# How many times does $8z^4+4z^3+2z^2+z^{-1}$ wind around $0$?

I am a newbie to abstract topology and am working through some exercises regarding fundamental groups. Here is the problem. Consider $f: S^1 \to \mathbb{C} \setminus \{ 0 \}$ given by $z \mapsto 8z^4+4z^3+2z^2+z^{-1}$. What is the winding number of $f$ about the origin?

My idea is as follows. We can collapse $\{ a+bi \ | \ a \geq 0 \}$ to $1+0i \in S^1$ and deform the rest part of $\mathbb{C}$ to $S^1$ in a continuous way. Denote $z = e^{i\theta}$. When $\theta = 0, \pi/2, \pi, \pi/3$, $8z^4$ will be $8+0i$, and $\Re(f(z)) \geq 8 - 4 - 2 - 1 > 0$. Similarly, when $\theta = \pi/4, 3\pi/4, 5\pi/4, 7\pi/4$, $\Re(f(z)) < 0$. Intuitively, this tells us that $f$ passes through the point $1 + 0i$ for four times, and hence the winding number would be $4$.

My question is: 1. Am I correct, intuitively? 2. How to write a formal proof about this? I find it extremely uncomfortable to answer questions like this. Everything seems intuitively trivial, but I am not sure if I miss anything that demands mathematical rigor.

• @DonAntonio I think we can consider a base-point preserving map $g$ from complex plane without the origin to the circle. Then, it suffices to consider $g \circ f$'s induced homomorphism $\varphi$ on $\mathbb{Z}$. The winding number, then, can be defined as $\varphi(1)$. – Minsheng Liu Nov 20 '16 at 22:18
I am not completely sure what to say about your intuition. You seem to think that the term $8z^4$ is most important, which certainly is correct. Moreover, it is certainly correct to try to deform $\mathbb C\setminus\{0\}$ (and it is crucial that $0$ is excluded) to $S^1$. Formally, this is most easily done by the map $z\mapsto z/|z|$. At this point, your intuitive reasoning gets a bit problematic. There are results on determining the mapping degree by counting intersections of a curve with a ray, but in order to apply them, you have to make sure that all these intersections are transversal and, more importantly, you have to count intersections including a sign. (You have to subtract the number of "backwards" intersections from the number of "forwards" intersections, and transversality is needed in order to make sure that each intersection is either "forwards" or "backwards".)
But once you have the idea that the $z^4$-term dominates, you don't need any counting argument, because $z\mapsto z^4$ more or less by definition has winding number $4$. So it is better to make the idea of that term being dominant precise: For $z\in S^1$, $|8z^4|=8$ while $|4z^3+2z^2+z^{-1}|=7$. Observing this you see that $H(z,t):=8z^4+(1-t)(4z^3+2z^2+z^{-1})$ for $z\in S^1$ and $t\in [0,1]$ has values in $\mathbb C\setminus\{0\}$ (it parametrizes the segment from $f(z)$ to $8z^4$). Thus $H$ defines a homotopy from $f$ to $z\mapsto 8z^4$, which in turn is evidently homomotopic to $z\mapsto z^4$. Thus all these maps have the same winding number $4$.