Solve $\cos(z) =3/4+i/4$ I need to solve the complex trinometric equation
$$\cos(z) =\frac{3}{4}+\frac{i}{4} $$
What I've done so far is:
Using the cos formula I got $e^{iz} +e^{-iz} =\frac{3}{2}+\frac{i}{2}$
Making $t=e^{iz} $ we have $t+\frac{1}{t}=\frac{3}{2}+\frac{i}{2}$
Multiplying by $t^2$ we get 
$$t^2-\frac{3+i}{2}t+1=0$$
Solving that we get 
$$t=\frac{(\frac{3}{2}+\frac{i}{2}) \pm \sqrt{(\frac{3}{2}+\frac{i}{2}) ^2-4}} {2} = \frac{3+i \pm \sqrt{-8+6i} } {4} $$
Converting 3+i to polar we get $\sqrt{10} e^{0.3218i}$
Converting $\sqrt{-8+6i}$ to polar we get $\sqrt{10} e^{-0.3218i}$
So $t=\frac{\sqrt{10} e^{0.3218i} \pm \sqrt{10} e^{-0.3218i}} {4} $
Which means $e^{iz} =\frac{\sqrt{10} e^{0.3218i} \pm \sqrt{10} e^{-0.3218i}} {4} = \frac{\sqrt{10}} {2} (\frac{e^{0.3218i} \pm e^{-0.3218i} }{2})$
And I dont know where to go from there

Taking a hint from @ChiefVS, one solution is $\alpha\approx 0.785398-i\,  0.346574$
All solutions are then:
$\{\alpha+2k\pi, -\alpha+2k\pi\}, \quad k\in \mathbb{Z}$
