Question

Show that $-3-4i=5i^{i(\pi +{tan}^{-1}4/3)}$

Answer 1

We know that if $z=(r,0)$, then $-z=r,-(\pi -\theta )$

$z=3+4i=5,0$ where $\theta ={\tan }^{-1}\left(4/3\right).$

$-z=-3-4i=5,-(\pi -\theta )$

Now adding $2\pi ,-(\pi -\theta )$ becomes

$\pi +\theta =\pi +{\tan }^{-1}\left(4/3\right).$