Given that 1+2i is a root of the equation z2+(p+5i)z+q(2−i)=0, find the values of p and q and the other root of the equation.
Assuming that the roots are α+β=−b/a and αβ=c/a with α=1+2i
Substitution gives β=−p−1−7i
Substitution of α and β in the quadratic equation gives p(1+2i)+q(2−i)−13=0
Further substitution gave me p=11−8i1+2i and q=16−33i3
I have still not found the solution for p and q (ans p=−1, q=7, root is −7i)
Any help or comments appreciated.
Answer
Hint: If p and q are both real, then one complex equation becomes two simultaneous real equations.
Substitute one of the roots to obtain
(1+2i)2+(p+5i)(1+2i)+q(2−i)=0
(−3+4i)+(p−10)+(5+2p)i+(2q−qi)=0
(p+2q−13)+(2p−q+9)i=0
This leads to two equations since both the real and imaginary parts have to be 0
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