Question

Let \(z\) be a complex number such that \(|z + 2| = |z - 2|\) and \(\arg\left(\frac{z+3}{z-i}\right) = \frac{\pi}{4}\). Then \(|z|^2\) is equal to:

• A. 9
• B. 4
• C. 5
• D. 1

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The locus of a complex number \(z\) satisfying \(|z - z_1| = |z - z_2|\) is the perpendicular bisector of the line segment joining \(z_1\) and \(z_2\).
Step 2: Key Formula or Approach:
For \(|z+2| = |z-2|\), the points are \(z_1 = -2\) and \(z_2 = 2\).
The perpendicular bisector of the segment joining \((-2, 0)\) and \((2, 0)\) is the y-axis.
Therefore, \(z\) is purely imaginary. Let \(z = iy\), where \(y \in \mathbb{R}\).
Step 3: Detailed Explanation:
Substitute \(z = iy\) into the second condition \(\arg\left(\frac{z+3}{z-i}\right) = \frac{\pi}{4}\):
\[ \arg\left(\frac{iy+3}{iy-i}\right) = \frac{\pi}{4} \] Simplify the argument by expressing the complex number in the form \(X + iY\):
\[ \frac{3+iy}{i(y-1)} = \frac{3+iy}{i(y-1)} \times \frac{-i}{-i} = \frac{-i(3+iy)}{y-1} = \frac{y - 3i}{y-1} \] So, the complex number is \(X + iY = \frac{y}{y-1} - i\frac{3}{y-1}\).
Since the argument is \(\frac{\pi}{4}\), the real and imaginary parts must be equal and strictly positive (\(X = Y>0\)):
\[ \frac{y}{y-1} = \frac{-3}{y-1} \] Since \(y \neq 1\), we can cancel the denominators to get:
\[ y = -3 \] Let's verify if \(X>0\) and \(Y>0\):
\[ X = \frac{-3}{-3-1} = \frac{-3}{-4} = \frac{3}{4}>0 \] \[ Y = \frac{-3}{-4} = \frac{3}{4}>0 \] Since both are positive, \(y = -3\) is the correct value.
Thus, \(z = -3i\).
Step 4: Final Answer:
We need to find \(|z|^2\):
\[ |z|^2 = |-3i|^2 = 3^2 = 9 \]