Question

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

• A. \(\frac{2}{3}\)
• B. \(\frac{3}{2}\)
• C. \(\frac{3}{4}\)
• D. \(\frac{4}{3}\)

Answer 1

The Correct Option is C

Given:

\( z = x + iy \), \( \frac{2z - 3i}{2z + i} \) is purely imaginary. This implies:

\( \text{Re} \left( \frac{2z - 3i}{2z + i} \right) = 0 \)

• Step 1: Substitute \( z = x + iy \):

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

• Rationalize the denominator:

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

• Step 2: Simplify the numerator and denominator:
  • The denominator is:

\( (2x)^2 + (2y + 1)^2 = 4x^2 + (2y + 1)^2 \)

• The numerator is:

\( 4x^2 + i[(2y - 3)(2x) - (2y + 1)(2x)] \)

• The real part is:

\( 4x^2 + (2y - 3)(2y + 1) \)

• Set the real part to 0:

\( 4x^2 + (2y - 3)(2y + 1) = 0 \)

• Step 3: Solve for \( x \) and \( y \):
  • Expand:

\( 4x^2 + [4y^2 + 2y - 6y - 3] = 0 \)

• Simplify:

\( 4x^2 + 4y^2 - 4y - 3 = 0 \)

• Substitute \( x = -y^2 \) (from \( x + y^2 = 0 \)):

\( 4(-y^2)^2 + 4y^2 - 4y - 3 = 0 \)

• Simplify further:

\( 4y^4 + 4y^2 - 4y - 3 = 0 \)

Final Answer: The equation is:

\( 4y^4 + 4y^2 - 4y - 3 = 0 \)

Correct Option: (3)