Question

Let \( z = x + iy \), where \( y>0 \). If \( z + \overline{z} = 6 \) and \( |z| + | \overline{z} | = 10 \), then \( z = \)

• A. \( 3 + 2i \)
• B. \( 3 + 5i \)
• C. \( 3 + 3i \)
• D. \( 3 + 4i \)
• E. \( 3 + i\sqrt{5} \)

Hint:

To solve for complex numbers, separate real and imaginary parts, and use the modulus property \( |z| = \sqrt{x^2 + y^2} \).

Answer 1

The Correct Option is D

Let \( z = x + iy \) and \( \overline{z} = x - iy \). From the first equation \( z + \overline{z} = 6 \), we get: \[ (x + iy) + (x - iy) = 6 \] \[ 2x = 6 \quad \Rightarrow \quad x = 3 \] Now, from the second equation \( |z| + |\overline{z}| = 10 \), we have: \[ |z| = \sqrt{x^2 + y^2} \quad {and} \quad |\overline{z}| = \sqrt{x^2 + y^2} \] Thus: \[ 2\sqrt{x^2 + y^2} = 10 \quad \Rightarrow \quad \sqrt{x^2 + y^2} = 5 \quad \Rightarrow \quad x^2 + y^2 = 25 \] Substituting \( x = 3 \) into the equation: \[ 9 + y^2 = 25 \quad \Rightarrow \quad y^2 = 16 \quad \Rightarrow \quad y = 4 \] Thus, \( z = 3 + 4i \). Therefore, the correct answer is (D).