If \( z \) is a complex number such that \( |z + 2| = |z - 2| \) and \( \arg\left( \frac{z-3}{z+i} \right) = \frac{\pi}{4} \), then the value of \( z \) is:
Show Hint
The equation \( |z - a| = |z - b| \) always represents the perpendicular bisector of the line segment joining points \( a \) and \( b \) in the complex plane.
Step 1: Understanding the Concept:
The condition \( |z + 2| = |z - 2| \) represents the locus of points equidistant from \( (-2, 0) \) and \( (2, 0) \), which is the perpendicular bisector of the segment joining them. The condition \( \arg\left( \frac{z-3}{z+i} \right) = \frac{\pi}{4} \) represents an arc of a circle passing through \( (3, 0) \) and \( (0, -1) \). Step 2: Key Formula or Approach:
1. From \( |z+2| = |z-2| \), we conclude \( z \) is purely imaginary (\( z = iy \)).
2. Substitute \( z = iy \) into the argument equation and solve for \( y \). Step 3: Detailed Explanation:
Let \( z = x + iy \).
Condition 1: \( |(x+2) + iy| = |(x-2) + iy| \implies (x+2)^2 + y^2 = (x-2)^2 + y^2 \).
Simplifying, we get \( x^2 + 4x + 4 = x^2 - 4x + 4 \implies 8x = 0 \implies x = 0 \).
Thus, \( z = iy \).
Condition 2: \( \arg\left( \frac{iy-3}{iy+i} \right) = \frac{\pi}{4} \).
The expression inside the argument is \( \frac{-3 + iy}{i(y+1)} \).
Multiplying numerator and denominator by \( -i \):
\[ \frac{(-3 + iy)(-i)}{i(y+1)(-i)} = \frac{3i + y}{y+1} = \frac{y}{y+1} + i\frac{3}{y+1} \]
We know \( \arg(A + Bi) = \tan^{-1}(B/A) \):
\[ \tan^{-1}\left( \frac{3/(y+1)}{y/(y+1)} \right) = \frac{\pi}{4} \implies \tan^{-1}\left( \frac{3}{y} \right) = \frac{\pi}{4} \]
\[ \frac{3}{y} = \tan\left(\frac{\pi}{4}\right) = 1 \implies y = 3 \]
So, \( z = 3i \). Step 4: Final Answer:
The value of \( z \) is \( 3i \).
