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If $$z = \frac{1}{2} - 2i$$, is such that $$|z + 1| = \alpha z + \beta(1 + i)$$, $$i = \sqrt{-1}$$ and $$\alpha, \beta \in R$$, then $$\alpha + \beta$$ is equal to
$$z = 1/2 - 2i$$. $$|z+1| = |3/2 - 2i| = \sqrt{9/4+4} = \sqrt{25/4} = 5/2$$.
$$\alpha z + \beta(1+i) = \alpha/2 - 2\alpha i + \beta + \beta i = (\alpha/2+\beta) + (-2\alpha+\beta)i$$.
This must equal 5/2 (real): $$-2\alpha+\beta = 0$$ and $$\alpha/2+\beta = 5/2$$.
$$\beta = 2\alpha$$. $$\alpha/2 + 2\alpha = 5/2$$, $$5\alpha/2 = 5/2$$, $$\alpha = 1, \beta = 2$$.
$$\alpha + \beta = 3$$. Option (2).
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