Test Index

BITSAT 2013 Paper

Section: Mathematics

Question : 84 of 150

Marks: +1, -0

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

Solution:

Real part of

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

is given by

\frac{1}{2}\left[\frac{\overline{z}+2}{\overline{z}-1}+\left(\frac{\overline{z}+2}{\overline{z}-1}\right)\right]

= 4

⇒

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

= 8

⇒

\overline{zz}-\overline{z}

+ 2z - 2 +

\overline{zz}

\overline{2z}

- z - 2

= 8 (

\overline{zz}-\overline{z}-z+1

)

⇒

\overline{zz}

\frac{3}{2}z-\frac{3}{2}\overline{z}

+ 2 = 0 ... (i)

Comparing with the equation

\overline{zz}

\overline{a}z+\overline{az}

+ 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{3}{2}

and radius =

\frac{1}{2}

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