For $f:X\subseteq\mathbb{R}\to\mathbb{R}$ and $g:X\subseteq\mathbb{R}\to\mathbb{R}$, we may say that $f$ and $g$ are equivalent if $\forall x\in{X}.f(x)=g(x)$.
But for many complex functions $f:Z\subseteq\mathbb{C}\to\mathbb{C}$ and $g:Z\subseteq\mathbb{C}\to\mathbb{C}$, there may be more than one $w\in\mathbb{C}$ such that $f(z)=w$ , so the statement $f(z)=g(z)$ may not be true even if $f$ and $g$ are the same thing.
For example, suppose $f(z)=\left(z^2+2i\right)^\frac{1}{2}$ and $g(z)=\left(z^2+2i\right)^\frac{1}{2}$. Depending on how you evaluate $f$ and $g$, you might arrive at the conclusion that $f(z)=-g(z)\implies f\not\equiv g$. But then, $f$ and $g$ have the same domain, the same codomain, and are defined by the same equation.
How is the "sameness" of $f$ and $g$ expressed given that the equivalence of the values at $z$ is not always guaranteed? Is there a 'proper' way to describe the de-facto of equivalence of complex, multivalued functions that eliminates this sort of confusion?
Best Answer
Usually by "multivalued" function one means a set-valued function. In this case, it make sense to use set equality (inclusion in both ways) to have a consistent definition of pointwise equality of functions.