Question

If a complex number z satisfies $|z^2-1|=|z{|}^2+1$, then z lies on

• A. The real axis
• B. The imaginary axis
• C. y = x
• D. a circle

Answer 1

The correct option is B The imaginary axis

$|z^2-1|=|z{|}^2+1$ ...(1)

Let $z=x+iy$

$(x+iy{)}^2=x^2-y^2+2ixy$

from equation (1)

$|x^2-y^2+2ixy-1|=(x^2+y^2)+1$

$|(x^2-y^2-1)+2ixy|=x_1+y^2+1$

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

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

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

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

$\Rightarrow -4x^2-4x^2{y}^2=-4x^2{y}^2$

$\to x=0$ this is imaginary axis

$\therefore$ z lies on imaginary axis.