Let `z` be a complex number satisfying the equation `z^2-(3+i)z+m+2i=0,w h e r em in Rdot` Suppose the equation has a real root. Then find non-real root.

Explanation

Let`alpha` be the root . Then,
`alpha^(2) -(3+ i)alpha + m + 2i = 0`
`rArr (alpha ^(2) - 3alpha + m) + i(2-alpha)=0`
`rArr alpha^(2) - 3 alpha +m = 0 and 2-alpha =0`
`rArr alpha = 2 and m =2`
Product of the roots is 2(1+i) with one roots as 2. Hence, the nonreal roots is 1+i.