Verified Solution • 2025–26 Session moderate

QJEEM18Ma2dee: JEE Main 2018 Mathematics - Class 11 JEEM18Ma2dee Solution: The set of all R , for which w = is purely imaginary number,…

The set of all α\alpha ∈\in R, for which w = 1+(1−8α)z1−z{{1 + \left( {1 - 8\alpha } \right)z} \over {1 - z}} is purely imaginary number, for all z ∈\in C satisfying |z| = 1 and Re z ≠\ne 1, is :

Multiple Choice Options

  • A an empty set
  • B {0}
  • C {0,14,−14}\left\{ {0,{1 \over 4}, - {1 \over 4}} \right\}
  • D equal to R

Step-by-Step Verified Solution

As w = 1+(1−8α)z1−z{{1 + \left( {1 - 8\alpha } \right)z} \over {1 - z}} , w is purely imaginary

∴w\therefore w + wˉ\bar w = 0

⇒\Rightarrow 1+(1−8α)z1−z{{1 + \left( {1 - 8\alpha } \right)z} \over {1 - z}} + 1+(1−8α)zˉ1−zˉ{{1 + \left( {1 - 8\alpha } \right)\bar z} \over {1 - \bar z}} = 0

⇒\Rightarrow [1 + (1 - 8 α\alpha )][1 - zˉ\bar z ] + [1 + ( 1 - 8 α\alpha )][1 - z] = 0

⇒\Rightarrow 2 - (z + zˉ\bar z ) + (1 - 8 α\alpha )(z + zˉ\bar z ) - 2(1 - 8 α\alpha ) = 0

⇒\Rightarrow 2 - (z + zˉ\bar z ) + (z + zˉ\bar z ) - 8 α\alpha (z + zˉ\bar z ) - 2 + 16 α\alpha = 0

⇒\Rightarrow 16 α\alpha = 8 α\alpha (z + zˉ)\bar z)

z + zˉ\bar z = 2 or α\alpha = 0

but z + zˉ\bar z = 2 is not possible as Re(Z) ≠\ne 1

∴\therefore α\alpha = 0

∴\therefore α\alpha ∈\in {0}

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