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Question Number 152091 by rexford last updated on 25/Aug/21

Given tbat Arg(z+1)=(Π/6) and Arg(z−1)=((2Π)/3).Find z. please help me

$${Given}\:{tbat}\:{Arg}\left({z}+\mathrm{1}\right)=\frac{\Pi}{\mathrm{6}}\:{and}\: \\ $$$${Arg}\left({z}−\mathrm{1}\right)=\frac{\mathrm{2}\Pi}{\mathrm{3}}.{Find}\:{z}. \\ $$$${please}\:{help}\:{me} \\ $$

Answered by Olaf_Thorendsen last updated on 25/Aug/21

Let z = x+iy Arg(z+1) = (π/6) ⇒ (y/(x+1)) = tan(π/6) = (1/( (√3))) (1) Arg(z−1) = ((2π)/3) ⇒ (y/(x−1)) = tan((2π)/3) = −(√3) (2) (1) : y = ((x+1)/( (√3))) (2) : ((x+1)/( (√3))) =−(√3)(x−1) x = (1/2) ⇒y = (((1/2)+1)/( (√3))) = ((√3)/2) z = x+iy = (1/2)+((√3)/2)i = e^(i(π/3))

$$\mathrm{Let}\:{z}\:=\:{x}+{iy} \\ $$$$\mathrm{Arg}\left({z}+\mathrm{1}\right)\:=\:\frac{\pi}{\mathrm{6}}\:\Rightarrow\:\frac{{y}}{{x}+\mathrm{1}}\:=\:\mathrm{tan}\frac{\pi}{\mathrm{6}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Arg}\left({z}−\mathrm{1}\right)\:=\:\frac{\mathrm{2}\pi}{\mathrm{3}}\:\Rightarrow\:\frac{{y}}{{x}−\mathrm{1}}\:=\:\mathrm{tan}\frac{\mathrm{2}\pi}{\mathrm{3}}\:=\:−\sqrt{\mathrm{3}}\:\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)\::\:{y}\:=\:\frac{{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\left(\mathrm{2}\right)\::\:\frac{{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:=−\sqrt{\mathrm{3}}\left({x}−\mathrm{1}\right) \\ $$$${x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{y}\:=\:\frac{\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${z}\:=\:{x}+{iy}\:=\:\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\:=\:{e}^{{i}\frac{\pi}{\mathrm{3}}} \\ $$

Commented by rexford last updated on 27/Aug/21

thank you

$${thank}\:{you} \\ $$

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