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  • If the real part of (z ̅+2)/(z ̅-1) is 4, then show that the locus of the point representing z in the complex plane is a circle.

If the real part of \frac{(\bar{z}+2)}{(\bar{z}-1)}  is 4, then show that the locus of the point representing z in the complex plane is a circle.

Answers (1)

Let \: \: z=x+iy \: \:

\frac{\bar{z}+2}{\bar{z}-1}= \frac{x-iy+2}{x-iy-1}

=\frac{\left [ \left ( x+2 \right )-iy \right ]\left [ \left ( x-1 \right )+iy \right ]}{\left [ \left ( x-1 \right )-iy \right ]\left [ \left ( x-1 \right )+iy \right ]}

=\frac{(x-1)(x+2)+y^2+i[(x+2)y-(x-1)y]}{(x-1)^2+y^2}

real \: \: part=4 \Rightarrow \frac{(x-1)(x+2)+y^2}{(x-1)^2+y^2}=4

x^2+x-2+y^2=4(x^2-2x+1+y^2 )

3x^2+3y^2-9x+6=0

The equation obtained is that of a circle. Hence, locus of z is a circle.

 

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infoexpert21

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