One interesting phenonmenon in complex analysis is the following,
If $f:\mathbb C\to\mathbb C$ is complex differentiable at point $a$ ($\equiv$derivative is a spiral similarity), and a local homeomophism with inverse $g$ near $a$, then $g$ is complex differentiable at point $b=f(a)$.
Its proof is as the page from Ahlfors' Complex Analysis, https://i.stack.imgur.com/UTdED.png
Same argument applies to usual one variable differentiation and probably any normed field, since we are allowed to invert quotient before taking limits on their norm, hence showing an implcit linkage between analysis and algebra.
However in multivariable calculus, one cannot invert quotient to prove similar theorem. By taking $f(x+h)=y+k, f(x)=y$ as in usual proof of inverse function theorem, one is required to show that for some real $\lambda>0$, $|k|\ge \lambda |h|$, so that one can bound the usual quotient $\dfrac{|h-f'(x)^{-1}k|}{|k|}$ by $\dfrac{|k-f'(x)h|}{|h|}$ up to some multiplicative constant. This is far from inverting quotients.
Is there any explicit explanation of this interplay between algebra and analysis?
P.S. One interesting corollary found is that if a homeomorphism $f:U\to V$, where both are subset of $\mathbb R^2$, and has invertible differential at point $A$, then its inverse is differentiable at point $f(A)$. (by normalising function so that it is complex differentiable)
P.S.2 Its generalisation (not verified): If $f:\mathbb C^n\to\mathbb C^n$ is local homeomorphism (from $U$ to $V$) and differentiable at $a\in U$ with invertible differential, then its local inverse is differentiable at $b$.
By this, if $f:\mathbb R^{2n}\to\mathbb R^{2n}$ satisfies similar condition, its inverse is differentiable at $b$.
P.S.3 Maybe an interesting question is whether one can define some algebraic structure on $\mathbb C^n$ like bicomplex number such that one can invert quotients for proof. (But it just need not be commutative, causing more problem.)
The question stems from the curious phenomenon that above version of inverse function theorems hold for complex numbers or normed fields that can be proven by inverting quotient. Hence if one can prove the theorem in general for $R^n$, then the question is $90\%$ solved, what remains unsolved is simply why algebra affects analysis by inverting quotient.
Proposition: If $f:\mathbb R^n\to\mathbb R^n$ is local homeomorphism (from $U$ to $V$) and differentiable at $a\in U$ with invertible differential, then its local inverse $g$ is differentiable at $b=f(a)$.
Proof:
WLOG $a=b=0$ and $f'(a)=I$, note that a $(n-1)$-sphere $S_R$ of radius $R$ centered $0$ is mapped to an annulus (region between two concentric sphere).
Let $\epsilon_R$ be the constant in differentiation ($|f(x)-f(0)-I(x-0)|\le \epsilon_R(x-0)|$ for all $x$ in ball of radius $R$, and $\epsilon_R\to0$ as $R\to0$ and $\epsilon_R$ as increasing function.) (Here we assume $R$ sufficiently small so that the supremum of $\epsilon_R$ is $<1$)
So $S_R$ is mapped into the annulus of two radii $R(1\pm\epsilon_R)$. By this, a sphere $S_R'$ in $V$ must be mapped from the annulus in $U$ of outer radius $\dfrac{R(1+\epsilon_R)}{1-\epsilon_R}$ and inner radius $\dfrac{R(1-\epsilon_R)}{1+\epsilon_R}$. If viewed in individual points $c$ (in the sphere of radius $R_c$) of $V$, its preimage must be in the ball included in the corresponding annulus in $U$ of diameter $\dfrac{{R_c}(1+\epsilon_{R_c})}{1-\epsilon_{R_c}}-\dfrac{{R_c}(1-\epsilon_{R_c})}{1+\epsilon_{R_c}}$.
By that, we know that $|g(c)-g(0)-I(c-0)|\le\dfrac{{R_c}(1+\epsilon_{R_c})}{1-\epsilon_{R_c}}-\dfrac{{R_c}(1-\epsilon_{R_c})}{1+\epsilon_{R_c}}$.
The term $R_c=|c-0|$ and the factor there is decreasing to 0 uniformly on each sphere. Hence the theorem is proven.
Next, the magic of complex differentiation here is not plainly inversion of quotient, one point that I miss is the common sense that if $\lim\frac{f(x)}{g(x)}=a\neq0$, then $\lim\frac{g(x)}{f(x)}=\frac1a\neq0$. If we are more concerned with this facts, one actually see that its proof includes items like $\frac{1}{a\pm\epsilon}$ as $\epsilon\to0$, the error range of it ($\frac{1}{a-\epsilon}-\frac{1}{a+\epsilon}$) is exactly the same as the second factor in above proof, they also embrace the same idea. This actually shows that algebra doesn't affect analysis in a macroscopic way, its effect must lie at least below the 'atom' (limit of inverse) of the analysis.