How to show that $\frac{1}{w-z} = \frac{1}{w-a} + \frac{z-a}{(w-a)^2} + ... $, $w,z \in \mathbb{C}$ In the book of The theory of Functions by Titchmarsh, at page 83, in the proof of the fact that analytic functions can be expressed as power series, it is stated that
$$\frac{1}{w-z} = \frac{1}{w-a} + \frac{z-a}{(w-a)^2} + \dots \frac{(z-a)^n}{(w-a)^{n+1}} \dots   ,$$
where $w$ is on the circle $\Gamma$ with centre $a$ and radius $\sigma < \delta$, where $|z-a| < \delta$ .
However, I cannot see how the RHS is equal to LHS ? I mean this is basically the Taylor's theorem for complex functions, but this is what we are trying to show in here, so it is confusing.
 A: \begin{align}\frac1{w-a}+\frac{z-a}{(w-a)^2}+\frac{(z-a)^2}{(w-a)^3}+\cdots&=\frac1{w-a}\sum_{n=0}^\infty\left(\frac{z-a}{w-a}\right)^n\\&=\frac{\frac1{w-a}}{1-\frac{z-a}{w-a}}\\&=\frac1{w-a-z+a}\\&=\frac1{w-z}.\end{align}
A: Hint. We have that
$$\frac{1}{w-z}=\frac{1}{w-a-(z-a)}=\frac{1}{w-a}\cdot \frac{1}{1-\frac{z-a}{w-a}}.$$
Now consider the case when $\left|\frac{z-a}{w-a}\right|<1$ and recall the main properties of the geometric series.
A: Suppose we have $S=\frac 1p+\frac {q}{p^2}+\frac {q^2}{p^3}+\dots$ and we multiply by $\frac pq$ then $$\frac pq S=\frac 1q+\frac 1p+\frac q{p^2}+\dots=\frac 1q+S$$ so that $$S=\frac 1{p-q}$$
Now apply with $p=w-a, q=z-a, p-q=w-z$
We are looking at simple geometric progression - but this technique of multiplying or dividing by the common ratio can help to get the calculations right. Also, even with quite simple expressions in the terms, using $p$ and $q$ here to express the shape of the expression saves some algebra, and I find it helps to reduce mistakes.
Obviously, in the context, attention also needs to be given to the conditions for convergence.
