Solve $\cos(z) =3/4+i/4$

Question

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

Answer 1

I guess the following helps you more: $$\cos z=\cos(x+iy)=\cos x\cosh y-i\sin x \sinh y$$

Answer 2

Taking a hint from @ChiefVS, one solution is $\alpha\approx 0.785398-i\, 0.346574$

All solutions are then:

$\{\alpha+2k\pi, -\alpha+2k\pi\}, \quad k\in \mathbb{Z}$

Answer 3

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

\begin{align} &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)} \end{align}

Thus, the solutions are $z=\pm (\frac\pi4+2\pi n-\frac i2\ln2)$.