Complex root of equation

Question

Suppose we have $$z^2 + kz + m=0$$ where $k,m$ are real and $z$ is complex such that two distinct roots of this equation lie on $Re(z)=1$ so what will be range of m? Since root of this this equation will be in conjugation hence they maybe expressed as $$z_1=1+ i.\alpha $$ and $$z_2=1- i.\alpha$$ therefore sum of roots that is $$-k= 1+ i.\alpha + 1- i.\alpha=2 $$ now nice we need distinct root therefore $$4-4m>0$$ hence $$m<1$$ but answer is incorrect , why??

Answer 1

In the quadratic formula, you want the discriminant negative to get complex roots; that's $k^2 - 4m < 0$.