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 2 years ago
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.
 2 years ago
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.

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hellow
 2 years ago
Best ResponseYou've already chosen the best response.1I think I found the answer online. It is posted here: http://www.freemathhelp.com/forum/threads/58401Usecos%28Z%29%28eizeiz%292tosolvecos%28Z%294. 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.

pjcappaert
 2 years ago
Best ResponseYou've already chosen the best response.0Hellow, how exactly do you make e^iz+e^iz=8 into a quadratic equation?

hellow
 2 years ago
Best ResponseYou've already chosen the best response.1We 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^28x+1=0 \[x=((8)\pm \sqrt{644})/2=4\pm2\sqrt{15}\] 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.)

pjcappaert
 2 years ago
Best ResponseYou've already chosen the best response.0Damn that was too easy. Thanks
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