac{1}{3}}=\frac{3}{2}  \left( 1- \left( \frac{1}{3} \right)^n \right)

    I'm not really sure where to go next
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by fardeen_gen View Post
    If |Z| < \frac{1}{3}, then prove that the equation 1 + 2\cos^2 \theta_{1}Z + 2\cos^2\theta_{2}Z^2 + ... + 2\cos^2\theta_{n}Z^n = 0, will not be satisfied by any complex number z.
    Because the Cauchy lower bound on the roots of this polynomial is greater than or equal to 1/3.

    (see here)

    CB
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  4. #4
    Super Member Showcase_22's Avatar
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    aww, shoot!

    I was way off!!
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