If z = 1/2 - 2i, is such that |z + 1| = αz + β(1 + i), i = √−1 and α, β ∈ R,​ then α + β is equal to

Question

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

(1) -4

(2) 3

(3) 2

(4) -1

Answer 1

Correct option is (2) 3

\(z=\frac{1}{2}-2 \mathrm{i}\)

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

\(\left|\frac{3}{2}-2 \mathrm{i}\right|=\frac{\alpha}{2}-2 \alpha \mathrm{i}+\beta+\beta \mathrm{i}\)

\(\left|\frac{3}{2}-2 \mathrm{i}\right|=\left(\frac{\alpha}{2}+\beta\right)+(\beta-2 \alpha) \mathrm{i}\)

\(\beta=2 \alpha\text{ and }\frac{\alpha}{2}+\beta=\sqrt{\frac{9}{4}+4}\)

\(\alpha+\beta=3\)