HOW DO I DO THIS?!
Solve $\displaystyle cos(z) = 5$
Help... please ...
need someone to go through slowly with me lol :S
You should note that $\displaystyle \displaystyle \cos{(z)} = \cos{(x + iy)} = \cos{(x)}\cosh{(y)} - i\sin{(x)}\sinh{(y)}$, so
$\displaystyle \displaystyle \cos{(x)}\cosh{(y)} = 5$ and $\displaystyle \displaystyle -\sin{(x)}\sinh{(y)} = 0$.
From the second equation, we can see that either $\displaystyle \displaystyle \sin{(x)} = 0 \implies x = n\pi, n \in \mathbf{Z}$ or $\displaystyle \displaystyle \sinh{(y)} = 0 \implies y = 0$.
If $\displaystyle \displaystyle x = n\pi$ then
$\displaystyle \displaystyle \begin{align*} \cos{(x)}\cosh{(y)} &= 5 \\ \cos{(n\pi)}\cosh{(y)} &= 5 \\ (-1)^n\cosh{(y)} &= 5 \\ \cosh{(y)} &= \frac{5}{(-1)^n} \end{align*}$
and since the hyperbolic cosine is always positive, that means we can only accept even values for $\displaystyle \displaystyle n$, so $\displaystyle \displaystyle n = 2m, m\in \mathbf{Z}$ and $\displaystyle \displaystyle y = \textrm{arcosh}\,{(5)}$.
So the first solution is $\displaystyle \displaystyle z = 2m\pi + \textrm{arcosh}\,{(5)}, m \in \mathbf{Z}$.
See if you can get the second solution, for where $\displaystyle \displaystyle y = 0$.