Solve the equation $\cos z = 7$ Solve the equation $\cos z = 7$
I know that $\cos z = \frac{e^{iz}+e^{-iz}}{2}$, but I think I can't use this identity.
Is anyone could help me at this point?
 A: You could transform the identity to 
cos(x + iy) = cos x cosh y - i sin x sinh y.
You've stated, cos(x + iy) = 7. So, sin x sinh y must be zero. Consequently, either sin x or sinh y must be zero.
Case 1: sin x = 0.
So cos x must be +/- 1. It follows that cosh y = +/- 7 since the real part of the identity is 7. Now, get your job done.
Case 2: sinh y = 0.
The sinh function expressed as a logarithm gives us y = 0. So cosh y = e^y - sinh y = 1 - 0 = 1. Hence, cos x = 7, which is simply not possible.
So you get your answer in case 1 itself.
Sorry for the lack of math typesetting. My current device doesn't support it. :-P
A: The identity you wrote is perfectly legitimate. Given that, we know that $w = e^{iz}$ satisfies $w + w^{-1} = 14$, or equivalently $w = 7 \pm 4 \sqrt{3}$. Thus we get $z = -i \log (7 \pm 4 \sqrt{3}) = \pm i \log (7 + 4\sqrt{3})$. (Note that the logarithm is only defined modulo $2\pi i$, which makes sense because $z$ itself is only defined mod $2\pi$. Also, $7 \pm 4 \sqrt{3} > 0$, so $z$ is pure imaginary.)
