Does the Fundamental Theorem of Algebra hold true for infinite polynomials?

Question

I know that by the Fundamental Theorem of Algebra, every polynomial of positive degree has a zero in $\mathbb{C}$. Does this also hold for polynomials of infinite degree? Like, for example, the Taylor expansion of $e^z$ or $\cos z$?

Answer 1

The fundamental theorem does not hold for infinite polynomials.

Your own example of $\mathrm e^z$ is a good one. We have $\mathrm e^z \neq 0$ for all $z \in \mathbb C$.

This raises a really important question.

All of the non-constant, finite Taylor Polynomials for the exponential function satisfy the fundamental theorem. For example:

• $1+z = 0$ has, when counted with multiplicity, one solution over $\mathbb C$.
• $1+z+\frac{1}{2}z^2 = 0$ has, when counted with multiplicity, two solutions over $\mathbb C$.
• $1+z+\frac{1}{2}z^2+\frac{1}{3!}z^3 = 0$ has, when counted with multiplicity, three solution over $\mathbb C$.
• $1+z+\frac{1}{2}z+\cdots+\frac{1}{n!}z^n = 0$ has, when counted with multiplicity, $n$ solution over $\mathbb C$.

For the Taylor Polynomial of degree $n$, let $R_n$ be the set of its roots, for example:

• $R_1 = \{-1\}$
• $R_2 = \{-1-\mathrm i, \ -1+\mathrm i\}$

What happens to these roots as $n \to \infty$?

I've calculated (using my computer) $R_1,$ $R_2$, $R_3$ and $R_4$ and found that all of the elements of $R_4$ have larger moduli than the elements of $R_3$, and similarly for $R_3$ and $R_2$, etc.

It seems that the roots are getting bigger in terms of their modulus. It is tempting to think that in the limit, they all retreat to infinity and that is why there are no finite solutions to $\mathrm{e}^{z} = 0$.

Sadly, $\mathrm e^z$ does not seem to behave well at complex infinity - at the North Pole of the Riemann Sphere. If we write $z = x+\mathrm i y$ then we get $\mathrm e^z = \left(\mathrm e^x\cos y\right) + \mathrm i \left(\mathrm e^x \sin y\right)$.

If we let $z \to \infty$ along the positive real axis, i.e. $y=0$, $x>0$ and $x \to +\infty$ then $\mathrm e^z \to + \infty$.

If we let $z \to \infty$ along the negative real axis, i.e. $y=0$, $x<0$ and $x \to -\infty$ then $\mathrm e^z \to 0$.

If we let $z \to \infty$ along either the positive, or the negative imaginary axis, i.e. $x=0$ and $y \to \pm \infty$ then there is no well-defined limit; the value of $\mathrm e^z$ lies in $\{\cos y +\mathrm i \sin y: y \in \mathbb R\}$, but it keeps spiralling around and never settles down.

I would be tempted to say that all of the roots have retreated to infinity and imposed some irreconcilable conditions that cause a discontinuity.

Don't believe a word of this though; it's all personal speculation.

Answer 2

Well, the original poster mentions the Taylor series for exp (z) which has no zeroes. So. The Fundamental Theorem of Algebra fails for infinite degree.

Oh well.