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

 To solve this problem, we need to determine when the expression \( \frac{2z - 3i}{2z + i} \) is purely imaginary. This means that the real part of this expression should be zero.

Let \( z = x + iy \), where \( x \) and \( y \) are real numbers.

1. Substituting \( z = x + iy \) into the given expression, we have: \(\frac{2(x+iy) - 3i}{2(x+iy) + i}\) = \frac{2x + 2iy - 3i}{2x + 2iy + i}\)
2. Simplifying further, the numerator becomes: \[ 2x + (2y - 3)i \] and the denominator becomes: \[ 2x + (2y + 1)i \]
3. To find when this expression is purely imaginary, we need to remove the real part from the expression. We will do this by multiplying both numerator and denominator by the complex conjugate of the denominator: \(\overline{2x + (2y + 1)i} = 2x - (2y + 1)i\).
4. After multiplying and simplifying, only the imaginary parts remain. This yields: \(\text{Numerator: } [(2x + (2y - 3)i) \cdot (2x - (2y + 1)i)]\)
   • Expanding both sides separately results in:
     • Numerator: \( 2x(2x) + 2x(-2y - 1)i + (2y - 3)i(2x) + (2y - 3)(-2y - 1)i^2 \)
     • Simplifying this expression, with \( i^2 = -1 \), gives:
   • \([4x^2 + (2y + 1)^2 ] = 4x^2 + 4y^2 + 4y + 1\)\)

For pure imaginary, the real part of the fraction becomes zero. That is:

• \(4x^2 - (2y - 3)(2y + 1) = 0\).
• Since it's given that \(x + y^2 = 0\), then \(4x^2 = (2y^2 + 3)(2y^2 - 1)\)
• Solving this gives us \(y = \pm \frac{1}{2}\).

We are required to evaluate \(y^4 + y^2 - y\):

1. Substituting \(y = \frac{1}{2}\) into \(y^4 + y^2 - y\), we get:
   • \(y^4 = \left(\frac{1}{2}\right)^4 = \frac{1}{16}\)\)
   • \(y^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\)\)
   • \(y = \frac{1}{2}\)\)
2. Therefore, \(\frac{1}{16} + \frac{1}{4} - \frac{1}{2}\)
3. Finding a common denominator of 16, we get: \[ \frac{1}{16} + \frac{4}{16} - \frac{8}{16} = \frac{1 + 4 - 8}{16} = \frac{-3}{16} \]

Thus, the correct choice is \(\frac{3}{4}\).