The following is an exercise from my book:
We want to find the Taylor Series of $f(z) = \frac{1}{1-z}$ about $z = 0$ where $z\in \mathbb{C}$. Since $f$ is holomorphic at $z=0$, we can choose to calculate its derivatives at this point by taking the limit along the real axis (as the direction does not matter). This is convenient, since we can hen treat $f$ as a function of a real variable $x$ and calculate the derivatives of $f(x)$ in the usual manner $f'(z) = (1-z)^{-2}, f''(z) = (1-z)^{-3}$, and so on. We find that the standard binomial expansion applies: $$f(z) = (1-z)^{-1} = \sum_{n=0}^{\infty}z^n$$
We know that this converges along the real axis for $|x|<1$. Since the region of convergence must be a disk, we conclude that the expansion converges on the complex plane for $|z|<1$, hence 1 is the radius of convergence.
I might be dumb but I don't really understand what's going on here..
- How is that the binomial expansion? Using the formula $(x+a)^n = \sum_{k=0}^{\infty}\binom{-n}{k}x^ka^{-n-k}$ I get $(1-z)^{-1} = \sum_{k=0}^{\infty}\binom{1}{0}(-z)^{k}$
- I understand that the geometric series converges when $|x|<1$, however the author goes on saying "since the region of convergence must be a disk, we conclude that the expansion converges on the complex plane for $|z|<1$. How can we make all this implications? We know that we $\underline{want}$ the function to converge in a disk on the complex plane, however, all we know is that it converges for $|x|<1$, how to get explain the rest?
- Where is the Taylor Series? Wasn't it the aim of the exercise? Is it just the $\sum_{n=0}^{\infty}(-z)^n$? How can it be the Taylor series expansion for the function $f$ in the complex plane? We obtained this in the real axis, not in the complex plane.
Please someone give me an insight on this exercise, cause it seems such a mess to me