If z=frac{1}{2}-2 i , is such that |z+1|=α z+β(1+i), i=√{-1} and α, β ∈ R quad , then α+β

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4-1.Complex numbers

hard

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

$-4$

$3$

$2$

$-1$

(JEE MAIN-2024)

$ \mathrm{z}=\frac{1}{2}-2 \mathrm{i} $

$ |\mathrm{z}+1|=\alpha \mathrm{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 \text { and } \frac{\alpha}{2}+\beta=\sqrt{\frac{9}{4}+4} $

$ \alpha+\beta=3$

Standard 11

Mathematics

medium

medium

easy

hard

(JEE MAIN-2019)

easy