The question:

Find all solutions $\displaystyle z \in C$ of cos(z) = cos(2)

My attempt:

cos(z) = cos(x)cosh(y) - isin(x)sinh(y) = cos(2)

So we need:

sin(x)sinh(y) = 0 [1]

cos(x)cosh(y) = cos(2) [2]

[1] is zero when $\displaystyle x = k\pi$ or y = 0

Substitute into [2]:

$\displaystyle cos(k\pi)cosh(0) \ne cos(2)$

Try just y = 0:

So cos(x)cosh(0) = cos(x) which we want to equal cos(2), thus $\displaystyle x = \pm2$

Try just $\displaystyle x = k\pi$

So $\displaystyle cos(k\pi)cosh(y) = \pm cosh(y) = cos(2)$

Case when k is odd:

cosh(y) = -cos(2)

Case when k is even:

cosh(y) = cos(2)

Therefore, y = $\displaystyle cosh^-1(cos(2))$

So, $\displaystyle z = \pm 2 + icosh^{-1}(cos(2))$

But the solution is $\displaystyle \pm(2 + 2k\pi)$ :/

What am I doing wrong? Thanks.