Uniqueness of Powerseries in an arbitrarily small neighborhood of $0$ in $\mathbb C$

Question

I have troubles understanding a proof in "Visual Complex Analysis". What we want to show is that for two powerseries $c_0 + c_1z + c_2z^2 + ...$ and $d_0 + d_1z + d_2z^2 + ...$, equality in a neighborhood of 0 implies their coefficients must be equal. The proof in the book is the following(translated from german to english):

"Seting $z=0$ we get $ c_0 = d_0$ so we can cut them out of both equations. Dividing by z and setting $z=0$ it follows that $c_1 = d_1$. Repeating this argument we get that all the coefficients are equal".

Why can we divide by $z$? Where does this proof use that the two powerseries are equal in a neighborhood of 0?

Answer 1

Since $c-0=d_0=1$, we have$$c_1z+c_2z^2+\cdots=d_1z+d_2z^2+\cdots$$in the neighborhood of $0$. So, for each $z\neq0$ in that neighborhhod, we have, after dividing both sides by $z$:$$c_1+c_2z+\cdots=d_1+d_2z+\cdots$$Bu then\begin{align}c_1&=\lim_{z\to0}c_1+c_2z+\cdots\\&=\lim_{z\to0}d_1+d_2z+\cdots\\&=d_1.\end{align}So, $c_1=d_1$. And now, start all over again.

Answer 2

Those series are in the form $c_nz^n$ and $d_nz^n$, it means that they are series of functions.

Setting $z=0$ we get $c_0 +c_10+...+c_n0 = c_0$ and we get $d_0+d_10+...+d_n0 = d_0$. So we have $c_0=d_0$.

We have the series $c_0+c_1z+...c_nz^n$ and $d_0+d_1z+...+d_nz^n = c_0+d_1z+...+d_nz^n$.

Then, $\frac{1}{z}\sum c_nz^n = \frac{c_0}{z}+c_1+...+c_nz^{n-1} = \frac{c_0}{z}+\sum_{n=1} c_nz^{n-1}$.

In $z=0$ we get $\frac{c_0}{z}+(c_1+c_2z+...+c_nz^{n-1}) = \frac{c_0}{z} + c_1$ and $\frac{d_0}{z}+\sum_{n=1} d_nz^{n-1}$.

Also in $z=0$ we get $\frac{d_0}{z}+(d_1+d_2z+...+d_nz^{n-1}) = \frac{d_0}{z} + d_1 = \frac{c_0}{z}+d_1$. So, $c_1 = d_1$.

Doing this for every term we get that $\sum c_nz^n = \sum d_nz^n$

Answer 3

Going back (almost) to first principles, and proving only what is needed to fill the gap:

Let $K$ be either $\mathbb{R}$ or $\mathbb{C}.$

Suppose that $(a_n)_{n\geqslant0}$ is a sequence in $K$ such that for some $w \in K \setminus \{0\}$ the series $\sum_{n=0}^\infty a_nw^n$ converges.

Then there exists $A \geqslant 0$ such that $\left\lvert a_nw^n \right\rvert \leqslant A$ for all $n \geqslant 0.$

If $z \in K$ and $\left\lvert z \right\rvert < \left\lvert w \right\rvert,$ then $$ \left\lvert a_nz^n \right\rvert = \left\lvert a_nw^n \right\rvert\left(\frac{\left\lvert z \right\rvert}{\left\lvert w \right\rvert}\right)^n \leqslant A\left(\frac{\left\lvert z \right\rvert}{\left\lvert w \right\rvert}\right)^n, $$ therefore the series $f(z) = \sum_{n=0}^\infty a_nz^n$ converges absolutely, by the Comparison Test.

Fix some $u \in K \setminus \{0\}$ such that $\left\lvert u \right\rvert < \left\lvert w \right\rvert.$

If $z \ne 0$ and $\left\lvert z \right\rvert \leqslant \left\lvert u \right\rvert,$ then because the series $\sum_{n=1}^\infty a_nz^{n-1}$ is also absolutely convergent, $$ \left\lvert\frac{f(z) - f(0)}{z}\right\rvert = \left\lvert \sum_{n=1}^\infty a_nz^{n-1} \right\rvert \leqslant \sum_{n=1}^\infty \left\lvert a_n \right\rvert \left\lvert z \right\rvert^{n-1} \leqslant \sum_{n=1}^\infty \left\lvert a_n \right\rvert \left\lvert u \right\rvert^{n-1} = B, \text{ say.} $$

Therefore $\left\lvert f(z) - f(0) \right\rvert \leqslant B \left\lvert z \right\rvert,$ for all $z \in K \setminus \{0\}$ such that $\left\lvert z \right\rvert \leqslant \left\lvert u \right\rvert.$

So $f$ is continuous at $0.$ That is, $$ \lim_{z \to 0} \, \sum_{n=0}^\infty a_nz^n = a_0. $$

I don't think even this minimal proposition is "obvious", but am I missing something?