Help with $\cos(z)=3/2$

Question

We were doing solutions to trigonometric equations and pre-calculus and I stumbled across a problem that had no real solutions but I inquired about the complex ones. Please tell me if I have made any mistakes, my own answer differs from that of Wolframalpha's.

$$\cos(z)=\frac32 \\ \frac{e^{iz}+e^{-iz}}{2}=\frac32 \\ e^{iz}+e^{-iz}=3 \\ e^{2iz}-3e^{iz}+1=0 \\ Z=e^{iz} \\ Z^2-3Z+1=0 \\ Z=\frac{3\pm\sqrt{3^2-4(1)(1)}}{2} \\ e^{iz}=\frac{3\pm\sqrt{5}}{2} \\ z=-i\ln(3\pm\sqrt 5)+i\ln(2)$$

Answer 1

You can solve this by transforming it to solving $\;\cos(it)=\cosh(t)=3/2,\;$ where $\;t:=e^x\;$ satisfies $\;x+1/x=3\;$ whose solutions are $\;x = (3\pm\sqrt{5})/2.$ Thus $\;z = i\log(x)\;$ is the solution to the equation $\;\cos(z) = 3/2.$ Your answer is correct, but note that $\;\log(x)\;$ is a multivalued function so keep that in mind. Also note that if $\;x_1 = (3+\sqrt5)/2,\; x_2 = (3-\sqrt5)/2\;$ then $\;x_1x_2=1\;$ so that $\;\log(x_1)=-\log(x_2).$

In general, the solutions to $\;\cos(z) = w\;$ are $\;z = \pm z_0 + 2\pi n\;$ where $\,z_0 := i\log((3+\sqrt5)/2)\,$ and $\;n\;$ is any integer.

Answer 2

More directly, $$\cos z = \cos(z-2\pi n)= \cosh (-i (z+2\pi n))=\frac32 $$ which leads to $-i (z-2\pi n)=\pm \cosh^{-1} \frac32$ and the solutions $$z= 2\pi n \pm i \cosh^{-1} \frac32 = 2\pi n \pm 2 i\ln \frac{\sqrt5+1}2 $$