In the real analytic case the implicit and inverse function theorem are essentially equivalent, i.e. one can be deduced from the other (from what I know it is more usual to prove the inverse function theorem using the implicit function theorem; for the converse see for example Tao's Analysis II).

I am interested in the complex analytic case of this deduction. More precisely I would like to prove a particular case of the implicit function theorem using the inverse function theorem of complex analysis. For reference I will state the two theorems I have in mind.

Theorem. (Inverse function theorem)
Let $f\colon U\to\mathbb C$, $U\subseteq\mathbb C$ open, be holomorphic. Let $z\in U$ such that $f'(z)\ne0$. Then there is a (possibly smaller) neighborhood $U_0\subseteq U$ of $z$ such that the restriction $f|_{U_0}$ is injective. Moreover, $f(U_0)$ is open and the inverse function is holomorphic too.

Theorem. (Implicit function theorem for complex polynomials)
Let $P(z,w)$ be a complex polynomial in two variables. Let $(a,b)\in\mathbb C^2$ such that $P(a,b)=0\ne\partial_wP(a,b)$. Then there are open neighborhoods $U$ and $V$ of $a$ and $b$, respectively, and a holomorphic function $f\colon U\to V$ such that $f(a)=b$ and if $f(z)=w$ for $z\in U$ and $w\in V$ then $P(z,w)=0$.

First off two general remarks. I know how to deduce both theorems from their real analytic counterparts. However, I have seen the first theorem being called implicit function theorem at a few occasions while it really looks like an inverse function theorem (nLab does this too in the real case). I also know a completely complex analytic proof of the second theorem which is far more involved than what I am looking for (and does not use the first theorem directly). I am not interested in these two cases.

Due to first remark I thought it might be possible to deduce the implicit function theorem (at least in this particular case) from the first theorem. But I had now luck in doing so. Specifically, I am not sure how to "reduce the dimension": A multivariate polynomial gives rise to a function $\mathbb C^n\to\mathbb C$ to which the simple version of the first theorem does not apply (or does it coordinatewise?). I tried to fix the first variable in $P(z,w)$ and looking at $P_a(w)=P(a,w)$ instead but by doing this I lost some control over the second variabl. In particular, I could not any longer show that $P(z,w)=0$ if $w=f(z)$.

Is there a simple proof for Inverse function theorem$\implies$Implicit function theorem for complex polynomials which I am missing? Or do I have to really appeal to the real analytic case?

Thanks in advance!


1 Answer 1


This is too long for a comment, and not really an answer. Like you, I've also seen a few complicated proofs of these theorems relying purely on complex analysis (sometimes dealing with very special cases), and I've also seen them being derived using the real-analytic counterparts. However, I've always found this absurd because all this difficulty arises for one single reason: the notion of complex-differentiability/holomorphicity in higher dimensions is usually not introduced (or even if it is, it is introduced in what I find an adhoc manner). This I find peculiar since the definitions are almost identical to the real case (we just have to require the Frechet derivative be complex-linear rather than real-linear).

As such, many of the local/pointwise theorems you know in the real case follow easily in the complex case (eg sum rule, product rule, chain rule, mean-value inequality, all have statements and proofs which generalize verbatim). The inverse and implicit function theorems are really theorems which belong in the multivariable/Banach-space setting (doesn't matter real or complex). In this sense, the exact same statement and proof (eg using Banach's fixed point theorem) valid in the real case holds verbatim in the complex case.

In particular, to go from the inverse to implicit function theorem for a function $P(z,w)$, doesn't have to be a polynomial, define $\Phi(z,w)=(z,P(z,w))$ and apply the inverse function theorem to $\Phi$ about $(a,b)$, then $(P\circ \Phi^{-1})(z,\xi)=P(z,(\text{pr}_2\circ \Phi^{-1})(z,\xi))=\xi$, so the function $(\text{pr}_2\circ \Phi^{-1})(z,0)$ is the desired implicit function (this proof works in any number of dimensions; finite or infinite, real or complex).

The full strength of the theorem is realized after one notes the equivalence between

  • "once complex differentiable/holomorphic on $U$"
  • "$r$-times continuously complex differentiable on $U$",
  • "complex analytic on $U$",

because then one can replace all the $C^r$ assumptions in the statement of the theorems by the seemingly weaker hypothesis of "once complex differentiable/holomorphic".

  • $\begingroup$ (+1) That's actually very helpful. Thanks! It also explains the deeper issue with the way inverse and implicit function are usually presented in the complex setting. I realized too that my main problem for going from inverse to implicit is the missing notion of multidimensional complex differentiability. I suppose an almost identical version of my $1$-dimensional inverse function theorem holds for complex differentiable functions $\mathbb C^2\to\mathbb C$ (or $\mathbb C^n\to\mathbb C$ for that matter)? $\endgroup$
    – mrtaurho
    Commented May 25, 2021 at 19:15
  • 1
    $\begingroup$ @mrtaurho yes (once you have the right definition of complex-differentiability in higher dimensions). You can even generalize to $\Bbb{C}^k\times \Bbb{C}^n\to\Bbb{C}^n$ (where the $n\times n$ submatrix is assumed invertible at a given point), and again the exact same proof I wrote above holds; actually it even holds for $f:X\times Y \to Z$ where $X,Y,Z$ are Banach spaces (real or complex, finite or infinite dimensional), assuming the "partial derivative" $(\partial_2f)_{(a,b)}:Y\to Z$ is a Banach-space isomorphism. $\endgroup$
    – peek-a-boo
    Commented May 25, 2021 at 19:21
  • $\begingroup$ What is the right definition here? simply the Frechet derivative you mentioned? $\endgroup$
    – mrtaurho
    Commented May 25, 2021 at 19:43
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    $\begingroup$ @mrtaurho yes, keeping in mind that if you're dealing with Banach spaces over a field $\Bbb{F}\in\{\Bbb{R},\Bbb{C}\}$, then the Frechet derivative is required to be $\Bbb{F}$-linear. Perhaps this answer of mine might be useful. $\endgroup$
    – peek-a-boo
    Commented May 25, 2021 at 19:48
  • 1
    $\begingroup$ I'd like to thank you again. I used this in a talk I gave (some while ago actually) and your answer and subsequent comments were extremely helpful for gaining intuition. Thanks again! $\endgroup$
    – mrtaurho
    Commented Aug 9, 2021 at 22:57

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