"If $z=\frac{1}{2}-2 i$ 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"

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"

To begin with, let's analyze the given equation:

$|z+1|=\alpha z+\beta(1+i)$

First, we compute the modulus of the left side:

Let's take the given value of z,

$z = \frac{1}{2} - 2i$

Now, we find $z + 1$:

$z + 1 = \left(\frac{1}{2} - 2i\right) + 1 = \frac{3}{2} - 2i$

Then, the modulus of $z + 1$ is calculated as follows:

$$|z + 1| = \left|\frac{3}{2} - 2i\right| = \sqrt{\left(\frac{3}{2}\right)^2 + (-2)^2}$$

$|z + 1| = \sqrt{\frac{9}{4} + 4}$

$|z + 1| = \sqrt{\frac{9}{4} + \frac{16}{4}}$

$|z + 1| = \sqrt{\frac{25}{4}}$

$|z + 1| = \frac{5}{2}$

Now we need to equate the modulus to the right-hand side of the equation and solve for $\alpha$ and $\beta$. Let's rewrite the equation:

$\frac{5}{2} = \alpha z + \beta(1 + i)$

Substitute $z$ with its value:

$\frac{5}{2} = \alpha \left(\frac{1}{2} - 2i\right) + \beta(1+i)$

Rewrite the equation separating real and imaginary parts:

$\frac{5}{2} = \alpha \left(\frac{1}{2}\right) - 2\alpha i + \beta + \beta i$

$$\frac{5}{2} = \left(\alpha \frac{1}{2} + \beta\right) + \left(-2\alpha + \beta\right)i$$

For the above equality to hold, both real and imaginary parts must be equal.

Equating real parts:

$\alpha \frac{1}{2} + \beta = \frac{5}{2}$

Equating imaginary parts:

$-2\alpha + \beta = 0$

We now have a system of two linear equations:

1) $\frac{\alpha}{2} + \beta = \frac{5}{2}$

2) $-2\alpha + \beta = 0$

Let's solve the system by isolating $\beta$ from the second equation and then substituting it into the first one:

$\beta = 2\alpha$

Now substitute $\beta$ in the first equation:

$\frac{\alpha}{2} + 2\alpha = \frac{5}{2}$

$\alpha \left(\frac{1}{2} + 2\right) = \frac{5}{2}$

$\alpha \left(\frac{1}{2} + \frac{4}{2}\right) = \frac{5}{2}$

$\alpha \left(\frac{5}{2}\right) = \frac{5}{2}$

To find the value of $\alpha$, we divide both sides by $\frac{5}{2}$:

$\alpha = 1$

Now, we use the value of $\alpha$ to find $\beta$:

$\beta = 2\alpha$

$\beta = 2 \cdot 1$

$\beta = 2$

Finally, we add both $\alpha$ and $\beta$ to find $\alpha + \beta$:

$\alpha + \beta = 1 + 2 = 3$

The value of $\alpha + \beta$ is 3.

So, the correct answer is Option C) 3.

"

If $z=\frac{1}{2}-2 i$ 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

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If $z=\frac{1}{2}-2 i$ 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

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