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

Note that the solutions to $t^2-\frac{3+i}{2}t+1=0$ can be simplified as,
$$t=\frac{3+i \pm \sqrt{-8+6i} } {4} =\frac{3+i \pm (1+3i) } {4}$$
or,
$$e^{iz}=1+i=\sqrt2 e^{i(\frac\pi4+2\pi n)} = e^{\frac12\ln2 +i(\frac\pi4+2\pi n)}$$
$$e^{iz}=\frac12(1-i) =\frac1{\sqrt2} e^{-i(\frac\pi4+2\pi n)} = e^{-\frac12\ln2 -i(\frac\pi4+2\pi n)} $$
Thus, the solutions are $z=\pm (\frac\pi4+2\pi n-\frac i2\ln2)$.
