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In Stein and Shakarchi's book, Chapter 1, Exercise 7 asks us to show that $$\left|\frac{w-z}{1-\overline{w}{z}}\right|<1$$ if $|z|<1$ and $|w|<1$, with equality if either $|z|=1$ or $|w|=1$. I was able to show this much, and for the second part of the question, most of it is immediate from the above, except showing that for a fixed $w\in\mathbb{D}=\{z\in\mathbb{C}:|z|<1\}$ $$F(z)=\frac{w-z}{1-\overline{w}{z}}$$ is holomorphic. I can do it if I completely expand everything and get it to the form $F(x,y)=u(x,y)+iv(x,y)$ (I haven't actually written everything out, only assured myself that I could actually get it to such a form). From here I should be able to check the Cauchy-Riemann equations and verify the partial derivatives are continuous, which is enough to show that $F$ is holomorphic, but I was wondering if there was a better/more elegant way of showing it? Thanks!

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Quotients of holomorphic functions are holomorphic where they're defined. –  Jesse Madnick Dec 28 '12 at 5:30
    
What is the definition of holomorphic that you're using? –  Jonas Meyer Dec 28 '12 at 5:37
    
@JonasMeyer: The limit definition (mimics the definition of a derivative for real variables). –  anon271828 Dec 28 '12 at 5:44
    
@anon271828: That isn't quite an answer to my question, but I take it you mean that a function is defined to be holomorphic if it is everywhere (complex) differentiable? Sometimes continuity of the derivative is also assumed (although it is redundant). In calculus, you know how to find the derivative of $f(x)=\dfrac{3-x}{1-3x}$? I just want to point out that this is very similar. –  Jonas Meyer Dec 28 '12 at 5:49
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2 Answers

up vote 6 down vote accepted

Quotients of holomorphic functions are holomorphic where they're defined -- that is, wherever the denominator is nonzero.

In your case, the denominator is never zero because.... (details left to you)

Edit: Of course, you do have to mention (or prove) that both the numerator and denominator are, in fact, holomorphic!

And if this fact about quotients hasn't been proven... well, I guess you'll have to do that, too. The proof is exactly the same as the real-variable case.

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Excellent! Thank you. Sometimes I just overthink things... Yes, it was a previous theorem that quotients of holomorphic functions are holomorphic. I didn't stick around to think about it much more than to observe the proof for real variables should suffice. Again, thanks for clearing up the minor confusion I had! –  anon271828 Dec 28 '12 at 5:42
    
You're very welcome –  Jesse Madnick Dec 28 '12 at 5:44
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Try to use Morera's theorem as an alternative way to prove the function is holomorphic.

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It seems like that will probably work, so I'll spend some time thinking about it. On the other hand, I don't think Morera's theorem has officially been introduced in the textbook. Are there other more-elementary methods to proving the function is holomorphic? Is it possible what I did is, in fact, the intended method? –  anon271828 Dec 28 '12 at 5:35
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@anon271828 This is overkill. Do what Jesse Madnick suggests. –  Potato Dec 28 '12 at 5:38
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@MhenniBenghorbal I did not downvote you, but this answer is needlessly complicated. We would have to verify the integral over every triangle is zero, which is an order of magnitude more time consuming than just proving that quotients of holomorphic functions are holomorphic. –  Potato Dec 28 '12 at 18:32
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@MhenniBenghorbal Many people vote based on the helpfulness of the answer, and technical correctness is a necessary but not sufficient condition for being helpful. –  Potato Dec 29 '12 at 23:38
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@Mhenni: That link does not appear to be very relevant to this situation. At the other question, it is a useful technique to change the order of integration in applying Morera's theorem. Here, we are talking about a rational function, and it is beyond my imagination to see how Morera's theorem is a useful way to show that rational functions are holomorphic. –  Jonas Meyer Jan 6 '13 at 5:56
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