# What are the functions obtained by complex polynomials evaluated at complex numbers

Given a complex polynomial $$f \in \mathbb{C}[x]$$, we can induce a function $$ev_{\mathbb{C}}f: \mathbb{C} \to \mathbb{C}, x \mapsto ev(c)(f)$$, where ev stands fpr evaluation. Denote the set of all such evaluated function as $$EV_{\mathbb{C}[x]}^{\mathbb{C}}$$For any function $$g:\mathbb{C} \to \mathbb{C}$$, it naturally induces an unique function $$\overline{g}: \mathbb{R}^{2} \to \mathbb{R}^{2}$$, namely, by separating out the real and imaginary part. Now applying it to functions in $$EV^{\mathbb{C}}_{\mathbb{C}[x]}$$, we can basically obtain a map $$h: EV^{\mathbb{C}}_{\mathbb{C}[x]} \to \mathbb{R}[x_{1},x_{2}] \times \mathbb{R}[x_{1},x_{2}]$$.

Intuitively, if we write out the complex polynomial, replace $$x$$ by $$x_{1} +x_{2}i$$, and expand it out, we obtain a polynomial in the form of $$f_{1}+f_{2}i$$, where $$f_{1},f_{2} \in \mathbb{R}[x_{1},x_{2}]$$

So I wonder what property does $$h$$ hold. In particular, is it surjective? I felt it is not, but I cannot prove that there exists a concrete example.

I'm just thinking about this when learning some basics about algebraic geometry. I'm not very familiar with complex analysis, so please be forgiving if this is a stupid question. Also, feel free to modify my notation, I made them up.

• Is $x \mapsto ev(c)(f)$ a typo? I can't understand what $c$ is.
– chi
Commented Jul 20 at 21:02

No, $$h$$ is not surjective. If $$(p_1(x_1,x_2),p_2(x_1,x_2))$$ is in the range of $$h$$, then, for some polynomial function $$p(z)\in\Bbb C[z]$$, $$p_1(x_1,x_2)=\operatorname{Re}(p(x_1+x_2i))$$ and $$p_2(x_1,x_2)=\operatorname{Im}(p(x_1+x_2i))$$. But then, by the Cauchy-Riemann equations, $$\frac\partial{\partial x_1}p_1=\frac\partial{\partial x_2}p_2$$ and $$\frac\partial{\partial x_2}p_1=-\frac\partial{\partial x_1}p_2$$. So, for instance, if you take $$p_1(x_1,x_2)=x_1$$ and $$p_2(x_1,x_2)=0$$, then $$(p_1(x_1,x_2),p_2(x_1,x_2))$$ is not in the range of $$h$$.

Your title question has a useful positive answer if you are willing to bypass your attempted method.

In brief, $$f : \mathbb C \to \mathbb C$$ is a polynomial if and only if it is differentiable at each point of $$\mathbb C$$ and at infinity!

To make sense out of this silly statement, we have to set up some notation. Consider the so-called "extended complex plane" $$\mathbb C^* = \mathbb C \cup \{\infty\}$$ (also known as the Riemann sphere). We can define the extended function $$f^* : \mathbb C^* \to \mathbb C^*$$ by the formula $$f^*(z) = \begin{cases} f(z) & \quad\text{if z \in \mathbb C} \\ \infty &\quad\text{if z = \infty} \end{cases}$$ Now we want to change coordinates, rewriting the function $$f^*$$ using the "multiplicative inverse coordinate" $$w = z^{-1}$$ in both the domain and the codomain. Multiplicative inversion is usually defined, of course, only for $$\mathbb C - \{0\}$$. But we can extend it to all of $$\mathbb C^*$$ by setting $$0^{-1}=\infty$$ and $$\infty^{-1}=0$$.

Changing coordinates, we obtain the function $$F : \mathbb C^* \to \mathbb C^*$$ defined by the formula $$F(w) = (f^*(w^{-1}))^{-1}$$ and note that $$F(0)=0$$.

With this notation, the function $$f : \mathbb C \to \mathbb C$$ is a polynomial if and only if its complex derivative $$f'(z)$$ exists at each $$z \in \mathbb C$$, and the function $$F$$ satisfies the following property:

There exists an open set $$U \subset \mathbb C$$ such that $$0 \in U$$ and $$F(U) \subset \mathbb C$$ and the complex derivative $$F'(0)$$ exists.

Incidentally, the proof of forward direction of this equivalence runs in parallel to one of the proofs of the fundamental theorem of algebra. If $$f(z)$$ is a polynomial function of degree $$d$$ then from the formula for $$F(w)$$ one shows that $$F(0)$$ is a zero of order $$d$$. From this conclusion one can immediately obtain the property of $$F$$ written above. Also, from this conclusion one can continue on, using some differential topology, to conclude that $$f$$ has exactly $$d$$ zeroes counted with multiplicity.