Question Number 146235 by mathdanisur last updated on 12/Jul/21
$${if}\:\:\:{arg}\:{z}_{\mathrm{1}} \:=\:\frac{\mathrm{4}\pi}{\mathrm{3}}\:\:\:{and}\:\:\:{arg}\:{z}_{\mathrm{1}} ^{\mathrm{2}} \:\centerdot\:{z}\:=\:\frac{\mathrm{7}\pi}{\mathrm{6}} \\ $$$${find}\:\:\:{arg}\:{z}\:=\:? \\ $$
Answered by gsk2684 last updated on 12/Jul/21
$${arg}\:{z}_{\mathrm{1}} ^{\mathrm{2}} {z}=\mathrm{2}{arg}\:{z}_{\mathrm{1}} +{arg}\:{z}+\mathrm{2}{k}\Pi \\ $$
Answered by puissant last updated on 12/Jul/21
$$\mathrm{arg}\:\mathrm{z}_{\mathrm{1}} ^{\mathrm{2}} .\mathrm{z}\:=\:\mathrm{arg}\:\mathrm{z}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{arg}\:\mathrm{z}\:=\:\frac{\mathrm{7}\pi}{\mathrm{6}} \\ $$$$\Rightarrow\mathrm{arg}\:\mathrm{z}=\frac{\mathrm{7}\pi}{\mathrm{6}}−\mathrm{2arg}\:\mathrm{z}_{\mathrm{1}} =\frac{\mathrm{7}\pi}{\mathrm{6}}−\frac{\mathrm{8}\pi}{\mathrm{3}} \\ $$$$\Rightarrow\:\mathrm{arg}\:\mathrm{z}=−\frac{\mathrm{3}\pi}{\mathrm{2}}.. \\ $$
Commented by mathdanisur last updated on 12/Jul/21
$${cool}\:{Ser},\:{thanks}\: \\ $$