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4-1.Complex numbers
hard
જો $z=\frac{1}{2}-2 i$ એ એવી છે કે જેથી $|z+1|=\alpha z+\beta(1+i)$ થાય $i=\sqrt{-1}$ અને $\alpha, \beta \in \mathbb{R}$,તો $\alpha+\beta=$.....................
A
$-4$
B
$3$
C
$2$
D
$-1$
(JEE MAIN-2024)
Solution
$ \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
