TITLE: Can we prove that all equations can be solved via complex numbers?
QUESTION [19 upvotes]: $x^2+1=0$ 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.

REPLY [3 votes]: 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 $c=0$.