Q

# 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

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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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