If z = 1 2 - 2 i is such that | z + 1 | = z + ( 1 + i ) , i = - 1 and then is equal to [2024]

Question

If $z = \frac{1}{2} - 2 i$ is such that $| z + 1 |$ $= α z + β (1 + i),$ $i = \sqrt{- 1}$ and $α, β \in R,$ then $α + β$ is equal to [2024]

Smart Achievers · Accepted Answer

(3)

Given, $z = \frac{1}{2} - 2 i$ ...(i)

and $| z + 1 | = α z + β (1 + i)$ ...(ii)

From (i) and (ii), we get:

$α, β \in R,$ $= α (\frac{1}{2} - 2 i) + β (1 + i)$

$\Rightarrow$ $| \frac{3}{2} - 2 i |$ $= α (\frac{1}{2} - 2 i) + β (1 + i)$

$\Rightarrow \sqrt{\frac{9}{4} + 4} = \frac{α}{2} - 2 α i + β + β i$ $[∵$ $| z |$ $= \sqrt{x^{2} + y^{2}}]$

$\Rightarrow \sqrt{\frac{25}{4}} = \frac{α}{2} + β + i (- 2 α + β) \Rightarrow \frac{5}{2} = \frac{α}{2} + β + i (- 2 α + β)$

$\Rightarrow \frac{5}{2} = \frac{α}{2} + β$ ...(iii) and $- 2 α + β = 0 \Rightarrow β = 2 α$ ...(iv)

Solving (iii) and (iv), we get $α = 1$ and $β = 2$

$∴$ $α + β = 1 + 2 = 3$

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