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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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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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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