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AP EAMCET Engineering 2011 Solved Paper

Section: Mathematics

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Question : 118 of 160

Marks: +1, -0

The locus of the complex number

z

\arg \left(\frac{z-2}{z+2}\right)=\frac{\pi}{3}

is

a circle
a straight line
a parabola
an ellipse

Solution:

Let

z=x+iy

\therefore \arg \left(\frac{z-2}{z+2}\right)=\frac{\pi}{3}

\Rightarrow \arg (z-2)-\arg (z+2)=\frac{\pi}{3}

\Rightarrow \arg ((x-2)+iy)-\arg ((x+2)+iy)

=\frac{\pi}{3}

⇒

\tan^{-1}\left(\frac{y}{x-2}\right)-\tan^{-1}\left(\frac{y}{x+2}\right)=\frac{\pi}{3}

⇒

\tan^{-1}\left[\frac{\frac{y}{x-2}-\frac{y}{x+2}}{1+\frac{y}{x-2}\times\frac{y}{x+2}}\right]=\frac{\pi}{3}

⇒

\left[\frac{4y}{x^{2}-4+y^{2}}\right]=\tan\frac{\pi}{3}=\sqrt{3}

⇒

x^{2}-4+y^{2}=\frac{4y}{\sqrt{3}}

⇒

x^{2}+y^{2}-\frac{4y}{\sqrt{3}}-4=0

Hence, it represents a equation of circle.

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