Extension of $x^2$ to the complex plane.

Question

Take $f(x)=x^2, x \in \mathbf{R}$. We know that the analytic extension of $f$ to the complex plane is $g(z)=z^2, z \in \mathbf{C}$ , or also $f(x+iy)=(x^2-y^2)+2ixy$. This can be checked via the following steps:

1. Observing that $g$ restricted to the real axes is equal to $f$
2. Via the Cauchy-Riemann conditions we can prove analyticity (complex differentiability)

But the steps $1,2$ seem a bit too much to prove something that seems or should be "obvious". Here is my question:

QUESTION: is there an "obvious"/"trivial by construction" reason why $g(z)=z^2$ is the analytic extension of $f(x)=x^2$ ?

I just put the question as an example in term of a quadratic polynomial but this should apply also to higher order ones.