# Proof that a function is holomorphic

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?

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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

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).

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@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.

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