What is the degree of the following map? 
Let $a \notin S^1$, $f:S^1 \rightarrow S^1$, be given by $z \mapsto \frac{z-a}{|z-a|}$. What is the degree of $f$?

Where by degree we mean under the usual definition here
or general case  here .

My current thoughts are usuing the definition that it is the value of the unique lift, when $f$ is idenitfied with $\hat{f}:I \rightarrow S^1$, $\hat{f}(1)=\deg f$. But this requires me to explicitly construct a lift of the map
$$ \hat{f}:I \rightarrow S^1, s \mapsto \frac{e^{2 \pi i s} - a }{ |e^{2 \pi i s} - a | }.$$
which seems plausible but messy. I would be interested for an elegant proof.

This is the answer proposed by Steve D. 
By the homotopy axiom, the map $H(f):H_1(S^1) \rightarrow H_1(S^1)$ is the same for homotopic maps $g \simeq f$. 
When $|a|<1$, we give the homotopy 
$$ H_t(z):= \frac{z-ta}{|z-ta|}.$$
So the degree of $f$ is equivalent to that of $g(z)=z$, which is $1$. 
When $|a|>1$, we give the homotopy 
$$H_t(z):= \frac{tz-a}{|tz-a|}$$ 
so the degree of $f$ is equivalent to that of the constant map, which has degree $0$. 
