Specific homotopy between complex conjugation and the identity.

Consider the set $\mathcal{C} = C^{\infty}(\mathbb{C}^*, \mathbb{C}^*)$, where $\mathbb{C}^* = \mathbb{C}\backslash\{0\}$.

Both $f(z) = z$ and $g(z) = \bar{z}$ can be seen as elements in $\mathcal{C}$.

Question: Is there a (smooth, not necessarily analytic) homotopy $H : [0, 1] \times \mathbb{C}^* \rightarrow \mathbb{C}^*$ between $f$ and $g$?

I tried what seemed to me like natural choices, such as deforming the imaginary part, but the problem is to avoid producing some function which maps a non-zero complex number to zero.

Motivation: In case anyone is wondering, this problem arises in showing that two complex line bundles over the $2$-sphere are (smoothly) isomorphic. The bundles are $L_g^*$ and $L_{1/g}$, where $g : \mathbb{C}^* \rightarrow \mathbb{C}^*$ is the gluing cocycle (there is only one, since the $2$-sphere is covered by two stereographic projections).

Thanks.

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Maybe I'm being silly, but: wouldn't this yield a homotopy $h$ between the identity and $z \mapsto z^{-1} = \bar z$ on $S^1$ by setting $h(t,z) = \frac{H(t,z)}{\lvert H(t,z)\rvert}$? This couldn't be because $z \mapsto z$ and $z \mapsto z^{-1}$ are distinguished by the degree. Probably such a degree argument can be done directly on $\mathbb{C}^\ast$. – t.b. Sep 16 '12 at 19:25
No, there is no such homotopy since the fundamental group of $\mathbb{C}^*$ is $\mathbb{Z}$ and such a homotopy would deform the unit circle parametrized clockwise into the unit circle parametrized counter-clockwise. – Michael Sep 16 '12 at 19:31
@t.b.,Michael thanks, I guess you're right, there exists no such thing. My reference must be wrong, I'll check it again. – student Sep 16 '12 at 19:58

Community verdict (from comments by t.b. and Michael): $z$ and $\bar z$ are not homotopic in $C(\mathbb C^*,\mathbb C^*)$.
More precisely, the set of homotopy classes of continuous maps $h\in C(\mathbb C^*,\mathbb C^*)$ is $\{[z\mapsto z^n]:n\in\mathbb Z\}$, and all classes $[z\mapsto z^n]$ are distinct (distinguished by their action on the fundamental group). The complex conjugation belongs to $[z\mapsto z^{-1}]$.