Fundamental theorem of algebra avoiding complex analysis Suppose you have to give a proof of fundamental theorem of algebra to someone who did not know complex analysis. You can’t therefore use Liouville’s Theorem. How would you proceed avoiding complex analysis and assuming that that person know a little bit of complex numbers?
 A: The following proof, while not the simplest, uses only the intermediate value theorem, the fundamental theorem of symmetric polynomials and splitting fields. The following proof is the one given by Ben Moonen, which was based on a proof by Gauss, and was given in Moonens Dutch lecture notes about rings and fields (originally written by Lenstra and Oort, see https://www.math.ru.nl/~bmoonen/RingenLich/RingenLichamen2017.pdf, page 150-152).
We will first need several lemmas.
Lemma 1 If $f$ is a quadratic polynomial in $x$ with complex coefficients, then $f$ has a complex zero.
Proof. We can assume that $f$ can be written as $f=x^2+\beta x+\gamma$ (by dividing by the leading coefficient). By completing the square, we can write $f=(x-\frac{1}{2}\beta)^2-(\frac{1}{4}\beta^2-\gamma)$, so therefore it is sufficient to prove that $\frac{1}{4}\beta^2-\gamma$ has a square root.
We write $\frac{1}{4}\beta^2-\gamma=a+bi$. If $b=0$, then a square root is $\sqrt{a}$ if $a$ is positive or zero, and $i\sqrt{|a|}$ if $a$ is negative. (These square roots exist, as $g=x^2-|a|$ has a root by the intermediate value theorem, since $g(0)<0$ and $g(a+1)>0$.)
We can therefore assume $b$ to be nonzero. We thus seek $c,d\in\mathbb{R}$ such that
$$(c+di)^2=(c^2-d^2)+2cdi=a+bi$$
This is equivalent to
$$c^2-d^2=a,\quad 2cd=b.$$
Because $b\neq 0$, also $c,d\neq 0$, and thus we can write:
$$c=\frac{b}{2d},\text{ thus } \frac{b^2}{4d^2}-d^2=a.$$
So the real number $d$ therefore has to be a root of the polynomial
$$g=4x^4+4ax^2-b^2.$$
As the polynomial has real coefficients, we can use the intermediate value theorem to show that there is indeed a real root $d$ by noticing that $g(0)<0$ and $g(x)>0$ for $x$ large enough. We can also find $c$ by $c=\frac{b}{2d}$. So now we found that $a+bi$ has a root $c+di$ in $\mathbb{C}$. This proves lemma 1.

Lemma 2 Let $f$ be a polynomial of odd degree with real coefficients, then $f$ has a real root.
Proof. We can assume that the leading coefficient of $f$ is positive. Then $f(x)>0$ if $x$ is large enough, and $f(x)<0$ if $-x$ is large enough. By the intermediate value theorem $f$ therefore has a root. This proves lemma 2.

Lemma 3 Assume every non-constant real polynomial has a root in $\mathbb{C}$, then every non-constant complex polynomial also has a root in $\mathbb{C}$.
Proof.
Let $g=a_n x^n+a_{n-1}x^{n-1}+...+a_0$, and define $\bar{g}$ to be the conjugate polynomial $g=\bar{a}_n x^n+\bar{a}_{n-1}x^{n-1}+...+\bar{a}_0$. Define $f$ to be their product $f=g\cdot \bar{g}.$ Because $\overline{g\cdot h}=\bar{g}\cdot \bar{h}$, we have $\bar{f}=f$, thus $f$ has real coefficients.
Because $f$ has degree $2n$, it is not a constant and therefore has a root. So there is a $\alpha\in \mathbb{C}$ such that $$g(\alpha)\bar{g}(\alpha)=0.$$
If $g(\alpha)\neq 0$, then $\bar{g}(\alpha)=0$, and thus $g(\bar{\alpha})=0$.
So we find that $g$ has a root in $\mathbb{C}$. This proves lemma 3.

Now the only thing we need before the proof is the existence of a splitting field of a polynomial $f$:
there is a field $F$ containing $\mathbb{C}$ which also contains all roots of $f$ (so $f$ can be written as product of linear factors).
Now we are ready to prove the theorem itself.
Proof of the fundamental theorem of algebra
Because of the previous lemma, it suffices to only consider polynomials with real coefficients with leading coefficient $1$. So let $f=x^n+a_{n-1}x^{n-1}+...+a_0$ be such a polynomial of degree $n=2^k u$, where $u$ is odd and $k$ is a positive integer. We will prove the result by induction on $k$. If $k=0$, then $f$ has odd degree, so lemma 2 shows that it has a root.
So assume $k>0$, so $n$ is even. By using coefficients $\alpha_i$ from $F$, we can write
$$f=(x-\alpha_1)(x-\alpha_2)...(x-\alpha_n).$$
Take $c$ to be any real number and consider
$$g_c=\prod_{1\leq i<j\leq n}(x-(\alpha_i+\alpha_j+c\alpha_i\alpha_j))$$
(where $\prod$ means that the product of all the terms is taken).
Every coefficient of $g_c$ is a symmetric polynomial in $\alpha_1,...,\alpha_n$, so by the fundamental theorem on symmetric polynomials, $g_c$ has real coefficients.
The degree of $g_c$ is equal to the amount of choices of $i$ and $j$ such that $1\leq i<j\leq n$, and this is equal to $\frac{1}{2}n(n-1)=2^{k-1}\cdot u\cdot (n-1)$. $n-1$ is odd, so the number of factors of $2$ in the degree of $g_c$ is $k-1$. By the induction hypothesis we therefore find that $g_c$ has a root in $\mathbb{C}$. But the roots of $g_c$ are exactly the $\frac{1}{2}n(n-1)$ expressions $\alpha_i+\alpha_j+c\alpha_i\alpha_j$.
We conclude: for every real number $c$ there are $i$ and $j$ with $1\leq i<j\leq n$ such that $\alpha_i+\alpha_j+c\alpha_i\alpha_j\in \mathbb{C}$. But there are infinitely many choices for $c$, but only $\frac{1}{2}n(n-1)$ for $i$ and $j$. Therefore there are $c, c'\in \mathbb{R}$ such that there are $i, j$ such that both
$$\alpha_i+\alpha_j+c\alpha_i\alpha_j\in \mathbb{C}\text{ and }\alpha_i+\alpha_j+c'\alpha_i\alpha_j\in \mathbb{C}.$$
By considering their difference we thus also have $(c-c')\alpha_i\alpha_j\in \mathbb{C}$ so $\gamma=\alpha_i\alpha_j\in \mathbb{C}$ and also $\beta=\alpha_i+\alpha_j\in \mathbb{C}$.
So $$(x-\alpha_i)(x-\alpha_j)=x^2-\beta x+\gamma$$
is a quadratic polynomial with complex coefficients. Therefore by lemma 1, it has a complex root so $\alpha_i\in\mathbb{C}$ or $\alpha_j\in\mathbb{C}$. As these were roots of $f$, $f$ has a complex root. This completes the proof of the fundamental theorem of algebra.
A: I would explain that if $f\colon[a,b]\longrightarrow\mathbb C$ is a continuous map, then we can define$$\int_a^bf(x)\,\mathrm dx=\int_a^b\operatorname{Re}f(x)\,\mathrm dx+i\int_a^b\operatorname{Im}f(x)$$and that, with this definition, the following familiar properties of the integrals of real functions hold:


*

*If $F$ is a primitive of $f$ , then $\displaystyle\int_a^bf(x)\,\mathrm dx=F(b)-F(a)$.

*$\displaystyle\frac{\operatorname d}{\operatorname dx}\int_a^b\varphi(x,y)\,\mathrm dy=\int_a^b\frac{\partial\varphi}{\partial x}(x,y)\,\mathrm dy$.

*$\displaystyle\left\lvert\int_a^bf(x)\,\mathrm dx\right\rvert\leqslant\int_a^b\bigl\lvert f(x)\bigr\rvert\,\mathrm dx$.


Now, let $P\colon\mathbb C\longrightarrow\mathbb C$ be a non-constant polynomial. Assume that it has no zeros. Then you can define$$\begin{array}{rccc}q\colon&[0,\infty)&\longrightarrow&\mathbb C\\&r&\mapsto&\displaystyle\frac1{2\pi}\int_0^{2\pi}\frac1{P\bigl(r\cos(\theta)+ir\sin(\theta)\bigr)}\,\mathrm d\theta.\end{array}$$It is clear that $q(0)=\frac1{P(0)}\neq0$. On the other hand, it follows from the third property mentioned above and from the fact that $P(x)$ is a polynomial function with degree $n$, for some $n\in\mathbb N$, that $\lim_{r\to\infty}q(r)=0$.
Now, I shall prove that $\bigl(\forall r\in(0,\infty)\bigr):q'(r)=0$. In fact\begin{align}q'(r)&=\frac1{2\pi}\frac{\mathrm d}{\mathrm dr}\int_0^{2\pi}\frac1{P\bigl(r\cos(\theta)+ir\sin(\theta)\bigr)}\,\mathrm d\theta\\&=\frac1{2\pi}\int_0^{2\pi}-\frac{\bigl(\cos(\theta)+i\sin(\theta)\bigr)P'\bigl(r\cos(\theta)+ir\sin(\theta)\bigr)}{P^2\bigl(\cos(r\theta)+ir\sin(\theta)\bigr)}\mathrm d\theta\\&=\frac i{2\pi}\int_0^{2\pi}\frac{\bigl(-\sin(\theta)+i\cos(\theta)\bigr)P'\bigl(r\cos(\theta)+ir\sin(\theta)\bigr)}{P^2\bigl(r\cos(\theta)+i\sin(r\theta)\bigr)}\mathrm d\theta\text{ (because $-1=i\times i$)}\\&=\frac i{2\pi r}\int_0^{2\pi}\frac{\operatorname d}{\operatorname d\theta}\frac1{P\bigl(r\cos(\theta)+ir\sin(\theta)\bigr)}\,\mathrm d\theta\\&=\frac i{2\pi r}\left(\frac1{P\bigl(r\cos(2\pi)+ir\sin(2\pi)\bigr)}-\frac1{P\bigl(r\cos(0)+ir\sin(0)\bigr)}\right)\\&=0.\end{align}There is a contradiction here, because then $q$ should be a constant function. But, if $q$ is a constant function, then you could not possibly have simultaneously $q(0)\neq0$ and $\lim_{q\to\infty}q(r)=0$.
