Plotting polynomials over the complex field In here, page 3:

In general, if a polynomial f has a root of multiplicity $k$ at a point $z_0,$ then when $f(z)$ is expanded in powers of $z −z_0$ it will have the form $$f(z) = c(z −z_0)^k + (\text{higher degree terms}).$$ For $z$ close to $z_0,$ the first term in this expansion will dominate the higher degree terms, and therefore we have $f(z) ≈ c(z − z_0)^k .$ Thus, near the point $z_0,$ the picture of $f$ will be similar to the picture of the function $c^k$ near $0. $



*

*What is the explanation for the expansion
$$f(z) = c(z −z_0)^k + (\text{higher degree terms})?$$
In particular, what is the explanation for the sum? I was expecting $c(z −z_0)^k\cdot Q(x).$

*Shouldn't the proximity of $z$ close to $z_0,$ render $c(z − z_0)^k$ insignificant with $z − z_0 \to 0$? Or is it precisely because it tends to $0$?
 A: In the first place, what you were expecting is not wrong at all. In the expression
$$f(z)=c(z-z_0)^k+R(z),$$
where $R(z)$ is what in the text is mentioned as higher degree terms,
it is implied that it is the Taylor expansion what's represented there, and that means that $R$ is of the form
$$R(z)=c_{k+1}(z-z_0)^{k+1}+\cdots+c_{n}(z-z_0)^{n},$$
where $n$ is the degree of the polynomial.
1) If this is so, then
$$f(z)=c(z-z_0)^k+c_{k+1}(z-z_0)^{k+1}+\cdots+c_n(z-z_0)^n=(z-z_0)^k\left(c+c_{k+1}(z-z_0)+\cdots+c_n(z-z_0)^{n-k}\right),$$
and so you have what you expected, with $Q(z)=c+\frac{R(z)}{(z-z_0)^k}$, which happens to be indeed a polynomial.
2) If $$z-z_0\approx 0,$$ then yes, you can expect $$c(z-z_0)^k\approx c\cdot 0^k=0.$$
Anyway, while it is trivially true that $f(z)$ will be close to $zero$ when $z$ is close to $z_0$ (because this is a $zero$ of $f$, which happens to be a continuous function), what is mentioned in the text is that the term of order $k$ dominates the others. While this is not very precise, the usual meaning of this is that when $z$ is close to $z_0$, the higher order terms (that is, $R(z)$) go to zero much faster (that is, faster in an essential way) that $(z-z_0)^k$.
There's a precise statement for that. Remember that if both $g$ and $h$ tend to $zero$ as $z\to a$, then it is indeterminate the limit
$$\lim_{z\to a}\frac{g(z)}{h(z)}.$$
This just means that you cannot give an answer with only that information, since if $g(z)\to 0$ that would account for the quotient also going to $0$, but $h(z)\to 0$ would lead to the quotient going to infinity. In a sense, you need to know if one of them go to zero essentially faster than the other (and then 'wins' the contest), or if they go to zero at proportional rates and so the answer is finite but not zero (among other possibilities).
In our case, the precise statement (which is easy to prove) is
$$\lim_{z\to z_0}\frac{f(z)}{(z-z_0)^k}=c$$
(and $c\neq 0$ if the multiplicity of $z_0$ is $k$ exactly).
Another way to see that this implies that the first term dominates the others is to say that from the above limit one can say that if $z\approx z_0$, then
$$\frac{f(z)}{(z-z_0)^k}\approx c,$$
and so
$$f(z)\approx c(z-z_0)^k$$
(that is, the initial expression without $R(z)$).
It's important, anyway, to understand which statements are precise and formal and which ones are just intuitive ideas. In a math text, you wouldn't normally see an imprecise statement formally exposed as a theorem, lemma, corollary, property or like. And if you do, check it out again: maybe what looks like an imprecise statement or notation, actuallly has a precise definition somewhere else in a previous chapter or section.

More on question 1).
A Taylor polynomial of order $n$ around $z=z_0$ for the function $f(z)$ (assuming it is at least $n$ times differentiable at $z=z_0$) is a polynomial $P$ such that
$$\deg P \le n$$
and which satisfies the $n+1$ conditions
$$f(z_0)=P(z_0), \quad f'(z_0)=P'(z_0), \quad f''(z_0)=P''(z_0),\quad \ldots, \quad f^{(n)}(z_0)=P^{(n)}(z_0).$$
It's not hard to see that there is exactly one such polynomial (hence the Taylor polynomial of $f$ of order $n$ at $z=z_0$), and it has a relatively simple expression if it is written in terms of powers of $z-z_0$ instead of just powers of $z$ (although of course, there's no distinction if $z_0=0$). This expression is:
$$P(z)=f(z_0)+f'(z_0)(z-z_0)+\frac{f''(z_0)}{2!}(z-z_0)^2+\cdots+\frac{f^{(n)}(z_0)}{n!}(z-z_0)^n=$$
$$=\sum_{k=0}^n \frac{f^{(k)}(z_0)}{k!}(z-z_0)^k.$$
Given the coincidences between $f$ and $P$, it is expected that $P$ represents to some extent a good aproximation of $f$ for values of $z$ close enough to $z_0$. Even more, for many functions taking $P$ with greater and greater order provides better aproximations, even to the point that with $n\to\infty$ the error of the approximation tends to zero and so the result is $f$ itself. The limit of the Taylor polynomial for $n\to \infty$ is
$$\lim_{n\to\infty}\sum_{k=0}^n \frac{f^{(k)}(z_0)}{k!}(z-z_0)^k=\sum_{k=0}^\infty \frac{f^{(k)}(z_0)}{k!}(z-z_0)^k=$$
$$=f(z_0)+f'(z_0)(z-z_0)+\frac{f''(z_0)}{2!}(z-z_0)^2+\cdots+\frac{f^{(n)}(z_0)}{n!}(z-z_0)^n+\cdots,$$
and this becomes an alternative expression for $f(z)$ know as the Taylor series of $f$ or the Taylor [series] expansion.
Anyway, this coincidence between $f$ and it's Taylor series is not always the case. In the first place, sometimes the coincidence holds for $z$ in a certain set containing $z_0$. For holomorphic functions,this is the set of the $z$ values such that the series converges (it can be proven that it will always be an open disc with center at $z_0$ and perhaps some or all the points of its border). But if this is the case, it is still true that $f$ equals it's Taylor series at that specific set.
Nevertheless, when $f$ is a polynomial of order $n$, it's Taylor polynomial of order $n$ at any $z_0$ is $f$ itself since it satisfies al the conditions (check), and so in this case $P$ is also the Taylor series (which is a finite sum and hence convergent for every $z$, so that $f(z)=P(z)$, it's Taylor polynomial. But attention, this happens only for polynomials: for every other holomorphic function the Taylor series is an infinite sum.
Finally, it can be seen that if $z_0$ is a zero of the polynomial $f$ of multiplicity $m$, then $f(z_0)=f'(z_0)=f''(z_0)=\ldots=f^{(m-1)}(z_0)=0$ and $f^{(m)}(z_0)\neq 0$ (this can be proven from the expressión $f(z)=(z-z_0)^kg(z)$, with $g(z_0)\neq0$). Or the condition on the derivatives can be taken as definition for multiplicity of a root and then prove through Taylor expansion that the last expression is true.
Whatever the chosen definition, the condition on the derivatives implies that
$$f(z)=f(z_0)+f'(z_0)(z-z_0)+\frac{f''(z_0)}{2!}(z-z_0)^2+\cdots+\frac{f^{(n)}(z_0)}{n!}(z-z_0)^n=$$
$$=\frac{f^{(m)}(z_0)}{m!}(z-z_0)^m+\frac{f^{(m+1)}(z_0)}{(m+1)!}(z-z_0)^{m+1}+\cdots+\frac{f^{(n)}(z_0)}{n!}(z-z_0)^n.$$
The first term in the last expression would be $c(z-z_0)^m$ with $c\neq 0$, and the others would be the higher order terms.
A: Let the distinct roots of $f$ be $z_0, z_1, \ldots z_m$ with multiplicities $k_0, k_1, \ldots, k_m$. Then we have:
$$
f(z) = b(z - z_0)^{k_0}(z - z_1)^{k_1} \ldots (z - z_n)^{k_m}
$$
where $b$ is the leading coefficient of $f(z)$. If we define:
$$
g(z) = (z - z_1)^{k_1} \ldots (z - z_n)^{k_m}
$$
and write $g$ as a polynomial in $z - z_0$:
$$
g(z) = a_0 + a_1 (z - z_0) + a_2(z - z_0)^2 + \ldots
$$
then $a_0 \neq 0$ (because $z_0$ is not a root of $g$) and we have:
$$
f(z) = b(z - z_0)^{k_0}g(z) = ba_0(z - z_0)^{k_0} + ba_1 (z - z_0)^{k_0+1} + ba_2(z - z_0)^{k_0+2} + \ldots
$$
which has the form:
$$
c(z - z_0)^{k_0} + \mbox{terms of higher degree}.
$$
