Fundamental Theorem of Algebra
Theorem: let f(z) be a function that is analytic in the entire complex plane \mathbb{C} (i.e., an entire function). If there exists a real number M > 0 such that |f(z)| \le M for all z \in \mathbb{C}, then f(z) must be a constant function.
To prove it we combines Cauchy’s Integral Formula for derivatives with the ML-bound. We will show that the derivative of the function, f^\prime(z), must be identically zero throughout the complex plane.
Proof: let z be an arbitrary point in \mathbb{C}. From the result of higher-order derivatives for analytic functions here, the value of the derivative at z can be expressed by the integral formula:
f^{\prime}(z) = \frac{1}{2\pi i}\oint_C \frac{f(\zeta)}{(\zeta-z)^2}\, \mathrm{d}\zeta
Since f(z) is entire, this formula is valid for any simple closed contour C that encloses the point z.
For our purposes, we choose the contour C to be a circle of radius R centered at the point z, and for any point \zeta on this contour, we have |\zeta - z| = R.
We now seek to bound the magnitude of f^\prime(z) using the Estimation Inequality (ML-bound).
The length of our circular contour is L = 2\pi R.
By the theorem’s hypothesis, the function is bounded, so |f(\zeta)| \le M for all \zeta \in C.
The magnitude of the denominator in the integrand is |\zeta - z|^2 = R^2.
Now, we apply the ML-bound to the integral for f^\prime(z):
\begin{aligned} | f^{\prime}(z) | & = \frac{1}{2\pi i}\oint_C \frac{f(\zeta)}{(\zeta-z)^2}\, \mathrm{d}\zeta \le \frac{1}{2\pi i}\oint_C \frac{|f(\zeta)|}{R^2}\, \mathrm{d}z = \frac{1}{2\pi }\int_0^{2\pi} \frac{|f(\zeta)|}{R^2}\, \mathrm{d}s \\ & \le M \frac{1}{2\pi}\int_0^{2\pi} \frac{1}{R^2}\, R\mathrm{d}\theta = \frac{M}{2\pi R}\int_0^{2\pi} \,\mathrm{d}\theta = \frac{M}{R} \end{aligned} So we have established the inequality:
|f^\prime(z)| \le \frac{M}{R}
The left-hand side, |f^\prime(z)|, is a fixed value that does not depend on the choice of R.
The right-hand side, however, depends on R. Because f(z) is entire, we are free to let the radius R of our integration path become arbitrarily large.
Taking the limit as R \to \infty:
\lim_{R \to \infty} \frac{M}{R} = 0
This forces the conclusion that |f^\prime(z)| \le 0, and then we must have |f^\prime(z)| = 0, which implies f^\prime(z) = 0.
As the initial choice of z was arbitrary, this result holds for every point in the complex plane.
A function whose derivative is identically zero on a connected domain (here, all of \mathbb{C}) must be constant. Therefore, f(z) is a constant function, which completes the proof.
Liouville’s theorem reveals a difference between complex and real functions.
We have introduced trigonometric functions here and demonstrated that functions like \sin(z) and \cos(z) are entire.
In real analysis, these functions are bounded, with |\sin(x)| \le 1 and |\cos(x)| \le 1 for all x \in \mathbb{R}.
However, in the complex plane, these functions are not constant. By Liouville’s theorem, this implies that they cannot be bounded.
Consequently, there must exist values of the complex variable z for which quantities like |\sin(z)| are greater than 1.
For example, let’s evaluate |\sin(z)| for a purely imaginary number, such as z = 2i. Using the exponential definition of the sine function:
\sin(z) = \frac{e^{iz} - e^{-iz}}{2i} Substituting z = 2i:
\sin(2i) = \frac{e^{i(2i)} - e^{-i(2i)}}{2i} = \frac{e^{-2} - e^{2}}{2i} = -i \frac{e^{2} - e^{-2}}{2} = i \sinh(2) The modulus is therefore:
|\sin(2i)| = |i \sinh(2)| = \sinh(2) = \frac{e^2 - e^{-2}}{2} \approx \frac{7.389 - 0.135}{2} \approx 3.627
which gives a value larger than 1.
Liouville’s theorem offers a direct path to proving the fundamental theorem of algebra.
Theorem: any non-constant single-variable polynomial with complex coefficients has at least one root in \mathbb{C}.
Proof: the proof is by contradiction. Let us assume the contrary and let’s suppose there exists a non-constant polynomial:
p(z) = a_n z^n + \dots + a_1 z + a_0
with n \ge 1 and a_n \neq 0 that has no roots in the complex plane.
If p(z) is never zero, then the function:
f(z) = \frac{1}{p(z)}
is defined and analytic for all z \in \mathbb{C}; it is an entire function.
We now examine whether f(z) is bounded.
As |z| \to \infty, the behavior of p(z) is dominated by its leading term:
|f(z)| = \frac{1}{|p(z)|} = \frac{1}{|z|^n}\frac{1}{\left|a_n + \frac{a_{n-1}}{z} + \cdots + \frac{a_0}{z^n}\right|}
so |p(z)| \to \infty. Consequently:
|f(z)| = \frac{1}{|p(z)|} \to \frac{1}{\left|a_n\right||z|^n} \to 0
as |z| \to \infty.
This implies there exists a radius R sufficiently large, such that for all |z| > R, we have |f(z)| < 1.
On the closed disk |z| \le R, the continuous function |f(z)| is bounded by the extreme value theorem. Therefore, f(z) is bounded on the entire complex plane.
Now we can apply Liouville’s theorem: since f(z) is a bounded entire function, it must be a constant.
If f(z) is constant, then p(z)=1/f(z) must also be constant.
This, however, contradicts the initial assumption that p(z) was a non-constant polynomial.
The contradiction forces us to reject the initial assumption, thereby proving that every non-constant polynomial must have at least one root in \mathbb{C}.