Tuesday, 1 March 2016

Roots and coefficients of complex quadratic equation




Given that 1+2i is a root of the equation z2+(p+5i)z+q(2i)=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=p17i



Substitution of α and β in the quadratic equation gives p(1+2i)+q(2i)13=0

for both roots! no simultaneous equation solution possible.



Further substitution gave me p=118i1+2i and q=1633i3




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(2i)=0


(3+4i)+(p10)+(5+2p)i+(2qqi)=0

(p+2q13)+(2pq+9)i=0




This leads to two equations since both the real and imaginary parts have to be 0


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