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Math Help - complex trig Q

  1. #1
    Senior Member Danneedshelp's Avatar
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    complex trig Q

    Hi, I am not seeing the following step:

    We are given cos(z)=\frac{e^{iz}+e^{-iz}}{2}=\frac{1}{2}. Therefore,

    e^{2iz}-e^{iz}+1=0, and sp by the quadratic formula....

    I don't see how the got cos(z)=\frac{e^{iz}+e^{-iz}}{2}=\frac{1}{2} into the form e^{2iz}-e^{iz}+1=0. I can see the mult. the 2 and subtracted e^{iz} over, but I don't see how the got the e^{2iz} term.

    Thanks
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  2. #2
    Member Traveller's Avatar
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    In the original equation, multiply both sides by 2, bring the 1 to the LHS, and multiply both sides by e^{iz}. Can you see what happens?
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  3. #3
    Senior Member Danneedshelp's Avatar
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    e^{iz}+e^{-iz}-1=0
    e^{iz}e^{iz}+e^{iz}e^{-iz}-e^{iz}=0
    e^{iz+iz}+e^{-iz+iz}-e^{iz}=0
    e^{2iz}+e^{0}-e^{iz}=0
    e^{2iz}-e^{iz}+1=0

    ...bad post, sorry.
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