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paglia Group Title

considering the equation cos(z)=4, z being a complex variable, how can I solve it? sorry for posting here, but supposed I could get some help here.

  • one year ago
  • one year ago

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  1. hellow Group Title
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    I think I found the answer online. It is posted here: http://www.freemathhelp.com/forum/threads/58401-Use-cos%28Z%29-%28e-iz-e-iz-%29-2-to-solve-cos%28Z%29-4. It turns out you can use the definition of the cosine of a complex number, and then use the quadratic equation to solve for e^(iz). Once you have found e^(iz), you can find z.

    • one year ago
  2. pjcappaert Group Title
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    Hellow, how exactly do you make e^iz+e^-iz=8 into a quadratic equation?

    • one year ago
  3. hellow Group Title
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    We can set x=e^(iz) in \[e ^{iz}+(e ^{iz})^{-1}=8\] So then we have x+x^-1=8. So we get (multiplying both sides by x): x^2+1=8x (quadratic!) x^2-8x+1=0 \[x=(-(-8)\pm \sqrt{64-4})/2=4\pm2\sqrt{15}\] So e^(iz) = 4+/- 2sqrt(15). From there I think you can take the natural log of both sides, then divide both sides by i: iz=ln(4+/2sqrt(15)) z=1/iln(4+/-sqrt(15)) = 1/i(i/i)ln(4+/-sqrt(15)) = -iln(4+/-sqrt(15)) (If you don't want to convert the e^(iz) to x, you can also work with the coefficients of \[e ^{2iz}-8e ^{iz}+1=0\] and and use the quadratic equation to find e^iz.)

    • one year ago
  4. pjcappaert Group Title
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    Damn that was too easy. Thanks

    • one year ago
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