<s>[INST] value of a $\frac{ p_t(z)}{z^m}$ Fundamental Theorem of Algebra by using Algebraic topology
In wikipidea it is written that

let $p(z)$ be a polynomial in $\mathbb{C}$ ;


$p(z)=z^m +a_1z^{m-1} +\cdots+ a_m$


Define a homotopy: $p_t(z)=tp(z) + (1-t)z^m$
then
$$\frac{ p_t(z)}{z^m} =1  + t \left(\frac{a_1}{z} +\cdots+\frac{a_m}{z}\right)$$

My doubt :Im not getting why $\frac{p_t(z)}{z^m} =1  + t (\frac{a_1}{z} +\cdots+\frac{a_m}{z})\text{?}$
My thinking :
\begin{align}
& \frac{ p_t(z)}{z^m} =\frac{tp(z)}{z^m}  + \frac{(1-t)z^m}{z^m} \\[8pt]
= {} & \frac{tp(z)}{z^m}+(1-t)\neq1  + t \left(\frac{a_1}{z} +\cdots+\frac{a_m}{z}\right)
\end{align}
 [/INST] $p_t(z) = tp(z) + (1-t)z^m = z^m + t(p(z) - z^m)$ so $$\frac{p_t(z)}{z^m} = 1 + t(\frac{p(z)}{z^m} - 1) = 1 + t(1 + \frac{a_1}{z} + \frac{a_2}{z^2} + \cdots + \frac{a_m}{z^{m-1}} - 1) = \\ 1+t(\frac{a_1}{z} + \frac{a_2}{z^2} + \cdots + \frac{a_m}{z^{m-1}})$$
</s>