Let z=x+ iy be a non-zero complex number such that z 2= i | z |2, where i =√-1, then z lies on the :

Question

Let z=x+ iy be a non-zero complex number such that z 2= i | z |2, where i =√-1, then z lies on the : (A) imaginary axis (B) real axis (C) line, y=x (D) line, y=-x

Tardigrade Admin · Accepted Answer

z=x+i y z2=i|z|2 (x+i y)2=i(x2+y2) (x2-y2)-i(x2+y2-2 x y)=0 (x-y)(x+y)-i(x-y)2=0 ⇒ x=y z lies on y=x

Q. Let $z = x +$ iy be a non-zero complex number such that $z^{2} = i ∣ z ∣^{2}$ , where $i = - 1$ , then $z$ lies on the :

imaginary axis

real axis

line, $y = x$

line, $y = - x$

$z = x + i y$
$z^{2} = i ∣ z ∣^{2}$
$(x + i y)^{2} = i (x^{2} + y^{2})$
$(x^{2} - y^{2}) - i (x^{2} + y^{2} - 2 x y) = 0$
$(x - y) (x + y) - i (x - y)^{2} = 0$
$\Rightarrow x = y$
$z$ lies on $y = x$

Q. Let z=x+ iy be a non-zero complex number such that z2=i∣z∣2, where i=−1​, then z lies on the :

imaginary axis

real axis

line, y=x

line, y=−x

z=x+iy z2=i∣z∣2 (x+iy)2=i(x2+y2) (x2−y2)−i(x2+y2−2xy)=0 (x−y)(x+y)−i(x−y)2=0 ⇒x=y z lies on y=x

Q. Let $z = x +$ iy be a non-zero complex number such that $z^{2} = i ∣ z ∣^{2}$ , where $i = - 1$ , then $z$ lies on the :

line, $y = x$

line, $y = - x$

$z = x + i y$
$z^{2} = i ∣ z ∣^{2}$
$(x + i y)^{2} = i (x^{2} + y^{2})$
$(x^{2} - y^{2}) - i (x^{2} + y^{2} - 2 x y) = 0$
$(x - y) (x + y) - i (x - y)^{2} = 0$
$\Rightarrow x = y$
$z$ lies on $y = x$