Showing that certain function does not exist I am trying to prove that there is no a continuous map $ \phi : \mathbb{C} \times I \to \mathbb{C} $ ( $I = [0,1] $) $\phi_{\alpha}(z) = \phi(z,\alpha) $ satisfying the following: for each $\alpha \in I$, $\phi_{\alpha } : \mathbb{C} \to \mathbb{C} $ is $\mathbb{R}$-linear, $\phi_{\alpha}(1) \neq 0 $ and $\phi_{\alpha}(zw) = \phi_{\alpha}(z)\phi_{\alpha}(w) $; and $\phi_0(i) = i , \phi_1(i) = -i $.
Perhaps there is one such a function. If so, how can we construct it? 
 A: We don't need the $\mathbb{R}$-linearity to see that no such map exists. Suppose we have a continuous $\phi\colon \mathbb{C}\times I \to \mathbb{C}$ such that $\phi_\alpha \colon z \mapsto \phi(z,\alpha)$ is multiplicative with $\phi_\alpha(1) \neq 0$ for all $\alpha$ and $\phi(i,0) = i$.
From the multiplicativity of the $\phi_\alpha$ we obtain $\phi_\alpha(1) = \phi_\alpha(1^2) = \phi_\alpha(1)^2$, and hence $\phi_\alpha(1) \in \{0,1\}$. Since $\phi_\alpha(1) = 0$ was excluded, we have $\phi_\alpha(1) = 1$ for all $\alpha\in I$.
Now $1 = i^4$, and thus $1 = \phi_\alpha(1) = \phi_\alpha(i^4) = \phi_\alpha(i)^4$ by multiplicativity, thus $\phi_\alpha(i) \in \{ 1, i, -1 , -i\}$ for all $\alpha\in I$.
But $\gamma\colon\alpha \mapsto \phi(i,\alpha)$ is supposed to be continuous, and takes values in a discrete set. Since $I$ is connected it follows that $\gamma(\alpha) = \gamma(0) = i$ for all $\alpha\in I$, in particular $\phi_1(i) = i \neq -i$.
A: Since $\phi_\alpha:\mathbb C\to\mathbb C$ is $\mathbb R$-linear we can write
$$
\phi_\alpha(z)=\phi_\alpha(x+iy)=\begin{pmatrix}a(\alpha)&b(\alpha)\\c(\alpha)&d(\alpha)\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix},
$$
where $a,b,c,d:[0,1]\to\mathbb R$ are continuous functions. Now since it preserves complex multiplication we have 
$$
\begin{cases}
\phi_\alpha(1\cdot 1)=\phi_\alpha(1)\phi_\alpha(1),\\
\phi_\alpha (i\cdot i)=\phi_\alpha(i)\phi_\alpha(i),
\end{cases}
$$
which give
$$
\begin{cases}
a(\alpha)+ic(\alpha)=(a(\alpha)+ic(\alpha))^2,\\
-a(\alpha)-ic(\alpha)=(b(\alpha)+id(\alpha))^2,
\end{cases}
$$
The first equation gives $a(\alpha)=c(\alpha)=0$ or $a(\alpha)=1,c(\alpha)=0$. If $a(\alpha)=c(\alpha)=0$ then also $b(\alpha)=d(\alpha)=0$ which is excluded. Thus, $a(\alpha)=1,c(\alpha)=0$ and $-1=(b(\alpha)+id(\alpha))^2$, so that $b(\alpha)=0,d(\alpha)=\pm 1$. Consequently 
$$
\phi_\alpha(z)=\phi_\alpha(x+iy)=\begin{pmatrix}1&0\\0&d(\alpha)\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=x+id(\alpha)y,
$$ 
where $d(\alpha)=\pm1$. This $d$ cannot be continuous!
However the problem seems more interesting than this. What one is looking for here (it seems) is a homotopy $\phi$ between the identity and the symmetry in the complex plane (with perhaps some additional condition?). If such a homotopy exists, $\phi_1$ should preserve orientation as $\varphi_0=$Id does, and the symmetry doesn't. 
