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.

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 $[z\mapsto z^{-1}]$.
At least tangentially related reference: Stein Manifolds and Holomorphic Mappings: The Homotopy Principle in Complex Analysis by Franc Forstnerič.
