log of $e^z , z \in \mathbb{C}$

$= \log_2 \left( \ln \left| e^{\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}} \right| \right)$ $= \log_2 \left( \ln e^{\frac{\sqrt{2}}{2}} \right)$ $= \log_2 \sqrt{2} - 1 = \boxed{-0.5}$

To prove the second line, formally define $e^z = \displaystyle \sum_{0 \leqslant k } \dfrac{z^k}{k!}$ (it's CONVENIENTLY written exponentially, but the definition itself is not yet exponential until you prove that the additive and multiplicative rules apply ), prove that $e^{a + b} = e^a e^b$ , and obtain that $e^{\overline{z}} = \overline{e^z}$ . This obtains $|e^{iy}|^2 = 1$ , hence $|e^{x + iy} | = e^x$ . Here, $z, a, b \in \mathbb{C}, x, y \in \mathbb{R}$ .

Similar solution with @Hobart Pao 's, using Euler's formula as follows:

$\begin{aligned} x & = \log_2\left(\ln \left|e^{\color{#3D99F6}e^{\frac \pi 4 i}} \right| \right) & \small \color{#3D99F6} \text{By Euler's formula: }e^{\theta i} = \cos \theta + i \sin \theta \\ & = \log_2\left( \ln \left| e^{\color{#3D99F6}\frac 1{\sqrt 2}+ \frac i{\sqrt 2}} \right| \right) \\ & = \log_2\left( \ln \left| e^{\frac 1{\sqrt 2}} e^{\frac i{\sqrt 2}} \right| \right) \\ & = \log_2\left( \ln \left(e^{\frac 1{\sqrt 2}} \color{#3D99F6}\left|e^{\frac i{\sqrt 2}} \right|\right) \right) & \small \color{#3D99F6} \text{Note that } \left| e^{\theta i} \right| = |\cos \theta + i \sin \theta | = 1 \\ & = \log_2\left( \ln \left(e^{\frac 1{\sqrt 2}} \color{#3D99F6}\cdot 1 \right) \right) \\ & = \log_2\left( \frac 1{\sqrt 2} \right) \\ & = \log_2\left( 2^{-\frac 12} \right) \\ & = - \frac 12 = \boxed{-0.5} \end{aligned}$