Let the complex number $\mathrm{z}=\mathrm{x}+$ iy be such that $\frac{2 z-3 i}{2 z+i}$ is purely imaginary. If $x+y^{2}=0$, then $y^{4}+y^{2}-y$ is equal to :

JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Complex Numbers

Let the complex number

z = x +

iy be such that

\frac{2 z - 3 i}{2 z + i}

is purely imaginary. If

x + y^{2} = 0

, then

y^{4} + y^{2} - y

is equal to :

Answer: (D)

With

2 z = 2 x + 2 i y

, write

\frac{2 z - 3 i}{2 z + i} = \frac{2 x + i ( 2 y - 3 )}{2 x + i ( 2 y + 1 )}

. Multiplying numerator and denominator by

2 x - i (2 y + 1)

, the real part of the numerator is

(2 x) (2 x) + (2 y - 3) (2 y + 1)

. For the quotient to be purely imaginary this real part must vanish:

4 x^{2} + 4 y^{2} - 4 y - 3 = 0

. Using

x = - y^{2}

gives

4 y^{4} + 4 y^{2} - 4 y - 3 = 0

, hence

y^{4} + y^{2} - y = \frac{3}{4}

JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Single correct option