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Math Help - Complex Numbers

  1. #1
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    Complex Numbers

    If z = cis\theta, where \theta < \frac{\pi}{4} show that:

    \frac{1-z^2}{z+1} = 2\sin(\frac{\theta}{2})cis(\pi-\frac{3\theta}{2})

    Any help would be much appreciated as all my attempts at solving this haven't gone too well.
    Last edited by DekuTrombonist; March 7th 2010 at 08:22 PM. Reason: Forgot to add restrictions on theta
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  2. #2
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    Quote Originally Posted by DekuTrombonist View Post
    If z = cis\theta, where \theta < \frac{\pi}{4} show that:

    \frac{1-z^2}{z+1} = 2\sin(\frac{\theta}{2})cis(\pi-\frac{3\theta}{2})

    Any help would be much appreciated as all my attempts at solving this haven't gone too well.
    Either the answer is wrong or I'm wrong. You decide which:

    \frac{1-z^2}{1+z} = \frac{(1+z)(1-z)}{1+z} = 1-z = 1-\cos\theta - i\sin\theta.

    Next, 1-\cos\theta = 2\sin^2\tfrac\theta2 and \sin\theta = 2\sin\tfrac\theta2\cos\tfrac\theta2. Also, -i = \text{cis}\bigl(-\tfrac\pi2\bigr). Putting those things together, I get

    \begin{aligned}1-z &= 2\sin\tfrac\theta2\bigl(\sin\tfrac\theta2 - i\cos\tfrac\theta2\bigr) \\ &= 2\sin\tfrac\theta2(-i)\bigl(\cos\tfrac\theta2 + i\sin\tfrac\theta2\bigr) \\ &= 2\sin\tfrac\theta2\text{cis}\bigl(-\tfrac\pi2\bigr)\text{cis}\,\tfrac\theta2 \\ &= 2\sin\tfrac\theta2\text{cis}\bigl(\tfrac\theta2-\tfrac\pi2\bigr)\end{aligned}.
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  3. #3
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    Thanks for your help on this one. Obviosuly the answer in the question is wrong which probably explains why I couldn't do it.
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