If z = (1)/(2) - 2i , is such that | z + 1 | = + β ( 1 + i ) , i =...

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

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

$⇒ \left|\dfrac{3}{2} - 2i\right| = \dfrac{\alpha}{2} - 2\alphai + \beta + \betai$

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

Now, on comparing real and imaginary part we get,

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

$⇒ 2\alpha = \beta, \dfrac{5}{2} = \left(\dfrac{\alpha}{2} + \beta\right)$

$⇒ 2\alpha = \beta, \dfrac{5}{2} = \dfrac{5\alpha}{2}$

$⇒ \alpha = 1, \beta = 2$

$⇒ \alpha + \beta = 3$