# How can I solve the simultaneous equations that arise in solving $\cos(z)=2$.

If I have $\cos(z)=2$ I can say $\cos(a+ib)=2$

using double angle ideas $\cos(a)\cos(ib)+\sin(a)\sin(ib)=2$ using Euler's formula $\cos(a)\cosh(b)+i\sin(a)\sinh(b)=2$ equating real and imaginary parts $\cos(a)\cosh(b)=2$, $\sin(a)\sinh(b)=0$.

from here I'm unsure how to solve this set of simultaneous equations.

• Do you have to use the trig, hyperbolig trig formulae? I think there's a more direct way. – Simon S Apr 26 '15 at 16:45
• I'm not sure but if I say that z=arccos(2) I end up with a purely real solution which makes me think that It's missing something? – Goods Apr 26 '15 at 16:48
• If $\arccos 2$ is real, you have something wrong with your $\arccos$.. – GEdgar Apr 26 '15 at 16:50
• hmmm.. yes it actually gives an error. I must have done cos before... – Goods Apr 26 '15 at 16:52

Hint: A better way to do this. $\cos z = (e^{iz}+e^{-iz})/2$. Then your equation $\cos z = 2$ becomes a quadratic equation for $e^{iz}$.

• so e^i2z-4e^iz+1=o and solve that? – Goods Apr 26 '15 at 16:50
• ...so if you actually want real and imaginary parts in your answer, you will need to know how to do real and imaginary parts of the complex logarithm. – GEdgar Apr 26 '15 at 17:05

Alternatively, as $\cos z = {1 \over 2} (e^{iz} + e^{-iz})$, setting this equal to $2$, and writing $w = e^{iz}$, we have

$${1 \over 2} (w + 1/w) = 2 \ \ \text{ or alternatively } w^2 + 1 = 4w$$

$w = 2 \pm \sqrt 3$. Now solve for $z$.

$$e^{iz}=\cos z+i\sin z=2+i\cdot\pm\sqrt{1-2^2}=2\pm\sqrt3$$

$$\implies iz=\ln(2\pm\sqrt3)\iff z=-i\ln(2\pm\sqrt3)$$