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Help me please, - Complex numbers and quadratic equations - JEE Main-14

The set of all  \alpha \epsilon \mathbf{R},   for which  w=\frac{1+\left ( 1-8\alpha \right )z}{1-z}  is a purely imaginary number, for all z\: \epsilon\: \mathbf{C} satisfying \left | z \right |=1   and Re \: \: z\neq 1 , is :

  • Option 1)

    an empty set

  • Option 2)

    \left \{ 0 \right \}

  • Option 3)

    \left \{ 0,\frac{1}{4},-\frac{1}{4} \right \}

  • Option 4)

    equal to R

 
Answers (1)
110 Views
H Himanshu

As we learned 

For \alpha =0\: \: \: \: \: \omega= \frac{1+2}{1-2}\: \: \: \: \omega =\frac{1+\cos \theta +i\sin \theta }{1-\cos \theta -i\sin \theta }

=\frac{\left ( 1+\cos \theta +i\sin \theta \right )\left ( 1-\cos \theta +i\sin \theta \right )}{\left ( 1-\cos \theta \right )^{2}+\sin ^{2}\theta }

=\frac{\sin ^{2}\theta -\sin ^{2}\theta +2i\sin \theta }{\left ( 1-\cos \theta \right )^{2}+\sin ^{2}\theta }\; \; \; \; \; purely\: imaginary

For \alpha =\frac{1}{4}    omega=\frac{1+\left ( 1-2 \right )2}{1-2}= \frac{1-2}{1-2}= 1        purely \: real

so {0} is the only solution 


Option 1)

an empty set

Option 2)

\left \{ 0 \right \}

Option 3)

\left \{ 0,\frac{1}{4},-\frac{1}{4} \right \}

Option 4)

equal to R

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