The number of solutions of the system of equations Re (z2) = 0, | z | = 2 is

Question

The number of solutions of the system of equations Re (z2) = 0, | z | = 2 is (A) 4 (B) 3 (C) 2 (D) 1

Tardigrade Admin · Accepted Answer

Let z = x + iy. Then z2 =x2-y2 + 2 i xy Since Re (z)2 = 0 ∴ x2-y2=0 ∴ y=± x Now |z|=2 ⇒ x2+y2=4 ⇒ x2=2 ⇒ x =±√2 Thus the solutions are (√2, √2), (√2, -√2), (-√2, -√2), (-√2, √2)

Q. The number of solutions of the system of equations $R e (z^{2}) = 0, ∣ z ∣ = 2$ is

$4$

$3$

$2$

$1$

Let $z = x + i y$ . Then $z^{2} = x^{2} - y^{2} + 2 i x y$
Since $R e (z)^{2} = 0$
$∴ x^{2} - y^{2} = 0$
$∴ y = \pm x$
Now $∣ z ∣ = 2$
$\Rightarrow x^{2} + y^{2} = 4$
$\Rightarrow x^{2} = 2$
$\Rightarrow x = \pm 2$
Thus the solutions are
$(2, 2), (2, - 2), (- 2, - 2)$ ,
$(- 2, 2)$

Q. The number of solutions of the system of equations Re(z2)=0,∣z∣=2 is

4

3

2

1

Let z=x+iy. Then z2=x2−y2+2ixy Since Re(z)2=0 ∴x2−y2=0 ∴y=±x Now ∣z∣=2 ⇒x2+y2=4 ⇒x2=2 ⇒x=±2​ Thus the solutions are (2​,2​),(2​,−2​),(−2​,−2​), (−2​,2​)

Q. The number of solutions of the system of equations $R e (z^{2}) = 0, ∣ z ∣ = 2$ is

$4$

$3$

$2$

$1$

Let $z = x + i y$ . Then $z^{2} = x^{2} - y^{2} + 2 i x y$
Since $R e (z)^{2} = 0$
$∴ x^{2} - y^{2} = 0$
$∴ y = \pm x$
Now $∣ z ∣ = 2$
$\Rightarrow x^{2} + y^{2} = 4$
$\Rightarrow x^{2} = 2$
$\Rightarrow x = \pm 2$
Thus the solutions are
$(2, 2), (2, - 2), (- 2, - 2)$ ,
$(- 2, 2)$