Can we prove that all equations can be solved via complex numbers?

$x^2+1=0$ cannot be solved via real numbers.

Because of this, we extend the real numbers to complex numbers.We can solve $x^2+1=0$ and $x^2+x+1=0$ equations after we define complex numbers.

I wonder if we can solve all equations ( includes only the functions that are analytic.) via complex numbers or not? If It is yes, how can we prove that claim?

For example: Can $z^{100}-5z+2=e^{i.\operatorname{erf}(z)}$ be solved via complex numbers?

where $\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}}\int_{0}^x e^{-t^2}\,\mathrm dt$

Note: This is just an example, I am not wondering the solution for a special example, I am wondering if a general proof is possible or not.

Update: I mention the functions that are analytic. $\bar z$ or $\Re{(z)}$ are not analytic functions.

Thanks for answers.

Solution 1:

All polynomial equations with non-constant polynomials with complex coefficients can be solved with complex numbers. This is the fundamental theorem of algebra. Link here.

All equations in general can not. For example, $z\bar z = -1$ has no solutions in $\mathbb C$.

In general, if you are asking if every equation $f(z) = 0$ has a solution in $\mathbb C$, you are asking if every function $f:\mathbb C\to\mathbb C$ has $0$ in its range (codomain). This is of course not true. There are many many functions which do not have $0$ in their codomain, of which $z\bar z+1$ is only one. There exist much uglier functions with this property, for example $$f(z)=\begin{cases}z&\text{ if } z\neq 0\\1&\text{ if } z=0\end{cases}.$$

Even all analytic functions do not contain $0$ in their codomain. For example, $f(z) = e^z$ does not hit $0$ at any point, meaning $e^z=0$ has no solution. However, in some way, analytic functions are the correct way to go. Because of Picard's little theorem also mentioned in the comments (Link) you know that if $f$ is entire (analytic and everywhere defined) and non-constant, then $f(z) = w$ has at least one solution for all values of $w$ except perhaps one. For example, $e^z=w$ has a solution (infinitely many of them) for all values of $w$ except $0$.

Edit: The fact that $f$ is not constant is a valid demand to make, of course, since if a function is constant, the equation $f(z)=0$ translates to $C=0$ for the constant value $C$ of $f$, and such equations are of little interest.

Solution 2:

The idea of "all equations" is somewhat cloudy: What are admissible equations in the context of this question?

Consider the equation $$f(z):=\sqrt{4+z^2}- \log z=0\ .$$ Here $f$ is not uniquely defined in all of ${\mathbb C}$. It would be difficult to make general statements about the existence of solutions if such equations are admitted.

However, according to Picard's Theorem we can say the following: If $f:\ {\mathbb C}\to{\mathbb C}$ is a nonconstant entire analytic function then the equation $$f(z)=c$$ has at least one solution for every given $c\in{\mathbb C}$, with the exception of at most one $\>c\in{\mathbb C}$. As an example consider the exponential function $f(z):=e^z$, which does not take the value $c=0$.