Complex Analytic Proof of Fundamental Theorem of Algebra The following question appears in Rosenlicht's Introduction to Analysis, which is primarily a real analysis text, but contains some problems about complex analysis:

I believe I've managed to prove parts (a) and (b) by showing $|f(z)| > \frac{1}{2}|z|^n$ when $|z|$ is large enough. From there, you can show you can find a disk centered around the origin that contains the point $z \in \mathbb{R}$ corresponding to the minimum.
I'm stuck on parts (c) and (d). I don't understand the algebraic manipulations that allow one to express $f(z)$ in the manner shown in (c). Does it depend on the remainder theorem? I also don't quite follow the argument of part (d), particularly the inequality shown in that part.
Thanks.
 A: For question $c)$, remember that for polynomials, Taylors's formula is an exact formula, i.e. if the polynomial $f$ has degree $n$, 
$$f(z)=f(\zeta)+f'(\zeta)(z-\zeta)+\dots+\frac{f^{(n)}(\zeta)}{n!}(z-\zeta)^n. $$
Some $f^{(i)}(\zeta)$ maybe $0$, depending on the multiplicity of $\zeta$ as a root of $f(z)-f(\zeta)=0$. If $m$ is the first order of derivation such that $f^{(m)}(\zeta)\ne 0$, Taylor's formula is
\begin{align}
f(z)&=f(\zeta)+\frac{f^{(m)}(\zeta)}{m!}(z-\zeta)^m+\dots+\frac{f^{(n)}(\zeta)}{n!}(z-\zeta)^n\\
&=f(\zeta)+\frac{f^{(m)}(\zeta)}{m!}(z-\zeta)^m\biggl(1+\dots+\frac{m!\,f^{(n)}(\zeta)}{n!\,f^{(m)}(\zeta)}(z-\zeta)^{n-m}\biggr).
\end{align}
A: Assuming here that Bernard's answer is sufficient for part c), I shall go through part d). [Let me know if further guidance is needed for part c)]
We know that we can write $$f(z) = f(\zeta) a(z-\zeta)^m(1+(z-\zeta)g(z))$$
Where $m \in \Bbb{Z}_+, a \in \Bbb{C}\setminus\{0\}$ and $g(z)$ is a polynomial in $z$.  
Now following the hint we can choose $\alpha \in \Bbb{C}$ with $\alpha^m = -\frac{f(\zeta)}{a}$ as $a \ne 0$.
writing $z=\zeta + t\alpha$ and plugging into the formula for $f(z)$ that 3) gives us gives:
$$f(\zeta + t\alpha) = f(\zeta) + a(t\alpha)^m(1+(t\alpha)g(\zeta + t\alpha))$$
Which becomes, using $\alpha^m = -\frac{f(\zeta)}{a}$:
$$f(\zeta + t\alpha) = f(\zeta)(1-t^m+t^{m+1}\alpha g(\zeta + t\alpha))=f(\zeta)(1+t^m(t\alpha g(\zeta + t\alpha)-1))$$
However we now let $t$ tend to $0$, and note that as $g(z)$ is a polynomial;
$\lim \limits_{t \to 0} \ g(\zeta + t\alpha)=g(\zeta)$ by continuity of polynomials, which is a finite value; so:
$\lim \limits_{t \to 0} \ [t\alpha g(\zeta + t\alpha)]=0$  
Hence we can make $-1<t^m(t\alpha g(\zeta + t\alpha ) -1)<0$ and so;
$|f(\zeta + t\alpha)| =|f(\zeta)||(1 + t^m(t\alpha g(\zeta + t\alpha)-1))|<|f(\zeta)|$ as long as $f(\zeta) \ne 0$  
Which is the contradiction that was required.
A: You can also prove (c) this way: Let $p(z)= c_0 + c_1z + \cdots +c_nz^n$ be a polynomial of degree $n\ge 1.$ Let $a\in \mathbb C.$ Then there exist constants $d_1,\dots, d_n,$ $d_n\ne 0,$ such that
$$\tag 1 p(z) = p(a) + d_1(z-a) + d_2(z-a)^2 +\cdots + d_n(z-a)^n$$
for all $z\in \mathbb C.$ Conclusion (c) follows readily from this.
Proving $(1)$ is simple when $n=1.$ Suppose it holds for $n\ge 1.$ Let $p$ be a polynomial of degree $n+1.$ Then $p(z) - p(a)$ is a polynomial of degree $n+1$ that vanishes at $a.$ Hence $(z-a)$ divides $p(z) - p(a),$ i.e., $p(z)-p(a) = (z-a)q(z)$ for some polynomial $q$ of degree $n.$ Because $(1)$ holds for $n,$ $q(z) = q(a) + d_1(z-a) + \cdots + d_n(z-a)^n.$ Thus
$$p(z) - p(a) = (z-a)(q(a) + d_1(z-a) + \cdots + d_n(z-a)^n)$$ $$= q(a)(z-a) + d_1(z-a)^2 + \cdots + d_n(z-a)^{n+1}.$$
This proves $(1).$
