How many complex roots does a cubic equation have?

1 Answer
Aug 10, 2015

Since every real number can be thought of as a complex number, one way to answer this is to say the answer is 3, after "counting multiplicities". On the other hand, if you are after non-real complex roots, and if the coefficients of your cubic equation are real numbers, the answer will be 0 or 2.


A real number #a# can be thought of as the complex number #a+0i#. A cubic such as #(x-2)(x-5)^2=x^3-12x^2+45x-50# would be an example that would then have three "complex" roots. The number 2 would be a root of multiplicity 1 and the number 5 would be a root of multiplicity 2.

If you are after non-real complex roots in a situation with real number coefficients, the same example above would provide a situation where there are 0 complex roots. Examples where there are two non-real complex roots are situations where they occur in complex conjugate pairs . For example, #(x-(3+4i))(x-(3-4i))(x+3)=x^3-3x^2+7x+75# has roots #x=-3#, #x=3\pm 4i#, and the complex numbers #3+4i# and #3-4i# are called complex conjugates.

It can be proved that if a polynomial has real coefficients, then the non-real complex roots of such a polynomial, if there are any, must occur in these complex conjugate pairs. That's why there can't be such cubics with 1 or 3 non-real complex roots.

One other thing worth mentioning: by the Intermediate Value Theorem, all cubics with real coefficients must have at least 1 real root (giving another reason such a cubic can't have 3 non-real complex roots).