# For a polynomial $p(z)$ with real coefficients if $z$ is a solution then so is $\bar{z}$

I can "see" it intuitively, though I do not know how correct this is: in a complex conjugate we change the sign of all imaginary parts, and since the effect of all imaginary parts cancels out on the whole, this change of sign would not matter.

I have tried, but I am unable to prove it. I tried using the polar forms.

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Show that conjugation respects sums and products. –  Qiaochu Yuan Jul 14 '11 at 23:28
–  Aryabhata Jul 14 '11 at 23:32
For degree 2 it was asked here. –  Américo Tavares Jul 14 '11 at 23:40

For any $p=a_0+\cdots+a_nx^n\in\mathbb{C}[x]$, there is a conjugate polynomial $\overline{p}=\overline{a_0}+\cdots+\overline{a_n}x^n$. Note that if $z$ is a root of $p$, i.e. $$p(z)=a_0+\cdots+a_nz^n=0,$$ then $\overline{z}$ is a root of $\overline{p}$, i.e. $$\overline{p}(\overline{z})=\overline{a_0}+\cdots+\overline{a_n}\,\,(\overline{z})^n=0,$$ because $$\overline{p}(\overline{z})=\overline{a_0}+\cdots+\overline{a_n}\,\,(\overline{z})^n=\overline{a_0}+\cdots+\overline{a_nz^n}=\overline{a_0+\cdots+a_nz^n}=\overline{0}=0,$$ where we have used that $\overline{ab}=\overline{a}\,\overline{b}$ and $\overline{a+b}=\overline{a}+\overline{b}$ for all $a,b\in\mathbb{C}$.

Now it only remains to note that if $p\in\mathbb{R}[x]$, then $p=\overline{p}$.

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Here is an explanation that draws on some very basic ideas in Galois theory. The idea is to observe that the conjugation map, $$z \mapsto \overline{z},$$ is an $\textit{automorphism}$ of $\mathbb{C}$ that fixes everything in $\mathbb{R}$ pointwise. To say that conjugation is an automorphism means that it "passes through" all the operations we have on a field. Moreover, conjugation extends to a map on polynomials with complex coefficients, by acting on each of the coefficients. And since it is a field automorphism, it does this in such a way that if $f(z) = 0$, then $\overline{f}(\overline{z}) = 0$ as well. In the case where $f$ is a $\textit{real}$ polynomial, so that $\overline{f} = f$, this means exactly that if $z$ is a root, then so is $\overline{z}$.

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Here is a simple proof of the Complex conjugate root theorem.

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Your intuition is correct, by the way. It, and all of the above answers, depend on the basic facts that $\overline{ab}=\overline{a}\overline{b}$ and $\overline{a+b}=\overline{a}+\overline{b}$. You can verify both these facts with a straightforward calculation, breaking $a$ and $b$ into their real and imaginary parts, but here is an intuitive explanation of why these facts are true.

Everything about the arithmetic of $\mathbb{C}$ is determined by a) the arithmetic of the reals, and b) the fact that $i^2=-1$. Put another way, $\mathbb{C}$ is the system you get by adding to the reals a number "$i$" satisfying the equation $z^2=-1$. This equation governs everything there is about the behavior of $i$. But once you have built this system, you can see that $-i$ satisfies the exact same equation, since $(-i)^2=-1$. Since it obeys the same defining relation, $-i$ functions in exactly the same way $i$ does with respect to the arithmetic of the complex numbers. $i$ "might as well have been" $-i$. If you have any equation in the field of complex numbers and you replace $i$ with $-i$ everywhere in the equation (while leaving all the real numbers untouched), it will remain true. Taking conjugates is precisely replacing $i$ with $-i$ everywhere. This is what's behind the fact that conjugation is so nicely behaved.

(In case useful: this is really an informal version of a standard argument from elementary field theory / Galois theory that if $k$ is a field, and $k(\alpha)$ and $k(\beta)$ are simple extensions of $k$, and both $\alpha$ and $\beta$ have the same minimal polynomial over $k$, then there is an isomorphism $\phi: k(\alpha) \rightarrow k(\beta)$ that is the identity on $k$ and sends $\alpha \mapsto \beta$. If $\beta$ happens to be in $k(\alpha)$ as in this case with $k=\mathbb{R}$, $\alpha = i$, $\beta = -i$, then $k(\alpha)$ and $k(\beta)$ are the same field and $\phi$ is an automorphism.)

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HINT $\$ A ring homomorphism $\rm\:h\:$ preserves sums and products so compositions of such, i.e. polynomial forms. Hence if $\rm\:h\:$ fixes the coefficients of a polynomial then it preserves its roots.

In detail, if $\rm\ h(x+y) = h(x) + h(y),\ \ h(x\:y) = h(x)\ h(y)\:,\:$ then by induction we deduce $$\rm h(a_0 + a_1\ r+\:\cdots\:+ a_n\ r^n)\ =\ h(a_0) + h(a_1)\ h(r)+\:\cdots\:+h(a_n)\ h(r)^n$$

i.e. $\rm\ h(p(r)) = p_h(h(r))\:.\:$ If $\rm\ h\$ fixes all coefficients $\rm\: h(a_{\:k}) = a_{\:k}\:$ then $\rm\ p_h = p\$ therefore $\rm\: p(r) =\: 0\:\ \Rightarrow\:\ p_h(h(r)) = h(p(r)) = h(0) = 0\:,\:$ so $\rm\:\ r\:$ root of $\rm\:p\ \Rightarrow\ h(r)\:$ root of $\rm\:p_h = p\:.$

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Here is a related question I once thought about. (I'm submitting this as an answer even though it's a comment -- it seems I've exceeded the comment length limit.)

Suppose we turn things around and consider "real-conjugation" where, for real numbers $a$ and $b$, the real-conjugate of $a + bi$ is $-a + bi$, and we consider algebraic equations with pure imaginary coefficients. Is it true for every algebraic equation with pure imaginary coefficients that, whenever $a + bi$ is a solution, then $-a + bi$ is also a solution? There are easy examples (even linear equations) in which this fails, and I'll leave it as an exercise for the reader to come up with an explanation for what goes wrong (besides the fact that there is a counterexample).

Incidentally, I wrote up a short two paragraph note about this for the Reader Reflections column of The Mathematics Teacher and received the following referee comment: "It was felt that the mathematics contained in your letter was too obscure and inaccessible for high school students."

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