[Om Sharma]

If $\mathrm{z}=\frac{1}{2}-2 \mathrm{i}$ is such that $|\mathrm{z}+1|=\alpha \mathrm{z}+\beta(1+\mathrm{i})$ and $\alpha, \beta \in \mathrm{R}$, then $\alpha+\beta$ is equal to [JEE(Main) 29 Jan 2024 S I]
(A) -1
(B) -4
(C) 2
(D) 3

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[Om Sharma]

Solution

$|\mathrm{z}+1|=\alpha \mathrm{z}+\beta(1+\mathrm{i})$, put $\mathrm{z}=\frac{1}{2}-2 \mathrm{i} \Rightarrow\left|\frac{3}{2}-2 \mathrm{i}\right|=\frac{\alpha}{2}-2 \alpha \mathrm{i}+\beta+\beta \mathrm{i} \Rightarrow \sqrt{\frac{9}{4}+4}=\frac{5}{2}=\left(\frac{\alpha}{2}+\beta\right)+(\beta-2 \alpha) \mathrm{i}$
Comparing real and imaginary parts $\frac{\alpha}{2}+\beta=\frac{5}{2}$ and $\beta=2 \alpha \Rightarrow \frac{\alpha}{2}+2 \alpha=\frac{5}{2} \Rightarrow \alpha=1, \beta=2$. Now, $\alpha+\beta=3$

[Om Sharma]

Answer

D

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