Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How can i show that the function $$f\colon\mathbb{C}\setminus\{-i\}\rightarrow\mathbb{C}\quad \text{defined by}\quad f(z)= \frac{1+iz}{1-iz}$$ is an holomorphic function?

share|cite|improve this question
By verifying the Cauchy-Riemann equations, for example, or by appealing to the fact that a quotient $\frac{f}{g}$ of holomorphic functions is holomorphic wherever $g$ doesn't vanish. So: pay special attention to $z=-i$ where you'll have trouble. – t.b. Jun 29 '11 at 17:49
That function is not a function from $\mathbb{C}$ to $\mathbb{C}$, since it is not defined at $z=-i$. However, it is holomorphic on $\mathbb{C}\setminus\{-i\}$. – Zev Chonoles Jun 29 '11 at 17:50
@Zev Chonoles. This is absolutely true – Katy23 Jun 29 '11 at 19:04
up vote 4 down vote accepted

Elementary operations or compositions of holomorphic functions give holomorphic functions on the maximal domain where the functions are defined. This is a consequence of the rules of derivation for product, ratio and compositions of functions. In your case, you have a ratio of two holomorphic functions, and that is a holomorphic function on the domain where the denominator does not vanish (this is mentioned in the comment of Theo Buehler).

share|cite|improve this answer
@Katy23: Well, you could do it as Jonas Meyer does in the other examples, using only the definition of the complex derivation, or you could just see that these two are polynomials with complex coefficients, and therefore elementary functions, which are holomorphic. – Beni Bogosel Jun 29 '11 at 19:10
Katy asked me in a comment how to prove that the numerator and denominator are holomorphic functions, then the comment was removed. – Beni Bogosel Jun 29 '11 at 19:15

One way is by differentiating it. You have $f(z)=\frac{1+iz}{1-iz}=-1+2\cdot\frac{1}{1-iz}$, so when $iz\neq 1$,

$\begin{align*}\lim_{h\to0}\frac{f(z+h)-f(z)}{h}&=\lim_{h\to 0}\frac{2}{h}\left(\frac{1}{1-i(z+h)}-\frac{1}{1-iz}\right)\\ &=\lim_{h\to 0}\frac{2}{h}\cdot\frac{1-iz-(1-i(z+h))}{(1-i(z+h))(1-iz)}\\ &\vdots \end{align*}$

The next steps involve some cancellation, after which you can safely let $h$ go to $0$.

This is not a very efficient method, but it illustrates that it only takes a bit of algebra to work directly with the definition of the derivative in this case. More simple would be to apply a widely applicable tool, namely the quotient rule, along with the simpler fact that $1\pm iz$ are holomorphic.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.