Let $a \neq b$ be two non-zero real numbers. Then the number of elements in the set $X=\{z \in \mathbb{C}: \operatorname{Re}(az^{2}+bz)=a \text{ and } \operatorname{Re}(bz^{2}+az)=b\}$ is equal to

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Complex Numbers

Let

a \neq = b

be two non-zero real numbers. Then the number of elements in the set

X = {z \in C : Re (a z^{2} + b z) = a and Re (b z^{2} + a z) = b}

is equal to

Answer: (C)

Write

z = x + i y

, so that

Re (z^{2}) = x^{2} - y^{2}

and

Re (z) = x

. With

s = x^{2} - y^{2}

the two conditions become

a s + b x = a

and

b s + a x = b

. Subtracting the second from the first gives

(a - b) (s - x) = a - b

, and since

a \neq = b

s - x = 1

. Multiplying the first by

a

, the second by

b

, and subtracting gives

(a^{2} - b^{2}) s = a^{2} - b^{2}

, so

s = 1

, whence

x = s - 1 = 0

. Then

s = x^{2} - y^{2} = - y^{2} = 1

forces

y^{2} = - 1

, which has no real solution. Hence no such complex number exists, and the set contains

0

elements.

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Single correct option