JEE Advanced · 2007 · 08 Apr 2007 · Paper 2 · Complex Numbers

∣ z ∣ = 1

and

z \neq = \pm 1

, then all the values of

\frac{z}{1 - z ^{2}}

lie on

Answer: (D)

Solution

Let

z = cos θ + i sin θ

where

∣ z ∣ = 1

. Then

z^{2} = cos 2 θ + i sin 2 θ

. We have:

\frac{z}{1 - z ^{2}} = \frac{cos θ + i sin θ}{1 - ( cos 2 θ + i sin 2 θ )} = \frac{cos θ + i sin θ}{2 sin ^{2} θ - 2 i sin θ cos θ}

Factoring the denominator:

2 sin^{2} θ - 2 i sin θ cos θ = - 2 i sin θ (cos θ + i sin θ)

Therefore:

\frac{z}{1 - z ^{2}} = \frac{cos θ + i sin θ}{- 2 i sin θ ( cos θ + i sin θ )} = \frac{1}{- 2 i sin θ} = \frac{i}{2 sin θ}

This is purely imaginary, so all values lie on the

y

-axis.

JEE Advanced · 2007 · 08 Apr 2007 · Paper 2 · Single correct option