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

Answer: (B)

3

Answer 2

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: $\boxed{3}$