If z=\frac12-2 i is such that |z+1|=\alpha z+\beta(1+i), i=\sqrt-1 and \alpha, \beta \in \mathbbR , then \alpha+\beta is equal to

Complex Numbers PYQ — JEE Main 2024 · 29 Jan · Shift 1 · Q45 with Solution

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Correct answer: C

Given

$z=\frac12-2i$

and

$|z+1|=\alpha z+\beta(1+i), \qquad \alpha,\beta\in\mathbb R$

We need to find $\alpha+\beta$ .
Compute $|z+1|$

First,

$z+1=\frac12-2i+1=\frac32-2i$

So,
$=\sqrt{\frac94+4} =\sqrt{\frac{25}{4}} =\frac52$$ Hence the equation becomes $$\frac52=\alpha\left(\frac12-2i\right)+\beta(1+i)$$$
Expand the right-hand side
$=\left(\frac\alpha2+\beta\right)+i(-2\alpha+\beta)$$ Since the left-hand side $\frac52$ is a **real number**, its imaginary part is $0$. Therefore, equate real and imaginary parts: $$\frac\alpha2+\beta=\frac52$$ $$-2\alpha+\beta=0$$$
Solve the system

From

$-2\alpha+\beta=0$

we get

$\beta=2\alpha$

Substitute into the real-part equation:

$\frac\alpha2+2\alpha=\frac52$

$\frac{5\alpha}{2}=\frac52$

$\alpha=1$

Then

$\beta=2$
Find $\alpha+\beta$

$\alpha+\beta=1+2=3$
Compare with stored answer

Our derived answer is 3, which matches option C and agrees with the stored correct answer.