Complex Number Question 7
Question 7 - 2024 (29 Jan Shift 1)
If $z=\frac{1}{2}-2 i$, is such that $|z+1|=\alpha z+\beta(1+i), i=\sqrt{-1}$ and $\alpha, \beta \in R \quad$, then $\alpha+\beta$ is equal to
(1) -4
(2) 3
(3) 2
(4) -1
Show Answer
Answer (2)
Solution
$\mathrm{z}=\frac{1}{2}-2 \mathrm{i}$
$|\mathbf{z}+1|=\alpha z+\beta(1+\mathrm{i})$
$\left|\frac{3}{2}-2 \mathrm{i}\right|=\frac{\alpha}{2}-2 \alpha \mathrm{i}+\beta+\beta \mathrm{i}$
$\left|\frac{3}{2}-2 \mathrm{i}\right|=\left(\frac{\alpha}{2}+\beta\right)+(\beta-2 \alpha) \mathrm{i}$
$\beta=2 \alpha$ and $\frac{\alpha}{2}+\beta=\sqrt{\frac{9}{4}+4}$
$\alpha+\beta=3$
