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  • Solve! - Complex numbers and quadratic equations - JEE Main-4

If \left | z^{2}-1 \right |= \left | z \right |^{2}+1, then z lies on

  • Option 1)

    a circle

  • Option 2)

    the imaginary axis

  • Option 3)

    the real axis

  • Option 4)

    an ellipse.

 

Answers (2)

best_answer

As we learnt in

Definition of Modulus of z(Complex Number) -

\left | z \right |=\sqrt{a^{2}+b^{2}} is the distance of z from origin in Argand plane

- wherein

Real part of z = Re (z) = a & Imaginary part of z = Im (z) = b

 

 |z^{2}-1|=|z|^{2}+1

\therefore\ \; \left | (x+iy)^{2}-1\right |=x^{2}+y^{2}+1

\therefore\ \; \left | (x^{2}-y^{2}-1)+i 2xy\right |=(x^{2}+y^{2}+1)

    (x^{2}-y^{2}-1)^{2}+4x^{2}y^{2}=(x^{2}+y^{2}+1)^{2}

\therefore\ \; 4x^{2}y^{2}=(x^{2}+y^{2}+1)^{2}-(x^{2}-y^{2}-1)^{2}

    =(x^{2}+y^{2}+x+x^{2}-y^{2}-1)(x^{2}+y^{2}+1-x^{2}+y^{2}+1)=2x^{2}+2(y^{2}+1)

4x^{2}y^{2}=4x^{2}y^{2}+4x^{2}

\therefore\ \; x =0

Correct option is 2.

 


Option 1)

a circle

This is an incorrect option.

Option 2)

the imaginary axis

This is the correct option.

Option 3)

the real axis

This is an incorrect option.

Option 4)

an ellipse.

This is an incorrect option.

Posted by

divya.saini

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