The local mapping of complex squaring at zero and at one

Given: The entire function $f(z)=z^2$, the algebra of complex polynomial derivatives (Linearity, product, reciprocal, and quotient rules for complex derivatives), and the fact that a nonzero complex number has exactly two square roots (The $n$-th roots of a complex number and the $n$ distinct roots of unity for every $n\ge 1$).

If $\mathrm{\Omega }$ is a complex domain, $f:\mathrm{\Omega }\to \mathbb{C}$ is nonconstant and holomorphic, and $a\in \mathrm{\Omega }$, then the local degree is ${\mathrm{deg}}_{a}f={\mathrm{ord}}_a(f-f(a))$ (Local degree of a nonconstant holomorphic map).

If $\mathrm{\Omega }$ is a complex domain, $f:\mathrm{\Omega }\to \mathbb{C}$ is nonconstant and holomorphic, $a\in \mathrm{\Omega }$, and $m={\mathrm{deg}}_{a}f$, then every neighbourhood $N$ of $a$ contains an open neighbourhood $V$ for which some $\rho >0$ gives exactly $m$ preimages in $V$ for $0<\mathrm{\mid }w-f(a)\mathrm{\mid }<{\rho }^m$, while $f(a)$ has only the preimage $a$, counted with multiplicity $m$ (A local degree-m holomorphic map has m nearby sheets).

If $f$ is nonconstant and holomorphic on a complex domain $\mathrm{\Omega }$ and $a\in \mathrm{\Omega }$, then $f^{\mathrm{\prime }}(a)\mathrm{\ne }0$, ${\mathrm{deg}}_{a}f=1$, local injectivity at $a$, and biholomorphy between neighbourhoods of $a$ and $f(a)$ are equivalent (Holomorphic inverse function theorem and local-degree criterion).