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  • Show that the complex number z, satisfying the condition arg⁡((z-1)/(z+1))=π/4 lies on a circle.

Show that the complex number z, satisfying the condition arg\left ( \frac{z-1}{z+1} \right )=\frac{\pi}{4}  lies on a circle.

Answers (1)

Let z=x+iy  

arg\left ( \frac{z-1}{z+1} \right )=\frac{\pi}{4}

arg\left ( z-1 \right )-arg\left ( z+1 \right )=\frac{\pi}{4}

arg\left ( x+iy-1 \right )-arg\left ( x+iy+1 \right )=\frac{\pi}{4}

arg\left ( x-1+iy \right )-arg\left ( x+1+iy \right )=\frac{\pi}{4}

\tan^{-1}\frac{y}{x-1}-\tan^{-1}\frac{y}{x+1}=\frac{\pi}{4}

\tan^{-1}\frac{\frac{y}{x-1}-\frac{y}{x+1}}{1+\left (\frac{y}{x-1} \right )\left (\frac{y}{x+1} \right )}=\frac{\pi}{4}

\tan^{-1}\left(\frac{y\left ( x+1+-x+1 \right )}{x^2-1+y^2}\right) = \frac{\pi}{4}

\tan\frac{\pi}{4} = \frac{2y}{x^2+y^2-1}

x^2+y^2-1=2y

x^2+y^2-1-2y=0

 The equation obtained represents the equation of a circle

Posted by

infoexpert21

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