If the least and the largest real values of α , for which the equation z+αz−1+2i=0z∈C and i=−1 has a solution, are p and q respectively, then 4p2+q2 is equal to_______.

x+iy+αx−12+y2+2i=0⇒y+2=0 & x+αx−12+y2=0It gives y=−2 , x2=α2x−2x+1+4α2=x2x2−2x+5⇒ α2∈0,54Therefore, 4αmax2+αmin2=40+54=5

If the least and the largest real values of α , for which the equation z+αz−1+2i=0z∈C and i=−1 has a solution, are p and q respectively, then 4p2+q2 is equal to_______.

x+iy+αx−12+y2+2i=0⇒y+2=0 & x+αx−12+y2=0It gives y=−2 , x2=α2x−2x+1+4α2=x2x2−2x+5⇒ α2∈0,54Therefore, 4αmax2+αmin2=40+54=5

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If the least and the largest real values of $α$ , for which the equation $z + α |z - 1| + 2 i = 0 (z \in C a n d i = \sqrt{- 1})$ has a solution, are p and q respectively, then $4 (p^{2} + q^{2})$ is equal to_______.

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$x + i y + α \sqrt{{(x - 1)}^{2} + y^{2}} + 2 i = 0 \Rightarrow y + 2 = 0 & x + α \sqrt{{(x - 1)}^{2} + y^{2}} = 0$

It gives $y = - 2, x^{2} = α^{2} (x - 2 x + 1 + 4)$

$α^{2} = \frac{x^{2}}{x^{2} - 2 x + 5} \Rightarrow α^{2} \in [0, \frac{5}{4}]$

Therefore, $4 [{(α_{m a x})}^{2} + {(α_{m i n})}^{2}] = 4 [0 + \frac{5}{4}] = 5$

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If the least and the largest real values of α , for which the equation z+αz−1+2i=0z∈C and i=−1 has a solution, are p and q respectively, then 4p2+q2 is equal to_______.

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If the least and the largest real values of $α$ , for which the equation $z + α |z - 1| + 2 i = 0 (z \in C a n d i = \sqrt{- 1})$ has a solution, are p and q respectively, then $4 (p^{2} + q^{2})$ is equal to_______.