On the equation Z2 - (a + I) Z - (I + 2) = 0 (a ∈ R) of complex number Z, (1) if the equation has a real number solution, find the value of a; (2) prove with the method of counter proof that for any real number a, the original equation cannot have pure imaginary root

On the equation Z2 - (a + I) Z - (I + 2) = 0 (a ∈ R) of complex number Z, (1) if the equation has a real number solution, find the value of a; (2) prove with the method of counter proof that for any real number a, the original equation cannot have pure imaginary root

(1) If the equation has a real solution, let z = m ∈ R, then M2 - (a + I) m - (I + 2) = 0, that is, m2-am-2 + (- m-1) I = 0, m2-am-2 = 0, and - M-1 = 0, M = - 1, a = 1. (2) suppose that the original equation has a pure imaginary root, let z = Ni, n ≠ 0, then there is (Ni) 2 - (a + I) Ni - (a + 2) I = 0, and the entire equation can be - N2 + n-2 + (- an-1) I = 0, n − N2 & nbsp; + n & nbsp; − 2 & nbsp; = & nbsp; 0 & nbsp; &For ①, the discriminant △ 0, the equation ① has no solution, so the equation system has no solution, so the hypothesis is not tenable, so the original equation can not have pure imaginary roots