How can I solve this complex Log equation?

Question

|Log z| = Re (Log z)

z is complex

Log z denotes the principle value of log z. -pi< Arg z< pi. Step by step solution please. I'll need to sketch it as well.

Answer 1

$|\log z| = \Re (\log z) $

Let $z = re^{it} $. $\log z =\log(r)+it $, so $|\log z| =\sqrt{\log^2(r)+t^2} $ and $\Re (\log z) =\log r $.

If $\log r \ge 0 $, then $t = 0$ so $z$ must be real.

If $\log r \lt 0 $, letting $s = -\log r$, and choosing the negative square root, $s =\sqrt{s^2+t^2} $, and this implies that $t = 0$.

Therefore, the only solutions are real $z$.