Question Number 152091 by rexford last updated on 25/Aug/21
$${Given}\:{tbat}\:{Arg}\left({z}+\mathrm{1}\right)=\frac{\Pi}{\mathrm{6}}\:{and}\: \\ $$$${Arg}\left({z}−\mathrm{1}\right)=\frac{\mathrm{2}\Pi}{\mathrm{3}}.{Find}\:{z}. \\ $$$${please}\:{help}\:{me} \\ $$
Answered by Olaf_Thorendsen last updated on 25/Aug/21
$$\mathrm{Let}\:{z}\:=\:{x}+{iy} \\ $$$$\mathrm{Arg}\left({z}+\mathrm{1}\right)\:=\:\frac{\pi}{\mathrm{6}}\:\Rightarrow\:\frac{{y}}{{x}+\mathrm{1}}\:=\:\mathrm{tan}\frac{\pi}{\mathrm{6}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Arg}\left({z}−\mathrm{1}\right)\:=\:\frac{\mathrm{2}\pi}{\mathrm{3}}\:\Rightarrow\:\frac{{y}}{{x}−\mathrm{1}}\:=\:\mathrm{tan}\frac{\mathrm{2}\pi}{\mathrm{3}}\:=\:−\sqrt{\mathrm{3}}\:\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)\::\:{y}\:=\:\frac{{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\left(\mathrm{2}\right)\::\:\frac{{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:=−\sqrt{\mathrm{3}}\left({x}−\mathrm{1}\right) \\ $$$${x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{y}\:=\:\frac{\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${z}\:=\:{x}+{iy}\:=\:\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\:=\:{e}^{{i}\frac{\pi}{\mathrm{3}}} \\ $$
Commented by rexford last updated on 27/Aug/21
$${thank}\:{you} \\ $$