Question Number 70145 by Scientist0000001 last updated on 01/Oct/19
$${prove}\:{that}\:;\:{arg}\left(\boldsymbol{{z}}\mathrm{1}\boldsymbol{{z}}\mathrm{2}\right)={arg}\left({z}\mathrm{1}\right)+{arg}\left({z}\mathrm{2}\right). \\ $$$${arg}\left({z}\mathrm{1}/{z}\mathrm{2}\right)={arg}\left({z}\mathrm{1}\right)−{arg}\left({z}\mathrm{2}\right). \\ $$
Answered by MJS last updated on 01/Oct/19
$${z}_{\mathrm{1}} ={r}_{\mathrm{1}} \mathrm{e}^{\mathrm{i}\theta_{\mathrm{1}} } ;\:{z}_{\mathrm{2}} ={r}_{\mathrm{2}} \mathrm{e}^{\mathrm{i}\theta_{\mathrm{2}} } \\ $$$${z}_{\mathrm{1}} {z}_{\mathrm{2}} ={r}_{\mathrm{1}} {r}_{\mathrm{2}} \mathrm{e}^{\mathrm{i}\left(\theta_{\mathrm{1}} +\theta_{\mathrm{2}} \right)} \\ $$$$\frac{{z}_{\mathrm{1}} }{{z}_{\mathrm{2}} }=\frac{{r}_{\mathrm{1}} }{{r}_{\mathrm{2}} }\mathrm{e}^{\mathrm{i}\left(\theta_{\mathrm{1}} −\theta_{\mathrm{2}} \right)} \\ $$$$\mathrm{arg}\:\left({r}\mathrm{e}^{\mathrm{i}\theta} \right)=\theta \\ $$