If $z=\frac{1}{2}-2 i$ is such that $|z+1|=\alpha z+\beta(1+i), i=\sqrt{-1}$ and $\alpha, \beta \in \mathbb{R}$, then $\alpha+\beta$ is equal to [JEE Main 2024, 29 Jan (Shift 1)]

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Correct answer: d) 3. Given $z=\frac{1}{2}-2 i$ $|z+1|=\alpha z+\beta(1+i)$ $\left|\frac{3}{2}-2 i\right|=\frac{\alpha}{2}-2 \alpha i+\beta+\beta i$ $\left|\frac{3}{2}-2 i\right|=\left(\frac{\alpha}{2}+\beta\right)+(\beta-2 \alpha) i$ On comparing $\beta=2 \alpha$ and $\frac{\alpha}{2}+\beta=\sqrt{\frac{9}{4}+4}$ $\alpha+\beta=3$

If \(z=\frac{1}{2}-2 i\) is such that \(|z+1|=\alpha z+\beta(1+i), i=\sqrt{-1}\) and \(\alpha, \beta \in \mathbb{R}\), t…

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