If z = frac{1}{2} - 2i is such that |z+1| = a | vedclass

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(B) Given $z = \frac{1}{2} - 2i$. Substitute $z$ into the equation $|z+1| = \alpha z + \beta(1+i)$: $|(\frac{1}{2} - 2i) + 1| = \alpha(\frac{1}{2} - 2

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$3$

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(B) Given $z = \frac{1}{2} - 2i$. Substitute $z$ into the equation $|z+1| = \alpha z + \beta(1+i)$: $|(\frac{1}{2} - 2i) + 1| = \alpha(\frac{1}{2} - 2i) + \beta(1+i)$ $|\frac{3}{2} - 2i| = (\frac{\alpha}{2} + \beta) + (\beta - 2\alpha)i$ Calculate the modulus on the left side: $|\frac{3}{2} - 2i| = \sqrt{(\frac{3}{2})^2 + (-2)^2} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{25}{4}} = \frac{5}{2}$ Equating the real and imaginary parts: Imaginary part: $\beta - 2\alpha = 0 \implies \beta = 2\alpha$ Real part: $\frac{\alpha}{2} + \beta = \frac{5}{2}$ Substitute $\beta = 2\alpha$ into the real part equation: $\frac{\alpha}{2} + 2\alpha = \frac{5}{2}$ $\frac{5\alpha}{2} = \frac{5}{2} \implies \alpha = 1$ Then $\beta = 2(1) = 2$. Therefore,$\alpha + \beta = 1 + 2 = 3$.

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

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