# 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

• Take $log_{e}$ both sides.
– user721016
Commented Mar 3, 2020 at 17:46
• Can I just take log? Commented Mar 3, 2020 at 18:18
• Yes and then make it periodic
– user721016
Commented Mar 3, 2020 at 22:08

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 $$z=\pm (\frac\pi4+2\pi n-\frac i2\ln2)$$.

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

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

All solutions are then:

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