386 Quadratic Equations with Complex Roots
Here’s a challenge!
Prove the second of the above statements.
Solution
Consider once again the general polynomial equation
ax^2 +bx+c= 0
The discriminant is
d=b^2 − 4 ac
Suppose that d< 0 and b≠ 0. We can use the “abbreviated discriminant” version of the quadratic formula
for complex roots:
x= [−b±j(|d|1/2)] / (2a)
Applying the right-hand distributive rule for addition and subtraction “in reverse” to the right side, we can
split that expression into a sum or difference of two ratios with a common denominator, like this:
x=−b/(2a)±j(|d|1/2)/(2a)
Therefore, we can write the two roots as
x=−b/(2a)+j(|d|1/2)/(2a) or x=−b/(2a)−j(|d|1/2)/(2a)
The fact that these are complex conjugates is obscure because the expressions are messy. To bring things
into focus, we can change a couple of names. Let
p=−b/(2a)
and
q= (|d|1/2)/(2a)
Now we can rewrite the roots as
x=p+jq or x=p−jq
These are complex conjugates (but not pure imaginary numbers) for all nonzero real numbers p and q.
You might ask, “Are p and q really both nonzero?” The answer is “Yes.” We can be sure that p≠ 0 because
b≠ 0 and a≠ 0, so −b/(2a) can’t be 0. We can be sure that q≠ 0 because d≠ 0 and a≠ 0, so (|d|1/2)/(2a)
can’t be 0.