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It is known that one-point compactification of complex plane is homeomorphic to sphere $S^2$. So one can extend polynomial $f$ with complex coefficients using the following diagram.

\begin{array} A\mathbb{C} & \stackrel{f}{\longrightarrow} & \mathbb{C} \\ \downarrow{} & & \downarrow{} \\ S^2 & \stackrel{f'}{\longrightarrow} & S^2 \end{array}

Suppose that $f_1(z)=z^k$. I succeeded to show that the degree of $f_1'$ equals k.

Proof. Let polynomial $u:S^1 \rightarrow S^1$ be $u(z)=z^k$. (Hatcher 2.32) Then the degree of $u$ equals k. And the suspension of $u$ equals $f_1'$. It is well known that suspension preserves degree.

What I want to show is the following: Suppose that $f_2$ is polynomial of degree $k$. Is degree of $f_2'$ $k$?

I think it sufficients to show two functions are homotopic to each other but not easy. First, I tried to check local degree of $f_2'$ at their roots; $f_2(z)=0$ and failed. Is any good idea for this one?

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Hint: the degree of a map $f:N\rightarrow M$ can also be computed at every regular value $y\in M$. So you just have to take a number $z\in C$, $z\neq 0$ determine $f^{-1}(z)$ which has $k$-elements, $f$ preserves the orientation on the neighborhood of every element in the preimage. (The differential of $f$ sends the base $(1,i)$ to $(nz^{n-1},nz^{n-1}i)$) So the degree is $k$.

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  • $\begingroup$ Actually, I solved this by calculating local degree at infinity. Anyway, thanks for answering. $\endgroup$
    – Elk
    Commented Feb 20, 2016 at 16:14

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