I am trying to prove that given $p: \hat{\mathbb{C}} \to \hat{\mathbb{C}}$ a polynomial given by $p(z)=a_nz^n+...+a_0$ then, $p$ is smoothly homotopic to the polynomial $q(z)=a_nz^n$. I am using the homotopy $F: \hat{\mathbb{C}} \times [0,1] \to \hat{\mathbb{C}}$, given by $F(z,t)=a_nz^n+t(a_{n-1}z^{n-1}+...+a_0)$. I can prove that this is an homotopy (its continous), the smooth part is what troubles me, since it looks like a bunch of calculations, and a lot of ''dirty'' work.
Lets say I want to keep my hands clean...How can I go about proving that $F$ is smooth? I have thought about approximating the homotopy by smooth maps but really do not know if I actually can, and if I can, how do I assure the smooth map I took, ''connects'' $p$ to $q$?
If it helps: I want to use this to prove that the Brouwer degree of a polynomial at $\hat{\mathbb{C}}$ is its actual degree.
Thanks in advance
First write $z=x+iy$ and $a_k=b_k+ic_k$ for all $k$. Then we can rearrange $F(x,y,t)=a_n(x+iy)^n+t(a_{n-1}(x+iy)^{n-1}+...+a_0)$ so that $F(x,y,t)=u(x,y,t)+iv(x,y,t)$, where the component functions $u$ and $v$ are elements of $\mathbb{R}[x,y,t]$. Then, since $u$ and $v$ are just polynomials in several variables, they are infinitely differentiable in each variable. It follows that $F$ is also inifinitely differentiable in each variable. So $F:\mathbb{R}^3\to\mathbb{R}^2$ is smooth.