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  • Solve this problem - Ifis a purely imaginary number and, then a value of is : - Complex numbers and quadratic equations - JEE Main

If \frac{z-\alpha }{z+\alpha }\left ( \alpha \in {R} \right ) is a purely imaginary number and \left | z \right |=2, then a value of \alpha is : 

  • Option 1)

    \frac{1}{2}

     

     

     

  • Option 2)

    \sqrt{2}

  • Option 3)

    1

  • Option 4)

    2

Answers (1)

best_answer

 

Purely Imaginary Complex Number -

z=x+iy, \boldsymbol{x=0}, y\epsilon R

& i2=-1

- wherein

Real part of z = Re (z) = x & Imaginary part of z = Im (z) = y

 

 

Properties of Conjugate of Complex Number -

\bar{\bar{z}}=z

- wherein

\bar{z} denotes conjugate of z

\frac{z-\alpha }{z+\alpha }=-(\frac{\bar{z}-\alpha }{\bar{z}+\alpha })

z\bar{z}+z\alpha -\alpha \bar{z}-\alpha ^{2}=-z\bar{z}+\alpha \bar{z}-\alpha \bar{z}+\alpha ^{2}

\alpha ^{2}=\left | z \right |^{2}

\alpha =\pm z


Option 1)

\frac{1}{2}

 

 

 

Option 2)

\sqrt{2}

Option 3)

1

Option 4)

2
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