1.If z=12−2iz = \dfrac{1}{2} - 2iz=21−2i is such that ∣z+1∣=αz+β(1+i)|z + 1| = \alpha z + \beta(1 + i)∣z+1∣=αz+β(1+i), i=−1i = \sqrt{-1}i=−1 and α,β∈R\alpha, \beta \in \mathbb{R}α,β∈R, then α+β\alpha + \betaα+β is equal to:a.−4-4−4b.333c.222d.−1-1−1Login to continueOnly logged in users canattempt or see the solution.
