Solution to Complex Equation $\cos z = 2i$ I tried solving the equation $\cos{z}=2i$ where $z$ is a complex number. The solution $i$ have ended up with is $z$, $= (4k+1)\frac{\pi}{2} - i\ln{(2+\sqrt{5})}$. However the text book solution is  $z=(2k+1)\frac{\pi}{2} - (-1)^k\ln{(2+\sqrt{5})}i$. Are the results geometrically same or my solution is erroneous. 
 A: $$\cos z = \frac{1}{2}\left(e^{iz}+e^{-iz}\right).$$ Solve first for $w=e^{iz}$.
$$w^2-4iw + 1=0$$
so:
$$w=\frac{4i\pm\sqrt{-16-4}}{2} = (2\pm\sqrt{5})i$$
So we have to deal with two cases with care. If $w=(2+\sqrt{5})i$, you get your solution:
$$z = (4k+1)\frac{\pi}{2} - i\ln(2+\sqrt{5})$$
If $w=(2-\sqrt{5})i$, then using $2-\sqrt{5}=-(2+\sqrt{5})^{-1}$, we have:
$$iz = -\log(2+ \sqrt{5}) +\left(\frac{3\pi}{2}+2\pi k\right)i$$
So:
$$z = \left(\frac{3pi}{2}+2\pi k\right) + \log(2+ \sqrt{5})i$$
In the first case, let $m=2k$ and you have:
$$z = (2m+1)\frac{\pi/2} - (-1)^{m}\ln(2+\sqrt{5})i$$
and the second case $m=2k+1$ then:
$$ z = (2m+1)\frac{\pi}{2} - (-1)^{2m+1}\ln(2+\sqrt{5})$$
So the full answer merges the two answers.

You can probably write this carefully as one case. The two values of $w$ can be written as $$w=i(-1)^k(2+\sqrt{5})^{(-1)^k}$$. Then:
$$iz = \log w = (-1)^k \log(2+\sqrt{5}) + \frac{\pi}{2}i + k\pi i$$
So:
$$z = \left(k+1/2\right)\pi - (-1)^k \log(2+\sqrt{5}) i$$
A: Your equation (for $k$) holds, since it is the equation from your book for $\hat{k}=2k$.
If $\hat{k}$ is impair such that the equation from the book holds there can not be a $k$ such that your equation for $k$ becomes the equation from the book for $\hat{k}$ since that would imply that $-1=\left(-1\right)^\hat{k}=+1$.
(two complex numbers coincide iff their real and complex coincide, I considered the complex and devided by $-\ln\left(2+\sqrt{5}\right)$)
The conclusion is, that you found only half of the solutions.
It seems like you used $\cos\left(x\right)=\cos\left(2\pi+x\right)$ but not $\cos\left(x\right)=\cos\left(-x\right)$.
A: You found half the answers.
Hint; What I usually do to solve these wrong-proof is write
$\cos x=\frac{e^{ix}+e^{-ix}}{2}$
And let $y:=e^{ix}$
Hint#2: Quadratics can have 2 different solutions.
A: Here is the solution without explanations:
$$\eqalign{
  & \cos (z) = 2i  \cr 
  & \cos (x + iy) = \cos (x)\cos (iy) - \sin (x)\sin (iy) = 2i  \cr 
  & \cos (x)\left( {{{{e^{i(iy)}} + {e^{ - i(iy)}}} \over 2}} \right) - \sin (x)\left( {{{{e^{i(iy)}} - {e^{ - i(iy)}}} \over {2i}}} \right) = 2i  \cr 
  & \cos (x)\left( {{{{e^{ - y}} + {e^y}} \over 2}} \right) - \sin (x)\left( {{{{e^{ - y}} - {e^y}} \over {2i}}} \right) = 2i  \cr 
  & \cosh (y)\cos (x) + {1 \over i}\sinh (y)\sin (x) = 2i  \cr 
  & \cosh (y)\cos (x) - i\sinh (y)\sin (x) = 2i  \cr 
  & \left\{ \matrix{
  \cosh (y)\cos (x) = 0 \hfill \cr 
  \sinh (y)\sin (x) =  - 2 \hfill \cr}  \right.  \cr 
  & \left\{ \matrix{
  \cos (x) = 0 \to x = (2n + 1){\pi  \over 2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,n = 0,1,2,... \hfill \cr 
  \left\{ \matrix{
  \sinh (y)\sin (x) = \sinh (y){( - 1)^n} =  - 2 \to \sinh (y){( - 1)^{n + 1}} = 2 \to \sinh ({( - 1)^{n + 1}}y) = 2 \hfill \cr 
  {( - 1)^{n + 1}}y = {\sinh ^{ - 1}}(2) = \ln (2 + \sqrt 5 ) \to y = {( - 1)^{n + 1}}\ln (2 + \sqrt 5 ) \hfill \cr}  \right. \hfill \cr}  \right.  \cr 
  & z = (2n + 1){\pi  \over 2} + i{( - 1)^{n + 1}}\ln (2 + \sqrt 5 ) \cr} $$
In final steps, I used the fact that $\sinh(y)$ is an odd function of $y$.
