2.1.6 So­lu­tion math­c­plx-f

Ques­tion:

Ver­ify that $e^{-2{\rm i}}$ is still the com­plex con­ju­gate of $e^{2{\rm i}}$ af­ter both are rewrit­ten us­ing the Euler for­mula.

An­swer:

$e^{-2{\rm i}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\cos(2)-{\rm i}\sin(2)$ and $e^{2{\rm i}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\cos(2)+{\rm i}\sin(2)$.