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!

  • $\begingroup$ Quotients of holomorphic functions are holomorphic where they're defined. $\endgroup$ Dec 28, 2012 at 5:30
  • $\begingroup$ What is the definition of holomorphic that you're using? $\endgroup$ Dec 28, 2012 at 5:37
  • $\begingroup$ @JonasMeyer: The limit definition (mimics the definition of a derivative for real variables). $\endgroup$
    – anon271828
    Dec 28, 2012 at 5:44
  • $\begingroup$ @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. $\endgroup$ Dec 28, 2012 at 5:49

1 Answer 1


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.

  • $\begingroup$ 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! $\endgroup$
    – anon271828
    Dec 28, 2012 at 5:42
  • $\begingroup$ You're very welcome $\endgroup$ Dec 28, 2012 at 5:44

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