JEE Challenger
More from Complex Numbers and Quadratic Equations
MathMedAdv 2007 • Paper 2

Locus of z / (1 - z^2) for Unit Modulus Complex Numbers

If ∣z∣=1|z| = 1 and z≠±1z \neq \pm 1, then all the values of z1−z2\frac{z}{1 - z^2} lie on

Options

A

a line not passing through the origin

B

∣z∣=2|z| = \sqrt{2}

C

the xx-axis

D

the yy-axis

Correct

Topics & Concepts

Complex Numbers and Quadratic EquationsComplex NumbersLocusPurely Imaginary Numbers

Step-by-Step Solution

Let w=z1−z2w = \frac{z}{1 - z^2}.

Since ∣z∣=1|z| = 1, we have zzˉ=∣z∣2=1  ⟹  zˉ=1zz \bar{z} = |z|^2 = 1 \implies \bar{z} = \frac{1}{z}.

Taking the complex conjugate of ww:
wˉ=zˉ1−zˉ2=1z1−1z2=1zz2−1z2=zz2−1=−z1−z2=−w\bar{w} = \frac{\bar{z}}{1 - \bar{z}^2} = \frac{\frac{1}{z}}{1 - \frac{1}{z^2}} = \frac{\frac{1}{z}}{\frac{z^2 - 1}{z^2}} = \frac{z}{z^2 - 1} = -\frac{z}{1 - z^2} = -w

Since wˉ=−w\bar{w} = -w, the real part of ww is zero:
Re(w)=w+wˉ2=0\text{Re}(w) = \frac{w + \bar{w}}{2} = 0

Because ww is purely imaginary, all values of w=z1−z2w = \frac{z}{1 - z^2} lie on the imaginary axis, which corresponds to the yy-axis.

Correct Answer: D

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