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JEE Advanced 2007Paper 2
Exam
JEE Advanced
Year
2007
Paper
Paper 2
Subject
Mathematics
Chapter
Complex Number

JEE Advanced 2007 · Paper 2 · Mathematics · Complex Number

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

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

  1. Aa line not passing through the origin
  2. B∣z∣=2|z|=\sqrt{2}
  3. Cthe X-axis
  4. Dthe Y-axis

Correct answer

D. the Y-axis

Step-by-step solution

Let z=cos⁡θ+isin⁡θz=\cos \theta+i \sin \thetaz1−z2=cos⁡θ+isin⁡θ1−(cos⁡2θ+isin⁡2θ)=cos⁡θ+isin⁡θ2sin⁡2θ−2isin⁡θcos⁡θ=cos⁡θ+isin⁡θ−2isin⁡θ(cos⁡θ+isin⁡θ)=i2sin⁡θ\begin{aligned}\frac{z}{1-z^2}& =\frac{\cos \theta+i \sin \theta}{1-(\cos 2 \theta+i \sin 2 \theta)}\\ & =\frac{\cos \theta+i \sin \theta}{2 \sin ^2 \theta-2 i \sin \theta \cos \theta}\\ & =\frac{\cos \theta+i \sin \theta}{-2 i \sin \theta(\cos \theta+i \sin \theta)}\\ & =\frac{i}{2 \sin \theta} \end{aligned}$Hence,Hence,\frac{z}{1-z^2}liesontheimaginaryaxis,i.e.lies on the imaginary axis, i.e.x=0.ALITERLet. ALITER LetE=\frac{z}{1-z^2}=\frac{z}{z \bar{z}-z^2}=\frac{1}{\bar{z}-z}$ which is imaginary.
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