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Question Number 39708 by NECx last updated on 10/Jul/18
Find the complex number z if  arg(z+1)=(π/3)  arg(z−1)=((4π)/3)
$${Find}\:{the}\:{complex}\:{number}\:{z}\:{if} \\ $$$${arg}\left({z}+\mathrm{1}\right)=\frac{\pi}{\mathrm{3}} \\ $$$${arg}\left({z}−\mathrm{1}\right)=\frac{\mathrm{4}\pi}{\mathrm{3}} \\ $$
Answered by MrW3 last updated on 10/Jul/18
let z=a+bi  (1)  (b/(a+1))=tan (π/3)=(√3)  ⇒b=(√3)(a+1)  ⇒z=a+(a+1)(√3)i, a≥−1  (2)  (b/(a−1))=tan (π/3)=(√3)  ⇒b=(a−1)(√3)  ⇒z=a+(a−1)(√3)i, a≤1
$${let}\:{z}={a}+{bi} \\ $$$$\left(\mathrm{1}\right) \\ $$$$\frac{{b}}{{a}+\mathrm{1}}=\mathrm{tan}\:\frac{\pi}{\mathrm{3}}=\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{b}=\sqrt{\mathrm{3}}\left({a}+\mathrm{1}\right) \\ $$$$\Rightarrow{z}={a}+\left({a}+\mathrm{1}\right)\sqrt{\mathrm{3}}{i},\:{a}\geqslant−\mathrm{1} \\ $$$$\left(\mathrm{2}\right) \\ $$$$\frac{{b}}{{a}−\mathrm{1}}=\mathrm{tan}\:\frac{\pi}{\mathrm{3}}=\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{b}=\left({a}−\mathrm{1}\right)\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{z}={a}+\left({a}−\mathrm{1}\right)\sqrt{\mathrm{3}}{i},\:{a}\leqslant\mathrm{1} \\ $$
Commented by MrW3 last updated on 10/Jul/18
Commented by NECx last updated on 10/Jul/18
Thank you so much sir
$${Thank}\:{you}\:{so}\:{much}\:{sir} \\ $$

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