Question

If \( z = \frac{1}{2} - 2i \), is such that \( |z + 1| = \alpha z + \beta (1 + i) \), \( i = \sqrt{-1} \) and \( \alpha, \beta \in \mathbb{R} \), then \( \alpha + \beta \) is equal to:

• A. -4
• B. 3
• C. 2
• D. -1

Answer 1

The Correct Option is B

We are given:
\(z = \frac{1}{2} - 2i\)
and\(|z + 1| = \alpha z + \beta(1 + i)\)

First, we calculate \( |z + 1| \):
\(|z + 1| = \left| \frac{1}{2} - 2i + 1 \right| = \left| \frac{3}{2} - 2i \right|\)
Using the modulus formula for complex numbers:

\(\left| \frac{3}{2} - 2i \right| = \sqrt{\left( \frac{3}{2} \right)^2 + (-2)^2} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{9}{4} + \frac{16}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2}\)
Now, substituting into the equation:
\(\frac{5}{2} = \alpha \left( \frac{1}{2} - 2i \right) + \beta(1 + i)\)

Expanding both sides:
\(\frac{5}{2} = \frac{\alpha}{2} - 2\alpha i + \beta + \beta i\)

Equating real and imaginary parts:

Real part: \(\frac{\alpha}{2} + \beta = \frac{5}{2}\)

Imaginary part: \(-2\alpha + \beta = 0\)

From the imaginary part:
\(\beta = 2\alpha\)
Substituting into the real part:

\(\frac{\alpha}{2} + 2\alpha = \frac{5}{2}\)

\(\frac{5\alpha}{2} = \frac{5}{2}\)
\(\alpha = 1\)
Substituting \(\alpha = 1\) into \(\beta = 2\alpha\):
\(\beta = 2\)
Thus:
\(\alpha + \beta = 1 + 2 = 3\)
So, the correct answer is: \(\boxed{3}\)