# Complex polynomials.

1. Apr 28, 2009

### stringy

Hello, I'm working through Polynomials by Barbeau. I think I may have provided a decent proof for one of the exercises, but I'd like a second opinion. Here's the exercise:

Show that every quadratic equation with complex coefficients has at least one complex root, and therefore can be written as the product of two linear factors with complex coefficients.

My proof:

Assume for contradiction that $$p \left( t \right) = qt^2 +wt+z$$ is a complex polynomial with $$q \neq 0$$. Further, suppose that $$p$$ has two real roots. Since the discriminant must vanish we have $$w^2 = 4qz$$. Furthermore, if the discriminant vanishes we are left with $$-w/2q$$ where $$q$$ and $$w$$ must cancel. Hence, $$w = -2q$$. We now have two simultaneous equations; substitution yields $$\left( -2q \right)^2 = 4qz \Rightarrow q=z$$. Then, $$q=z \Rightarrow w = -2z$$. Therefore, $$p \left( t \right) =qt^2 +wt+z = zt^2 -2zt + z \Rightarrow t^2-2t+1$$, contradicting the assumption that $$p$$ is a complex polynomial.

It then follows that any complex polynomial may be written in the form $$q \left( t- r_1 \right)\left( t-r_2 \right)$$, where $$q \in \mathbb{C}$$ and $$r_1,r_2$$ are roots.

The critical assumption is that the discriminant must vanish. I'm nearly sure that it does, I'm having difficulty rigorously justifying it though. If the discriminant did not vanish then we'd be left with a complex number and its conjugate in the numerator of the quadratic formula and nothing would cancel. So, for the polynomial to have two real roots then it must vanish, right?

The author implies actually evaluating the square root of the discriminant in the quadratic formula, but if I define $$q = q_1 + q_2i$$, $$w = w_1 + w_2i$$, and $$z = z_1 + z_2i$$, that's going to lead to a huge mess...or am I missing something?

Any help would be appreciated. Thanks!

2. Apr 28, 2009

### Dick

i*x^2-3ix+2i has roots x=1 and x=2 but it's coefficients are complex. I don't think you are supposed to prove that the roots are NOT real. You are just supposed prove it has a root (real or complex). After all, reals are just a special case of 'complex'.

3. Apr 28, 2009

### stringy

I didn't quite interpret the problem statement that way. It specifically says "at least one complex root," which I take as implying that the imaginary part of the one of the roots must not equal zero. So, negating that means that I must show that if a complex quadratic with ALL real roots exists, it implies a contradiction.

Showing that the quadratic has a real root and then saying that the root is complex because $$\mathbb{R} \subset \mathbb{C}$$ seems to be a little weak, but I suppose it would work. After all, I often make things more complicated then need be.

And, the roots of $$i x^2 -3i x +2i = 0$$ are the same as $$x^2 -3x +2 =0$$ which is not a complex polynomial. I don't think the author had those types of complex polynomials in mind when he wrote the problem, otherwise it would be a counterexample to the statement of the exercise (of course, ignoring the fact that $$\mathbb{R} \subset \mathbb{C}$$).

I see what you mean though! Thanks for taking a look at it for me.

4. Apr 28, 2009

### n!kofeyn

Just to reiterate what Dick said, you just want to prove that there is one complex root. It makes no difference whether its imaginary part is zero or not, and this is not weak. A real number is a complex number and the complex numbers are not just just those with nonzero imaginary part.
Yes it is a complex polynomial. It just happens to be a real polynomial as well. Once again, real numbers are complex numbers!

5. Apr 28, 2009

### stringy

Indeed, your points have been taken. Since the complex numbers are a closed field, all polynomials with complex coefficients will have a root in the complex numbers. I was not debating that.

I misspoke. Let me not say that Dick's solution was weak, that was not my intent. You have my apologies.

Rather, I believe the author's statement of the problem contains ambiguous language. I know what he's going for, but he said it wrong. Considering the fact that this is a whole section devoted to complex numbers, I believe the author intended to show something with this problem other then a trivial example like all quadratics having roots. That would not fit with the tone of the other problems, and the author's hint points in a completely different direction then the "reals are complex" solution.

I also believe the author was not considering such complex polynomials that can be rewritten as a real polynomial with identical roots.

Let me make a related conjecture (with much more qualification) that I believe fits the author's intent:

Consider a polynomial $$p \left( t \right) = q t^2 + wt+ z$$ where $$q,w,z \in \mathbb{C}$$, $$q,w,z \notin \mathbb{R}$$, and $$q,w,z$$ are not constant multiples of each other. Show that all polynomials of this form have at least one root $$x+yi$$ where $$y \neq 0$$.

I suppose I had interpreted the problem this way and never looked back. I sometimes can't see the forest for the trees, thus making things more difficult then need be. That was entirely my error. May we reconsider this conjecture instead of the first one?

6. Apr 28, 2009

### Dick

That statement doesn't work either. Any nonzero complex number is a multiple of any other. If that's the sort of thing you think they want you to prove then I think you want to assume w/q and z/q are not both real. Then it's true.

7. Apr 28, 2009

### HallsofIvy

The problem with that interpretation is that you can't prove that statement- it's not true! The polynomial equation $ix^2- 7ix+ 12i= 0$ has non-real coefficients but it has real roots, 3 and 4. The only reasonable interpretation of this is that you are asked to prove the "fundamental theorem of algebra", that every polynomial with complex coefficients (which includes the real numbers) has at least one solution in the real numbers (which, again, includes the real numbers).