## paglia 3 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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1. hellow

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.

2. pjcappaert

Hellow, how exactly do you make e^iz+e^-iz=8 into a quadratic equation?

3. hellow

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.)

4. pjcappaert

Damn that was too easy. Thanks