JEE MAIN - Mathematics (2018 - 15th April Morning Slot - No. 5)
The set of all $$\alpha $$ $$ \in $$ R, for which w = $${{1 + \left( {1 - 8\alpha } \right)z} \over {1 - z}}$$ is purely imaginary number, for all z $$ \in $$ C satisfying |z| = 1 and Re z $$ \ne $$ 1, is :
an empty set
{0}
$$\left\{ {0,{1 \over 4}, - {1 \over 4}} \right\}$$
equal to R
Explanation
As w = $${{1 + \left( {1 - 8\alpha } \right)z} \over {1 - z}}$$, w is purely imaginary
$$ \therefore w$$ + $$\bar w$$ = 0
$$ \Rightarrow $$ $${{1 + \left( {1 - 8\alpha } \right)z} \over {1 - z}}$$ + $${{1 + \left( {1 - 8\alpha } \right)\bar z} \over {1 - \bar z}}$$ = 0
$$ \Rightarrow $$ [1 + (1 - 8$$\alpha $$)][1 - $$\bar z$$] + [1 + ( 1 - 8$$\alpha $$)][1 - z] = 0
$$ \Rightarrow $$ 2 - (z + $$\bar z$$) + (1 - 8$$\alpha $$)(z + $$\bar z$$) - 2(1 - 8$$\alpha $$) = 0
$$ \Rightarrow $$ 2 - (z + $$\bar z$$) + (z + $$\bar z$$) - 8$$\alpha $$(z + $$\bar z$$) - 2 + 16$$\alpha $$ = 0
$$ \Rightarrow $$ 16$$\alpha $$ = 8$$\alpha $$(z + $$\bar z)$$
z + $$\bar z$$ = 2 or $$\alpha $$ = 0
but z + $$\bar z$$ = 2 is not possible as Re(Z) $$ \ne $$ 1
$$ \therefore $$ $$\alpha $$ = 0
$$ \therefore $$ $$\alpha $$ $$ \in $$ {0}
$$ \therefore w$$ + $$\bar w$$ = 0
$$ \Rightarrow $$ $${{1 + \left( {1 - 8\alpha } \right)z} \over {1 - z}}$$ + $${{1 + \left( {1 - 8\alpha } \right)\bar z} \over {1 - \bar z}}$$ = 0
$$ \Rightarrow $$ [1 + (1 - 8$$\alpha $$)][1 - $$\bar z$$] + [1 + ( 1 - 8$$\alpha $$)][1 - z] = 0
$$ \Rightarrow $$ 2 - (z + $$\bar z$$) + (1 - 8$$\alpha $$)(z + $$\bar z$$) - 2(1 - 8$$\alpha $$) = 0
$$ \Rightarrow $$ 2 - (z + $$\bar z$$) + (z + $$\bar z$$) - 8$$\alpha $$(z + $$\bar z$$) - 2 + 16$$\alpha $$ = 0
$$ \Rightarrow $$ 16$$\alpha $$ = 8$$\alpha $$(z + $$\bar z)$$
z + $$\bar z$$ = 2 or $$\alpha $$ = 0
but z + $$\bar z$$ = 2 is not possible as Re(Z) $$ \ne $$ 1
$$ \therefore $$ $$\alpha $$ = 0
$$ \therefore $$ $$\alpha $$ $$ \in $$ {0}
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