Suppose $f(z) = z^0$ and $g(z) = 0$ where $z \in \mathbb{C}$
When $z \ne 0$ we have $f(z) = 1$ And $g(z) = 0$
When $z = 0$ the values of $f(z)$ and $g(z)$ is of indeterminate or undefined.
In such a case, can one find the contour integral on a circle, say $|z| = \epsilon$ And use the value of integral or use the residue theorem to find the values of $f(0)$ and $g(0)$? Or because the continuity of $f(z)$ at $z=0$ could be in question (not knowing for sure what the value of $f(0)$ will not make this method rigorous?
I found the answer of this question in discussion about Riemann’s theorem about removable singularity. Probably because this question is currently on hold I cannot add an answer. Therefore I will answer it here.
Riemann Singularity Theorem essentially states that when a function is holomorphic and bounded everywhere in a small punctured disk around $z=z_0$ Except at $z_0$ then such singularity can be removed and the function can be treated as holomorphic at $z=z_0$. In the case of $f(z) = z^0$, $f(0) = 1$ and in case of $g(z)=0^z$, $g(0) = 0$. Thus when it comes to removable singularity, the value at the indeterminate point can be determined by the limit and it is not left as a “subjective” choice! Thus it is not a matter of whether we “want” it to be holomorphic when it comes to removable singularity of a complex function. (However for real functions it is probably subjective).
I am speculating that the convention $0^0 = 1$ may be coming from the observation that most continuous functions we deal with are nonzero and when they are raised to $0$ They will resolve to $1$. (Examples: $z^z, sin(z)^z$, etc)
BTW, there are two other types of singularities:
If $z=z_0$ is a pole, then the function is unbounded in the disk near $z=z_0$, and we cannot use the above theorem.
If there is an essential singularity at $z=z_0$, then a path can be found for ANY complex value one desires as $z$ approaches $z_0$!!!
Singularities are explained very nicely by Prof. Steven J Miller in a lecture available on the YouTube here
If you say "I want $f:z \mapsto z^0$ to be a holomorphic function", then your argument with the contour integral (as well as the continuity argument) shows that "$0^0 = 1$" is the only possible definition in order for $f$ to be as you require.