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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 $f(z) = \frac{z}{|z|}$?


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!

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Are you sure that degree is defined for everything? Ok, what is then the degree of $\sin$ or $\arcsin$? – Ilya Jul 4 '11 at 11:00
What is $z$? Please define your function $r$ properly; what is its domain and codomain? – wildildildlife Jul 4 '11 at 11:33
As i know the dgree of a map is defined for maps from $S^1\to S^1$ and at your case $f(z)=z$ has degree 1. – omar Jul 4 '11 at 11:53
@omar: no, I'm interested in the case $f(z) = \frac{z}{|z|}$. – Rudy the Reindeer Jul 4 '11 at 11:58
$f(z) = z/|z| = z$ since any point on the circle has absolute valu 1. Am I missing something? – Lalit Jain Jul 4 '11 at 15:42
up vote 1 down vote accepted

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 $S^1$.

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Thank you!! Let me repeat to see if I understand correctly: $deg (i) = 1$ because the inclusion is the identity. $deg (i \circ r) = deg (r \circ i) = 1$ because both are homotopic to the identity functions (respectively). Therefore $deg (i \circ r) = deg(r) deg(i) = deg(r) \cdot 1 = 1$ implies $deg(r) = 1$. – Rudy the Reindeer Jul 4 '11 at 13:21
I think I have to add '@....' in order to appear in your inbox. – Rudy the Reindeer Jul 4 '11 at 13:33
No, it reached my inbox :) – wildildildlife Jul 4 '11 at 14:00
Oh well : ) Ok, I assume no comment means yes what I repeated is what you meant. Thanks for your help! – Rudy the Reindeer Jul 4 '11 at 14:07
@last comment: no, I edited my answer! – wildildildlife Jul 4 '11 at 14:37

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