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

(A) straight line parallel to $y$-axis.

(C) circle of radius $\sqrt{2}$.

(D) straight line parallel to $x$-axis.

JEE Main · 2015 · 04 Apr 2015 · Offline · Complex Numbers

A complex number z is said to be unimodular if

∣ z ∣ = 1

. Suppose

z_{1}

and

z_{2}

are complex numbers such that

\frac{z _{1} - 2 z _{2}}{2 - z _{1} z ˉ _{2}}

is unimodular and

z_{2}

is not unimodular. Then the point

z_{1}

lies on

a

Answer: (B)

Unimodularity means

\frac{z _{1} - 2 z _{2}}{2 - z _{1} z ˉ _{2}} = 1

, so

∣ z_{1} - 2 z_{2} ∣ = ∣2 - z_{1} \overset{z}{ˉ}_{2} ∣

. Squaring and expanding the products with conjugates, the cross terms

z_{1} \overset{z}{ˉ}_{2} + \overset{z}{ˉ}_{1} z_{2}

cancel, leaving

∣ z_{1} ∣^{2} + 4∣ z_{2} ∣^{2} = 4 + ∣ z_{1} ∣^{2} ∣ z_{2} ∣^{2}

. Group this as

∣ z_{1} ∣^{2} (1 - ∣ z_{2} ∣^{2}) = 4 (1 - ∣ z_{2} ∣^{2})

. Because

z_{2}

is not unimodular,

1 - ∣ z_{2} ∣^{2} \neq = 0

and divides out, giving

∣ z_{1} ∣^{2} = 4

. So

z_{1}

lies on a circle of radius

2

JEE Main · 2015 · 04 Apr 2015 · Offline · Single correct option