Question

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.

Answer 1

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.

Answer 2

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

Answer 3

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.)

Answer 4

Damn that was too easy. Thanks