A quadratic equation with complex coefficients

Factual Questions
1

As we all know, the discriminant of a quadratic equation with real components can be used to tell us whether we have two distinct real roots, a repeated real root, or a conjugate pair of complex roots.

There are three possible cases for a quadratic equation with complex, non-real coefficients: a distinct pair of non-conjugate complex roots, a repeated complex root, or one real root and one complex root. Can the discriminant of such an equation be used to get information about which case we have?

2

There are three possible cases for a quadratic equation with complex, non-real coefficients: a distinct pair of non-conjugate complex roots, a repeated complex root, or one real root and one complex root. Can the discriminant of such an equation be used to get information about which case we have?
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-b + SQRT(b^2 - 4 a c) / 2a

Either -b plus or minus the square root term would have to give you a real multiple of the complex number a for there to be one real root.

Mathematica doesn’t give me any answers when I try to solve for Imaginary part of the quadratic formula is 0 for the plus case.

Mathematica won’t even give Imaginary parts of a general complex number a + b i.

This kind of thing is interesting to me so I’m looking at it some more and looking at graphs but it takes four real variables just for complex b and c if a is scaled out.

3

Nitpick: this is not quite true. We can multiply a quadratic equation with real coefficients by an arbitrary non-real complex number to get a quadratic with non-real coefficients, but the solutions are still reals or complex conjugates. If you restrict to monic quadratics, though (the coefficient of x[sup]2[/sup] is 1), or at least make one coefficient real, then these are the three possible cases.

The case for a repeated root is clearly the same as in the real case (b[sup]2[/sup] - 4ac = 0), whether the quadratic is monic or not. The test for the other two cases doesn’t seem to be any easier than just computing the two roots explicitly, though. (For the monic case, when one of the roots is real, the discriminant will lie on one of a pair of parabolas {x[sup]2[/sup]-r[sup]2[/sup] +/- 2 i r x: x real}, where r = Im(b). I haven’t been able to come up with a simpler criterion.)

4

For purposes of talking about roots, let’s stick to monic quadratics. I forgot to mention that in the OP.

5

I still don’t have any simpler criteria involving only b[sup]2[/sup]-4ac in the case a=1. But if you allow more general equations there might be something…

Let 0 = z[sup]2[/sup]+bz+c = (z-r)(z-s), so that -b = r+s and c = rs. Let t = Im(c)/Im(-b). If r is real and nonzero (if r=0, then clearly one root is real) and s is not real, then t = Im(rs)/Im(r+s) = r, so c/t = s = -(b+t). But if c/t = -(b+t) for some t, then t and c/t are the roots of the quadratic; so if neither root is real then we will have c/t =/= -(b+t) for all real t. Thus: one root of z[sup]2[/sup]+bz+c is real iff c = -t(b+t), where t=Im(c)/Im(-b). (Special cases: if exactly one of b and c is real, the quadratic has no real roots unless c=0; if both are real, use the real discriminant.)

6

Interesting. I wasn’t really expecting the traditional discriminant to be useful, as it relies on ordering and we don’t have that in C.

So the question is: how can we distinguish between a repeated complex root and a pair of non-conjugate complex roots? I’ll play around with it today.

7

Oops, I missed the earlier statement that b[sup]2[/sup] - 4ac = 0 iff the roots are repeated, regardless of of their nature. Oh well, it was easy to confirm.

So given an arbitrary monic quadratic polynomial z[sup]2[/sup] + bz + c over C, we can classify the roots as follows:[ol][li]If b[sup]2[/sup] - 4c = 0, we have one repeated root. If b and c are both real, the root is real; otherwise, it’s complex.[]Otherwise, if b and c are both real, we have two real roots when b[sup]2[/sup] - 4c > 0 and a conjugate pair of complex roots when b[sup]2[/sup] - 4c < 0.[]If exactly one of b and c is real, there is a real root only if c = 0.[]If both b and c are complex, then we have one real root only if c = -t(b+t), where t=Im©/Im(-b).[]Otherwise, we have a distinct pair of non-conjugate complex roots.[/ol][/li]
It probably would be simpler to just use the quadratic formula, but this is interesting.