Intuitively this makes sense, since complex roots come in conjugate pairs, so if that's all there was, you'd expect your degree to be even.

So, if any$z_0$ is a root of $f(z)$, then $\overline{z_0}$ is also a root.

Supposing that $f(z_0) = a_0(z_0)^{2k+1} + ... + a_{2k}z_0 + a_{2k+1} = 0$


$f(\overline{z_0}) = a_0(\overline{z_0})^{2k+1} + ... + a_{2k}\overline{z_0} + a_{2k+1}$

Should also equal zero, but I can't think of a simple argument why. I know that $a_0(\overline{z_0})^{2k+1} = a_0(\overline{z_0})^{2k}\overline{z_0} = a_0(\overline{z_0^2})^{k}\overline{z_0}$

But it seems to me like things are getting messier than they should.

I think once I have shown this, the fact that there must be a real root will follow immediately, since in $\mathbb{C}$ a polynomial of degree $n$ has at most $n$ distinct roots.

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    $\begingroup$ Hint: Just look at the behavior as $x\to \pm \infty$. $\endgroup$ – lulu Jun 7 '16 at 17:52

You are on the right track. If $P$ is a polynomial with real coefficients, then $P(\overline{z}) = \overline{P(z)}$. Thus if $P(z) = 0$, $P(\overline{z}) = \overline{P(z)} = \overline{0} = 0$. Your only problem here is to show that if $w$ is a root of degree $n$, then so is $\overline{w}$. But this follows by differentiation, for if $P^{(n)}(w) = 0$, $P^{(n)}(\overline{w}) = 0$ (for $P^{(n)}$ is a real polynomial if $P$ is). $w$ is an $n'th$ root of $P$ if and only if $$P(w) = P'(w) = \dots = P^{(n)}(w) = 0$$ so we see from the above calculation that $w$ is an $n'th$ root of $P$ if and only if $\overline{w}$ is an $n$'th root as well.

Now applying the fundamental theorem of algebra, write $P = \prod (X - w_i)^{n_i}$. If $w_i$ is not real, then we have shown that $(X - \overline{w})^{n_i}$ divides $P$ as well. Thus since the number of non real divisors must work out to an odd ordered polynomial (since the factors appear in pairs), we must have a real root somewhere.

A simpler solution uses basic calculus. If $f$ is odd, you can show that as $z \to \infty$, either $f(z) \to \infty$, or $f(z) \to - \infty$ (this holds for all polynomials, regardless of the degree). In the other direction, you will have $\lim_{z \to -\infty} f(z) = - \lim_{z \to \infty} f(z)$ (which only holds for odd polynomials). Now apply the intermediate value theorem. The intuitive idea is that the leading coeffiencient of the highest order term of the polynomial governs the asymptotic growth of the polynomial, so for large values (negative or positive), the polynomial follows the direction of the leading coefficient.


If $f(z)$ is of odd degree, then

either $\lim_\limits{x\to\infty} f(x) = \infty$ and $\lim_\limits{x\to-\infty} f(x) = -\infty$ or

$\lim_\limits{x\to\infty} f(x) = -\infty$ and $\lim_\limits{x\to-\infty} f(x) = \infty$

In either case, there is a sign change.

And since polynomial functions are continuous. What does the intermediate value theorem have to say about that?

By the way, this is one of the ways to express the Fundamental Theorem of Algebra.

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    $\begingroup$ The Fundamental Theorem of Algebra states that a polynomial of degree $n$ with complex coefficients has $n$ complex roots, not that a polynomial of odd degree with real coefficients has a real root. What do you mean by your final statement? $\endgroup$ – Clayton Jun 7 '16 at 18:11

Suppose $f$ has no real roots. Then $f(\alpha) = 0$ if and only if $f(\overline{\alpha}) = 0$ (1). Furthermore, for any root $\alpha$ of $f$ we have $\alpha \neq \overline{\alpha}$. So we can write $$f(z) = \prod_{i=1}^{n}(z-\alpha_i)^{k_i} (z-\overline{\alpha_i})^{q_i},$$ where $n=\deg(f)$, $\alpha_i$ are distinct complex numbers and $k_i$ and $q_i$ are the multiplicities of $\alpha_i$ and $\overline{\alpha_i}$, respectively. If we show $k_i = q_i$ for every $i$ then the degree of $f$ is even, a contradiction. To show this it is sufficient to show $$\frac{f}{(z-\alpha_i)^j}(\overline{\alpha_i}) =0 \text{ for } 0 \leq j \leq k_i \text{ and } \frac{f}{(z-\alpha_i)^{k_i}}(\overline{\alpha_i})\neq 0. $$ However, this follows immediately from (1).


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