Find $z$

Relevant wiki: Euler's Formula

$\begin{aligned} \cos z & = 3 & \small \color{#3D99F6} \text{By Euler's formula }e^{i\theta} = \cos \theta + i \sin \theta \\ \frac {e^{iz}+e^{-iz}}2 & = 3 \\ e^{iz}+e^{-iz} & = 6 & \small \color{#3D99F6} \text{Multiply both sides by }e^{iz} \\ e^{2iz}-6e^{iz} + 1 & = 0 & \small \color{#3D99F6} \text{Solve for }e^{iz} \end{aligned}$

$\begin{aligned} \implies e^{iz} & = \frac {6 \pm \sqrt{6^2-4}}2 & \small \color{#3D99F6} \text{Let }z = x + iy \\ e^{i(x+iy)}& = 3 \pm 2\sqrt 2 \\ {\color{#3D99F6}e^{-y}}\color{#D61F06} e^{ix} & = {\color{#3D99F6}(3 \pm 2\sqrt 2)}\color{#D61F06}(1) \\ & = {\color{#3D99F6}e^{3 \pm 2\sqrt 2}}\color{#D61F06}e^{2k\pi i} & \small \color{#D61F06} \text{where }k \in \mathbb Z \\ \color{#D61F06} x & \color{#D61F06}= 2k \pi \\ \color{#3D99F6}y & \color{#3D99F6}= - \ln (3 \pm 2\sqrt 2) \\ \implies z & = \boxed{2k \pi - i\ln (3 \pm 2\sqrt 2)} \end{aligned}$

From the Euler's theorem: cos(z) = 1/2*(e^iz+e^-iz)

cos(z) = 1/2*(e^iz+e^-iz) = 3 e^iz+e^-iz = 6 (e^iz)^2 - 6(e^iz) + 1 = 0 e^iz = 3+/-2√2 ln(3+/-2√2) = iz z = -iln(3+/-2√2) + 2kpi, (k e Z)