x + i y + α ( x − 1 ) 2 + y 2 + 2 i = 0 x + iy + \alpha \sqrt {{{(x - 1)}^2} + {y^2}} + 2i = 0 x + i y + α ( x − 1 ) 2 + y 2 + 2 i = 0
∴ \therefore ∴ x + α ( x − 1 ) 2 + y 2 = 0 x + \alpha \sqrt {{{(x - 1)}^2} + {y^2}} = 0 x + α ( x − 1 ) 2 + y 2 = 0
y = − - − x 2 = α 2 ( x 2 − 2 x + 1 + 4 ) {x^2} = {\alpha ^2}({x^2} - 2x + 1 + 4) x 2 = α 2 ( x 2 − 2 x + 1 + 4 )
α 2 = x 2 x 2 − 2 x + 5 ⇒ x 2 ( α 2 − 1 ) − 2 x α 2 + 5 α 2 = 0 {\alpha ^2} = {{{x^2}} \over {{x^2} - 2x + 5}} \Rightarrow {x^2}({\alpha ^2} - 1) - 2x{\alpha ^2} + 5{\alpha ^2} = 0 α 2 = x 2 − 2 x + 5 x 2 ⇒ x 2 ( α 2 − 1 ) − 2 x α 2 + 5 α 2 = 0
x ∈ R ⇒ D ≥ 0 x \in R \Rightarrow D \ge 0 x ∈ R ⇒ D ≥ 0
4 α 4 − 4 ( α 2 − 1 ) 5 α 2 ≥ 0 4{\alpha ^4} - 4({\alpha ^2} - 1)5{\alpha ^2} \ge 0 4 α 4 − 4 ( α 2 − 1 ) 5 α 2 ≥ 0
α 2 [ 4 α 2 − 2 α 2 + 20 ] ≥ 0 {\alpha ^2}[4{\alpha ^2} - 2{\alpha ^2} + 20] \ge 0 α 2 [ 4 α 2 − 2 α 2 + 20 ] ≥ 0
α 2 [ − 16 α 2 + 20 ] ≥ 0 {\alpha ^2}[ - 16{\alpha ^2} + 20] \ge 0 α 2 [ − 16 α 2 + 20 ] ≥ 0
α 2 [ α 2 − 5 4 ] ≤ 0 {\alpha ^2}\left[ {{\alpha ^2} - {5 \over 4}} \right] \le 0 α 2 [ α 2 − 4 5 ] ≤ 0
0 ≤ α 2 ≤ 5 4 0 \le {\alpha ^2} \le {5 \over 4} 0 ≤ α 2 ≤ 4 5
∴ \therefore ∴ α 2 ∈ [ 0 , 5 4 ] {\alpha ^2} \in \left[ {0,{5 \over 4}} \right] α 2 ∈ [ 0 , 4 5 ]
∴ \therefore ∴ α ∈ [ − 5 2 , 5 2 ] \alpha \in \left[ { - {{\sqrt 5 } \over 2},{{\sqrt 5 } \over 2}} \right] α ∈ [ − 2 5 , 2 5 ]
then 4 [ ( q ) 2 + ( p ) 2 ] = 4 [ 5 4 + 5 4 ] = 10 4[{(q)^2} + {(p)^2}] = 4\left[ {{5 \over 4} + {5 \over 4}} \right] = 10 4 [ ( q ) 2 + ( p ) 2 ] = 4 [ 4 5 + 4 5 ] = 10
