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

$$GiventbatArg(z+1)=\frac{\mathrm{\Pi }}{6}and$$$$Arg(z-1)=\frac{2\mathrm{\Pi }}{3}.Findz.$$$$pleasehelpme$$

Answered by Olaf_Thorendsen last updated on 25/Aug/21

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

Commented by rexford last updated on 27/Aug/21

$$thankyou$$

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