If z = 1 2 - 2 i , is such that | z + 1 | = α z + β ( 1 + i ) , i = - 1 and α , β ∈ R , then α + β is equal to

Given: $\left|z+1\right|=\alpha z+\beta \left(1+i\right), z=\frac{1}{2}-2i$

$\Rightarrow \left|\frac{1}{2}-2i+1\right|=\alpha \left(\frac{1}{2}-2i\right)+\beta \left(1+i\right)$

$\Rightarrow \left|\frac{3}{2}-2i\right|=\frac{\alpha }{2}-2\alpha i+\beta +\beta i$

$\Rightarrow \sqrt{\frac{9}{4}+4}=\left(\frac{\alpha }{2}+\beta \right)+i\left(-2\alpha +\beta \right)$

Now, on comparing real and imaginary part we get,

$\Rightarrow \frac{5}{2}=\left(\frac{\alpha }{2}+\beta \right)+i\left(-2\alpha +\beta \right)$

$\Rightarrow 2\alpha =\beta , \frac{5}{2}=\left(\frac{\alpha }{2}+\beta \right)$

$\Rightarrow 2\alpha =\beta , \frac{5}{2}=\frac{5\alpha }{2}$

$\Rightarrow \alpha =1, \beta =2$

$\Rightarrow \alpha +\beta =3$