Placing complex polynomial into Taylor series form For the following complex polynomial: 
Write the following polynomials in Taylor form centered at $z=2$:
$z^{10}$
How come this simplifies to just a Taylor series of a binomial?
Detailed explanation will be greatly appreciated!
 A: On one hand, if $f(z)=z^{10}$, then $f(z)=a_0+a_1(z-2)+a_2(z-2)^2+\cdots +a_{10}(z-2)^{10}$, with $a_k=\dfrac{f^{(k)}(2)}{k!}$.  All higher order terms vanish because $f^{(11)}(z)\equiv 0$.  You can directly compute $f^{(k)}(2)$ for $k\in\{0,1,\ldots,10\}$.
On the other hand, $f(z)=z^{10}=(2+(z-2))^{10}=\sum\limits_{k=0}^{10}{10\choose k}2^{10-k}(z-2)^k$ by the binomial theorem.
The fact that these expansions are identical is a consequence of the uniqueness of Taylor series coefficients, but it is also straightforward to verify by induction, without reference to the series, that $f^{(k)}(z)=k!{10\choose k}z^{10-k}=\dfrac{10!}{(10-k)!}z^{10-k}$ for each $k\in \{0,1,\ldots,10\}$ and $z\in\mathbb C$.
A: $$
f(z) = f(z_0) + f'(z_0)(z - z_0) + \frac 12 f''(z_0)(z - z_0)^2 + \ldots
$$
or otherwise
$$
f(z) = \sum_{n = 0}^\infty \frac {f^{(n)}(z_0)}{n!} (z - z_0)^k
$$
Obviously, $f^{(n)}(z_0) = 0$ for all $n > 10$ so you end up with 
$$
f(z) = \sum_{n = 0}^{10} \frac {f^{(n)}(z_0)}{n!} (z - z_0)^k
$$
Next, $f^{(n)}(z) = \frac{10!}{(10-n)!}z^{10-n}$, so
$$
f(z) = \sum_0^{10} \frac {10!}{(10-n)!n!}z_0^{10-n} (z - z_0)^n
$$
which is binomial decomposition for $[z_0 - (z - z_0)]^{10}$, and it makes sense.
PS: here $z_0 = 2$
A: Another way to look at it is that $z^{10}=((z-2)+2)^{10}$ which is of the form $(a+b)^{10}$ and then just expand using your favorite method of obtaining the binomial coefficients. Since this is a finite binomial expansion it converges for all (complex) numbers.
