The Question:
This question is not actually about the complete proof for it, just one step. However, I still feel this is an appropriate title. The question is simply to show that given a polynomial $p(z)$ of degree at least 1, there exists a polynomial $h(z)$ of degree $n-1$ such that $p(z) = (z-z_0)h(z) + p(z_0)$.
My Attempt:
Write
$$
p(z) = a_0 + a_1z + a_2z^2 + \cdots + a_nz^n, \\
h(z) = b_0 + b_1z + b_2z^2 + \cdots + b_{n-1}z^{n-1}.
$$
Then we have
$$
(z-z_0)h(z) = b_0z + b_1z^2 + b_2z^3 + \cdots + b_{n-1}z^n – z_0(b_0 + b_1z + b_2z^2 + \cdots + b_{n-1}z^{n-1}).
$$
By comparing coefficients we find $a_k = b_{k-1} – z_0b_k$, with $b_n = 0$. Thus we can solve for each $b_k$ by working backwards from $b_{n-1} = a_n$.
Now, however, I am a bit stuck on the $p(z_0)$ part. Using the method described above I don't really end up with anything but $p(z_0) = a_0 + z_0b_0$, and I don't seem to have used the fact that $a_n \neq 0$ anywhere. That last remark really bothers me. I feel like I should definitely be using the fact that $p(z)$ is in fact of degree $n$.
Best Answer
First, note that if it is true for $p$ that $$p(z)=(z-z_0)q(z)-p(z_0)\tag1$$ then if $p$ is not constant, $p(z)-p(z_0)$ has the same degree as $p,$ and we get $\deg q=\deg p-1.$ (When $p(z)$ is constant, $q(z)=0.$)
So the degree of $q$ follows just from the existence of a solution to $(1).$
Then, it is easier to first prove it for the specific case $p(z)=z^k.$ $$ \begin{align} p(z)&=z^k\\&=(z^k-z_0^k)+z_0^k\\&=\left((z-z_0)\sum_{i=0}^{k-1}z_0^{i}z^{k-1-i}\right) +p(z_0)\tag2 \end{align} $$
Then note that if it is true for both $p_1(z)$ and $p_2(z)$ polynomials , then it is true for $p(z)=a_1p_1(z)+a_2p_2(z),$ for $a_1,a_2$ complex numbers.
Finally, prove by induction that it is true for any polynomial $p.$
We can use $(2)$ to get an explicit formula. If $$p(z)=\sum_{i=0}^na_iz^i$$
Then $$ \begin{align} p(z)-p(z_0)&=\sum_{i=1}^n a_{i}(z-z_0)\sum_{j=0}^{i-1}z_0^{i-1-j}z^j\\ &=(z-z_0)\sum_{j=0}^{n-1}\left(\sum_{i=j+1}^{n}a_iz_0^{i-1-j}\right)z^j \end{align} $$
So if $$b_j= \sum_{i=j+1}^{n}a_iz_0^{i-1-j}$$ then $$q(z)=\sum_{j=0}^{n-1} b_jz^j$$ is your quotient.
When $n>0,$ $b_{n-1}=a_n,$ this also gives the degree of $q$ is one less than the degree of $p$ $\deg p>0.$
This works for polynomials over any commutative ring of coefficients.
General division algorithm doesn’t over arbitrary rings, but general division algorithm does work over fields, like the fields of complex numbers, real numbers or rational numbers.