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JEE Mains · Maths · STD 11 - 4.1 complex nubers

The set of all \(\alpha  \in R\), for which \(w = \frac{{1 + \left( {1 - 8\alpha } \right)z}}{{1 - z}}\) is a purely imaginary number, for all \(z \in C\) satisfying \(\left| z \right| = 1\) and \({\mathop{\rm Re}\nolimits} \,z \ne 1\),  is

  1. A \(\left\{ 0 \right\}\)
  2. B an empty set
  3. C \(\left\{ {0,\frac{1}{4}, - \frac{1}{4}} \right\}\)
  4. D equal to \(R\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\left\{ 0 \right\}\)

Step-by-step Solution

Detailed explanation

\(\because|z|=1\) and \(\text { Re } z \neq 1\) Suppose \(z=x+i y\) \( \Rightarrow x^{2}+y^{2}=1\) Now, \(w = \frac{{1 + (1 - 8\alpha )z}}{{1 - z}}\) \( \Rightarrow w = \frac{{1 + (1 - 8\alpha )(x + iy)}}{{1 - (x + iy)}}\)…
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