# 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.

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You can solve any equation $p(z)=0$ , where $p(z)$ is a polynomial with Complex coefficients, and $z$ is a Complex variable, but not all functional equations. How do you define $erf(z)$ ? –  user99680 May 5 at 7:03
@user99680 erf(z) is an just example that equation can be very hard to see solution. I could write much more longer , I want to see general proof for analytic function equations . –  Mathlover May 5 at 7:08
Consider $e^z=0$ –  leonbloy May 5 at 16:15
There are some nice proofs of Picard's little theorem here –  Mike Miller May 8 at 16:03
Maybe that is work opening a new question. Something along the lines of "What is the most elementary proof of little Picard's theorem?" or similar. –  5xum May 9 at 13:47

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.

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if we do not use $\bar z$???. I mention we will use just analytic functions. –  Mathlover May 5 at 7:04
"has $0$ in its range" (rather than domain). –  Andres Caicedo May 5 at 7:11
Well, if the equation is analytic and everywhere defined (which are sensible conditions) we have Picard's theorem, which tells us that essentially all equations do have solutions! –  Mariano Suárez-Alvarez May 5 at 7:12
You've left out the vital restriction that the function not be constant. $f(z) = 1$ only has solutions to $f(z)=w$ for one value of $w$. –  user2357112 May 5 at 10:38
@Taemyr No, user2357112 was correct as I should point out that $f$ is not constant when quoting Picards theorem. I edited my answer. –  5xum May 5 at 10:49

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 $c=0$.

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