Mathematics desk
< July 21	<< Jun \| July \| Aug >>	July 23 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

July 22

Showing inverse of a bijective holomorphic function on open set is holomorphic

I am pretty sure I proved this in detail, following a proof in my book. But, I am trying to get a quick proof in case I need to prove it on a qualifying exam just to use the result. By quick, I mean I want to explain it a bit but not show much of the detail, basically name a couple theorems. Is the following enough?

First, f(z) is never 0 since it is one-to-one. By the inverse function theorem, $(f^{-1})'(z)={\frac {1}{f'(f^{-1}(z))}}$ for each point. Since f(z) is holomorphic, f'(z) is also, so 1/f'(z) is also as the quotient of holomorphic functions is holomorphic as long as the denominator is not 0.

Thanks. StatisticsMan (talk) 01:04, 22 July 2009 (UTC)[reply]

Which inverse function theorem are you using, exactly? And how are you showing that injective holomorphic functions have nonzero derivative? Algebraist 01:08, 22 July 2009 (UTC)[reply]

First, I had the derivative formula incorrect and have changed it. As to which inverse function theorem I'm using, I do not know I guess. Are you asking because the formula was wrong? The detailed proof I got from the book mostly starts by showing the inverse is continuous at each point. Then, it shows the derivative is equal to 1 / f'(f^{-1}(z)) at each point. This is the result I want really, though I don't see the exact thing in the article on Inverse function theorem here. If I consider the total derivative, that is the complex derivative, and consider

\mathbb {C} =\mathbb {R} ^{2}

, then it says that the derivative of the inverse exists and is continuously differentiable. But, it does not give a formula for it. Is it always the same formula? As far as showing that an injective holomorphic function have nonzero derivatives, this is something that is tricky to me. My book gives a theorem that says if a nonconstant holomorphic function on a connected open set which has f(P) = Q with order k, then there exists a small disk around Q such that every point in the disk (other than Q) has exactly k distinct preimages in a small disk around P. But, I don't see how this implies a nonzero derivative implies that the function is locally one-to-one, as they say. StatisticsMan (talk) 01:45, 22 July 2009 (UTC)[reply]

I asked about your inverse function theorem because there is a complex inverse function theorem which contains the result you're trying to prove, and presumably would not be accepted as part of a proof of it in an exam. Algebraist 01:48, 22 July 2009 (UTC)[reply]

Well, I would like to know how to prove this directly and in a nice way. But, for now I'm working on a problem asking if there is a bijective holomorphic function which maps the unit disk onto the complex plane. The easy solution is, no, because its inverse is entire and bounded and thus constant. If I take some time to explain that, including using Liouville's theorem and the inverse function theorem and such, then perhaps it would not be trivial and would be good enough to get credit. If I have time later, I can come back to try to prove more. But, the point is, I want a quick proof of the fact to use in cases like this. StatisticsMan (talk) 01:56, 22 July 2009 (UTC)[reply]

Also, is the complex version just the same basic thing? I looked up "inverse" in my 3 main complex analysis books and found nothing basically. StatisticsMan (talk) 01:57, 22 July 2009 (UTC)[reply]

I would call the following result an inverse function theorem: let f be a holomorphic function from an open set U to the complex plane, let x be in U, and suppose f'(x)≠0. Then there is a neighbourhood V of x such that f is injective of V, the inverse map is holomorphic, and the derivative of the inverse is given by [whatever the formula is]. Algebraist 02:17, 22 July 2009 (UTC)[reply]

Here's how the book I have proves it:

First they prove an injective holomorphic function f on an open set U has non-zero derivative. Suppose f'(z₀) = 0. Then

f(z)-f(z_{0})=a(z-z_{0})^{k}+G(z)

for some a ≠ 0, k ≥ 2, and G(z) a function that vanishes to order k+1 at z₀, so then we can choose a radius δ > 0 around z₀ small enough so that |G(z)| < (a/2)δ^k = c for all z within δ of z₀ and so that f'(z) ≠ 0 for z ≠ z₀. Let

F(z)=a(z-z_{0})^{k}-c

, so that f(z) - f(z₀) - c = F(z) + G(z). |F(z)| = c > |G(z)| along the circle of radius δ, so by Rouché's theorem, f(z) - f(z₀) - c has k (not necessarily distinct) zeros in the circle. z₀ is not a root, so f' is non-zero at each root, so the roots are distinct, so f is not injective. So I guess the basic idea is that if a holomorphic function has a multiple root at some point, then by adding a small constant we get a function that has distinct roots near the point, which shows that neither function is injective in a neighborhood of the point.

Then they prove the inverse function part just using the definition of the derivative. Let g = f^-1 on f(U) and w = f(z).

\lim _{w\to w_{0}}{\frac {g(w)-g(w_{0})}{w-w_{0}}}=\lim _{f(z)\to f(z_{0})}{\frac {z-z_{0}}{f(z)-f(z_{0})}}

which converges to 1/f'(z). You don't need to show that 1/f' is holomorphic, just that the complex derivative of g exists everywhere in f(U).

(I hope I didn't screw any of this up.) Rckrone (talk) 03:54, 22 July 2009 (UTC)[reply]

What book is this from? StatisticsMan (talk) 00:58, 30 July 2009 (UTC)[reply]