Fundamental Theorem of Algebra

Theorem of the week

This week's theorem of the week is the fundamental theorem of algebra, and the picture is the proof!

Theorem

Every non degree-zero polynomial $p(z) = a_nz^n + ...+ a_1z +  a_0$ has a root in $\mathbb{C}$.

Picture proof

To see how the picture proves this, write $z$ as $Re^{i\theta}$, then for all $k$
$$z^k = R^ke^{k i \theta}$$
So for sufficiently large $R$ the $a_nz^n$ term dwarfs all the others and so the image of $\{z\in\mathbb{C}: \lvert z \rvert = R\}$ must go around the origin $n$ times, like the rubber band in the photo.  But when $R= 0$ the image is just $\{a_0\}$ which goes around the origin zero times.  So, for some $0 < r < R$ the image of $\{z\in\mathbb{C}: \lvert z \rvert = r\}$ must cross the origin.  QED.

Less handwavy proof

In order to obtain a contradiction assume $p(z)$ has no zeros.  Then $\frac{z^{n-1}}{p(z)}$ is everywhere differentiable, which in turn means that its closed loop integrals are zero$^{(\dagger)}$.  So, letting $\gamma_R$ be the circle $\lvert z \rvert = R$ we have
$$\begin{align} 0 &= \int_{\gamma_R}\frac{z^{n-1}}{p(z)}dz\\ &= \int_{0}^{2\pi}\frac{R^{n-1}e^{i(n-1)\theta}}{\sum_{k=0}^na_kR^ke^{ik\theta}}Rie^{i\theta}d\theta\\ &= i\int_{0}^{2\pi}\frac{R^ne^{in\theta}}{\sum_{k=0}^na_kR^ke^{ik\theta}}d\theta\\ &= i\int_{0}^{2\pi} \frac{d \theta}{\sum_{k=0}^na_kR^{k-n}e^{i(k-n)\theta}}  \\ &\underset{R \to \infty}{\longrightarrow} \frac{2\pi i}{a_n} \\ \end{align}$$
which was the contradiction we were seeking.

Footnotes

$(\dagger)$ Let's assume $f(z)$ is differentiable in some open set containing $\Omega$ and see if we can show that the closed loop integral around $\Omega$ is zero.  It's sufficient to show the real part is zero since $if(z)$ is also  differentiable.  The following uses Gauss' divergence theorem, and the definition $f(x+iy) = u(x,y)+iv(x,y)$:

$$\begin{align} Re\left(\int_{\partial\Omega} f(z) dz\right)   &= \int_{\partial\Omega} udx - vdy \\  &=\int_{\partial\Omega} (v,u) \cdot (-dy,dx) \\  &=\int_{\partial\Omega} (v,u) \cdot \hat{\textbf{n}} dl \\  &=\int_\Omega v_x + u_y dA \\  &= \int_\Omega Im\left(\frac{\partial f}{\partial x}\right) + Im\left(i\frac{\partial f}{\partial y}\right) dA \\  &= \int_\Omega Im\left(\frac{df}{dz}\right) + Im\left(-\frac{df}{dz}\right) dA \\  &= 0 \end{align}$$