JEE-Mains 2015 (2/30)

Algebra Level 5

A complex number $z$ is said to be unimodular if $|z|=1.$ Suppose $z_1$ and $z_2$ are complex numbers such that $\dfrac{z_1-2z_2}{2-z_1\bar{z_2}}$ is unimodular and $z_2$ is not unimodular. Then the point $z_1$ lies on a :

straight line parallel to

x

-axis circle of radius 2 circle of radius

\sqrt{2}

straight line parallel to

y

-axis

1 solution

Anandhu Raj
Jun 15, 2015

Given $\left| \frac { z_{ 1 }-2z_{ 2 } }{ 2-z_{ 1 }\bar { z_{ 2 } } } \right| =1\quad \quad$

$\Rightarrow \left| z_{ 1 }-2z_{ 2 } \right| =\quad \left| 2-z_{ 1 }\bar { z_{ 2 } } \right|$

$\Rightarrow (z_{ 1 }-2z_{ 2 })(\bar { z_{ 1 } } -2\bar { z_{ 2 } } )=(2-z_{ 1 }\bar { z_{ 2 } } )(2-\bar { z_{ 1 } } z_{ 2 })$

Solving, ${ \left| z_{ 1 } \right| }^{ 2 }+4{ \left| z_{ 2 } \right| }^{ 2 }=4+{ \left| z_{ 2 } \right| }^{ 2 }{ \left| z_{ 1 } \right| }^{ 2 }$

$\Rightarrow ({ \left| z_{ 1 } \right| }^{ 2 }-4)(1-{ \left| z_{ 2 } \right| }^{ 2 })=0$

but, $(1-{ \left| z_{ 2 } \right| }^{ 2 })\neq 0$ because given that $z_{ 2 }$ not unimodular.

$\Rightarrow ({ \left| z_{ 1 } \right| }^{ 2 }-4)=0$

${ \Longrightarrow \left| z_{ 1 } \right| }=2$

$Which\quad represents\quad a\quad circle\quad of\quad radius\quad 2$

JEE-Mains 2015 (2/30)

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