# Can a cubic equation have three complex roots?

If $x+ iy$ is a root of a cubic equation $ax^3+bx^2+cx+d=0$, then $x-iy$ is also a root of this equation.

My question is:

Can $k-il$ be the other root of this equation if $l\neq 0$?

If yes, then it would seem that $k+ il$ should also be a root, but in that case the cubic equation would have four roots which is impossible.

Where is the mistake in my logic?

($i$ denotes $\sqrt {-1}$, and is sometimes pronounced as $iota$.)

• $k-il$ won't be the third imaginary root.. the third root will be real if the first two are imaginary – Shreyas S Dec 10 '16 at 17:32
• It depends, if your coefficients are non-real complex numbers then it can happen $(x-i)^3$. But if the polynomial is over the field of real numbers and by complex roots we mean non-real roots then it cannot happen. – Anurag A Dec 10 '16 at 17:32
• I have never heard of the imaginary unit $i$ pronounced as "iota". To me, "iota" is the greek letter $\iota$. – Arthur Dec 10 '16 at 17:32
• Depends. Are $a,b,c,d$ real numbers? If not, then it is possible, because in that case it is no longer true that $x-iy$ is a root if and only if $x+iy$ is. But, if the coefficients are all real, then your argument works fine. Another way of seeing the same thing will become obvious if you plot the graph of the cubic. You will see (and it follows from intermediate value theorem) that a cubic with real coefficients always has a real root. – Jyrki Lahtonen Dec 10 '16 at 17:33
• In my high school , while teaching about complex numbers , teachers pronounce it iota , and I was not sure that the same symbol is used everywhere to define a complex number, So I explained, Should I edit? – Atul Mishra Dec 11 '16 at 13:02

If the coefficients $a$, $b$, $c$ and $d$ are real, then the equation $ax^3+bx^2+cx+d=0$, assuming $a\ne0$, has at least a real root: indeed, writing $$f(x)=x^3+\frac{b}{a}x^2+\frac{c}{a}x+\frac{d}{a}$$ we have $$\lim_{x\to-\infty}f(x)=-\infty \qquad \lim_{x\to\infty}f(x)=\infty$$ so the intermediate value theorem provides one real value $r$ such that $f(r)=0$.

Since the equation has at most three distinct roots, it follows that it cannot have three distinct complex nonreal roots.

If $a$, $b$, $c$ and $d$ are not supposed to be real, but just complex, then it is surely possible: consider $$(x-i)(x-2i)(x-3i)=0$$

• Small point to note: the property that is mentioned at the start of the question implies we are talking of polynomials with real coefficients. – leonbloy Dec 10 '16 at 20:52
• @leonbloy Yes, that's why I started with examining the case $a,b,c,d\in\mathbb{R}$. – egreg Dec 10 '16 at 20:53
• @leonbloy, the question doesn’t so much imply that the coefficients are real, as assume it. Not quite the same thing. And precisely because OP is a relative beginner, he was thrown off by the omission of this essential assumption. – Lubin Dec 10 '16 at 23:09

Your argument is an almost correct proof of the fact that a cubic equation cannot have three complex (non-real) roots. The only thing you fail to account for is multiple roots: you do not treat the case in which the third root is $x+iy$ or $x-iy$; that is, it coincides with one of the other two roots.

This case is impossible for a cubic equation with real $a,b,c,d$, but the theorem you quote is not sufficient to prove this. You need a different argument in this case. The simplest result you can use to exclude it is the following (a special case of Viète's formulas):

if $z_1,z_2,z_3$ (counted with multiplicity) are the roots of a cubic equation $ax^3+bx+cx+d=0$, then $z_1+z_2+z_3 = -\frac{b}{a}$.

I’d like to relate something that might be useful to your current level of study, in addition to the other answers.

Start with an observation that might be made to sound pithy but turns out to be quite profound: given that i and −i are both square roots of −1, how do you know which is which? After all, they share a property that is their defining characteristic!

The answer is that you can’t. Not only is the labeling arbitrary, but you can switch them without changing anything.

So, in general, any time you start with pure real numbers (e.g. the coefficients and constants) and perform normal famialiar arithmetic operations, then any time you wind up with complex numbers, you must have a symmetry such that you can switch all the (i)s and (−i)s without changing anything.

A special case of this is what you’re remembering: that any complex roots of a polynomial must come in pairs (a±bi).

Let $z^*$ denote the complex conjugate of z. We have $(z_1z_2)^*=z_1^*z_2^*$ and $(z_1+z_2)^*=z_1^*+z_2^*.$ And $z^*=z$ for real $z.$ Therefore, for real $a,b,c,d$ we have $$0=az^3+bz^2+cz+d\iff 0=0^* =(az^3+bz^2+cz+d)^*=$$ $$=a(z^*)^3+b(z^*)^2+c(z^*)+d.$$ So if $z_,z_2$ are non-real zeroes of $az^3+bz^2+cz+d ,$ with real $a,b,c,d$ then so are $z_1^*$ and $z_2^*.$ If $z_2\ne \{z_1,z_1^*\},$ this give at least 4 zeroes for the cubic.