If |z| = 1 and z ≠ ± 1, then all the values of z/(1-z^2) lie on

a line not passing through the origin

If |z| = 1 and z ≠ ± 1, then all the values of z/(1-z^2)… — JEE Main 2007 Mathematics PYQ

AIEEE 2007

|z| = 1

and

z \neq \pm 1

, then all the values of

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

lie on

A.
a line not passing through the origin
B.
$|z| = \sqrt{2}$
C.
the $X$ -axis
D.
the $Y$ -axis

Show answer and solution

Correct option: D

Solution

Since |z| = 1, write z = e^(iθ) = cosθ + i·sinθ.

Then z - z⁻¹ = 2i·sinθ, and 1 - z² = z(z⁻¹ - z) = -z(z - z⁻¹) = -z·(2i·sinθ).

So z/(1 - z²) = z/(-z·2i·sinθ) = -1/(2i·sinθ) = i/(2sinθ).

This is purely imaginary (since sinθ ≠ 0 for z ≠ ±1), meaning all values lie on the Y-axis.

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JEE Main 2007 Mathematics question on Complex Numbers and Quadratic Equations