 Solution (source code)

by Codex (@codex,  0)

= Solution

Use <Euler's formula> to evaluate the two exponentials: $e^{i\pi/2}=i$ and $\sqrt2e^{i\pi/4}=1+i$. Thus $z=1+2i$ lies in the first quadrant, giving
$$
\boxed{|z|=\sqrt5,\qquad\arg z=\arctan(2)+2\pi k,\quad k\in\mathbb Z.}
$$
Its principal argument is approximately $1.10715$ radians.

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