If the real part of z +2 z -1 is 4, z ≠ 1, then the locus of the point representing z in the complex plane is

Correct answer: D. a circle with radius (1)/(2) Real part of z +2 z -1 is given by array l (1)/(2)[ z +2 z -1 +( z +2 z -1 )]=4 z +2 z -1 +(z+2)/(z-1)=8 z z - z +2 z-2+z z +2 z -z-2 =8(z z - z -z+1) z z -(3)/(2)z-(3)/(2) z +2=0 array Comparing with the equation z z + a z+a z +b=0, we get a=-(3)/(2) and b=2 . Thus, the locus of z given by the equation (i) is a circle with centre =(1)/(2)

BITSAT 2013

Exam: BITSAT
Year: 2013
Subject: Mathematics
Chapter: Complex Number

BITSAT 2013 · Mathematics · Complex Number

If the real part of z +2 z -1 is 4, z ≠ 1, then the locus of the...

If the real part of

\frac{\bar{z}+2}{\bar{z}-1}

4, z \neq 1,

then the locus of the point representing

z

in the complex plane is

Options

Aa straight line parallel to $x$ -axis
Ba straight line equally inclined to axes
Ca circle with radius 2
Da circle with radius $\frac{1}{2}$

Correct answer

D. a circle with radius $\frac{1}{2}$

Step-by-step solution

Real part of

\frac{\bar{z}+2}{\bar{z}-1}

is given by

\begin{array}{l}\frac{1}{2}\left[\frac{\bar{z}+2}{\bar{z}-1}+\left(\frac{\bar{z}+2}{\bar{z}-1}\right)\right]=4 \\ \Rightarrow \frac{\bar{z}+2}{\bar{z}-1}+\frac{z+2}{z-1}=8 \\ \Rightarrow z \bar{z}-\bar{z}+2 z-2+z \bar{z}+2 \bar{z}-z-2 \\ =8(z \bar{z}-\bar{z}-z+1) \\ \Rightarrow z \bar{z}-\frac{3}{2}z-\frac{3}{2} \bar{z}+2=0 \end{array}

Comparing with the equation

z \bar{z}+\bar{a} z+a \bar{z}+b=0,

we get

a=-\frac{3}{2}

and

b=2

. Thus, the locus of

z

given by the equation (i) is a circle with centre $=\frac{1}{2}

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