Question

If \( f(z) = \frac{1 - z^3}{1 - z} \), where \( z = x + iy \) with \( z \neq 1 \), then \( \mathrm{Re}(f(z)) = 0 \) reduces to

• A. \( x^2 + y^2 + x + 1 = 0 \)
• B. \( x^2 - y^2 + x - 1 = 0 \)
• C. \( x^2 - y^2 - x + 1 = 0 \)
• D. \( x^2 - y^2 + x + 1 = 0 \)
• E. \( x^2 - y^2 + x + 2 = 0 \)

Hint:

Always simplify complex fractions using algebraic identities before substituting \(z = x+iy\).

Answer 1

The Correct Option is D

Concept: Use identity: \[ \frac{1 - z^3}{1 - z} = 1 + z + z^2 \quad (z \neq 1) \]

Step 1: Simplify the function
\[ f(z) = 1 + z + z^2 \]

Step 2: Substitute \( z = x + iy \)
\[ z^2 = (x+iy)^2 = x^2 - y^2 + 2ixy \] \[ f(z) = 1 + (x+iy) + (x^2 - y^2 + 2ixy) \]

Step 3: Separate real part
Real part: \[ 1 + x + x^2 - y^2 \]

Step 4: Apply condition
\[ \mathrm{Re}(f(z)) = 0 \] \[ 1 + x + x^2 - y^2 = 0 \]

Step 5: Rearrangement
\[ x^2 - y^2 + x + 1 = 0 \] \[ \boxed{x^2 - y^2 + x + 1 = 0} \]