How to prove or disprove surjectivity of a complex function?

Question

I have the following complex function:

$f:\mathbb{C}\rightarrow\mathbb{C}$ where $f(z)=z^2+2z+3$

This function is trivially not injective since it's not injective in the real subsets.

Now, I am very inclined to think that this function is surjective and I am trying to prove it.

The way I am trying to do it the typical, assume $z_0=f(z)$ for some $z\in \mathbb{C}$ show that this equation always has a solution.

Since these are complex numbers we want the real part to equal the real part and the imaginary part to equal the imaginary part in both sides of the equation. However, trying to solve the above system of 2 nonlinear equations results in some very long expressions and I am having trouble with the algebra.

I want to know if there is a simpler way to show surjectivity or if I am wrong and this function is not surjective.

Answer 1

You seek a zero of $$f(z)-z_0= (z+1)^2+(2-z_0)$$ To find a zero of this, you need to solve $$(z+1)^2=z_0-2$$ So, suppose that $z_0-2$ is an arbitrary complex number. Can you find a square root, i.e. $y\in\mathbb{C}$ such that $y^2=z_0-2$? If so, then $y=z+1$, so you seek $z=y-1$.

Answer 2

Here is a proof that works for rational functions which I though it might be worth mentioning.

The rational functions are the complex analytic maps $\mathbb{P}^1 \to \mathbb{P}^1$, by compactness the image is closed, by the open mapping theorem the image is open, so it must be surjective by connectedness. (this in particular proves fundamental theorem of algebra)

For an elementary proof for this particular case see vadim123's answer.