1
$\begingroup$

I am wondering about how to compute the degree of r(z) := \frac{z}{| z |}. I know that $deg f_n = n$ where $f_n (z)= z^n$ i.e. the degree is how many times it goes around $0$. I also know that $deg(f \circ g) = deg (f) \cdot deg (g)$. Now with $f(z)=z$, $deg f = 1$, but what about $f(z) = \frac{z}{|z|}$?

Edit:

Where $r: \mathbb{C} \backslash \{ 0 \} \rightarrow S^1$. The degree is defined for continuous functions $f:S^1 \rightarrow S^1$. $ \mathbb{C} \backslash \{ 0 \}$ is homotopy equivalent to $S^1$ so I was thinking $r$ can be viewed as function $r: S^1 \rightarrow S^1$. I'm asking this question because I think $deg (r)$ is used in a proof of the fundamental theorem of algebra.

Many thanks for your help!

  • 0
    You're not missing anything! Thank you! Maybe you could post this as an answer, I'll accept it.2011-07-05

1 Answers 1

1

Note that $r:\mathbb{C}-\{0\}\to S^1$ is itself a (/ the standard) homotopy equivalence, with homotopy inverse just the inclusion. So if you want to view $r$ as function $S^1\to S^1$ via the standard homotopy equivalence, i.e. $S^1\to \mathbb{C}-\{0\}\to S^1$, then this is homotopic equivalent to the identity map, which has degree 1!

Edit, in reply to your comment:

I don't know what $deg(r)$ means. You said you wanted to view $r$ as a map $S^1\to S^1$ using the homotopy equivalence of $S^1$ and $\mathbb{C}-\{0\}$. I can only interpret this to mean that instead of $r$ we consider $r\circ i$. Or, if you will, we define $\deg(r):=\deg(r\circ i)$. Now the latter has degree 1 because $r\circ i$ is homotopic to the identity map on $S^1$.

  • 0
    ? $r$ is a map $\mathbb{C}-\{0\}\to S^1$. It is not a map $S^1\to S^1$. $r\circ i$ *is* a map $S^1\to S^1$.2011-07-05