solve the equation: $\cos (z) = 1 + i$ , $z \in ℂ$. I want to solve the equation:  $\cos (z) = 1 + i$  , $z \in ℂ$.
I started by saying $\cos (z) = \frac {e^{iz} + e^{-iz}}{2}$.
I am therefore led to solve:
$$e^{iz} + e^{-iz} = 2 + 2i $$ this implies that $$e^{2iz}+1 =(2 + 2i) e^{iz}$$
By setting $x = e^{iz}$
Consider the equation:
$$x^2-(2 + 2i) x + 1 = 0$$
which is equivalent to solving the equation $$ x^2-2 \sqrt{2} e^{i \frac {\pi}{4}} x + 1 = 0$$
then $$ (x- \sqrt{2} e^{i \frac{\pi}{4} })^2 + 1 + 2i = 0$$
We set $$w = x- \sqrt {2} e^{i \frac{\pi}{4}}$$ so $$w^2 = -1-2i$$ But I can't seem to continue. An idea please.
 A: Try to use the complex definition of the inverse cosine and its analytic extension to simplify:
$$\cos^{-1}(z)=\frac\pi 2+i \ln\left(iz+\sqrt{1+z}\sqrt{1-z}\right)+2k\pi\Bbb Z$$
$$\mathop \implies^\text{z=1+i}\cos^{-1}(1+i)=\frac\pi2+i\ln\left(i(1+i)+\sqrt{1-(1+i)^2}\right)=\frac\pi2+i\ln\left(i-1+\sqrt{1-2i}\right)$$
To take a radical of a complex number, use De Moivre’s theorem and $n\in \Bbb Z$:
$$\sqrt{1-2i}=\sqrt[4]{1^2+(-2)^2}e^{\frac12i\left(\tan^{-1}\left(-\frac21\right)+2\pi n\right)}=\sqrt[4]5\cos\left(\frac12 \tan^{-1}(2)\right) + \sqrt[4]5i\sin\left(\frac12 \tan^{-1}(2)\right)=\pm\sqrt[4]5 \sqrt{\frac12+\frac{\sqrt{5}}{10}}\pm\sqrt[4]5\sqrt{\frac{2}{5+\sqrt5}}i$$
Putting it all together gives the general solution of the following. The signs can be chosen to be +,- or -,+:
$$\cos^{-1}(1+i)=i\ln\left(\pm\sqrt[4]5 \sqrt{\frac12+\frac{\sqrt{5}}{10}}-1\mp\sqrt[4]5\sqrt{\frac{2}{5+\sqrt5}}i+i\right)+ \frac\pi2+ 2\pi k, k\in \Bbb Z.$$
Proof of result for a case. Please correct me and give me feedback!
A: At this point
$x^2-(2 + 2i) x + 1 = 0.$
Applying the quadratic formula:
$x = 1+i \pm \sqrt {2i - 1}:$
$(1+i + \sqrt {2i - 1})(1+i - \sqrt {2i- 1}) = 2i - (2i-1) = 1$
This means we can let $e^{iz} = 1+i + \sqrt {2i - 1}$ and $e^{-iz} = 1+i - \sqrt {2i - 1}$
$iz = \ln (1+i + \sqrt {2i - 1})\\
z = -i\ln (1+i + \sqrt {2i - 1})$
