Requirement for a given function to be smooth I have quite a basic question about the derivatives. My uncertainty comes mainly from the fact that I don't know how the complex logarithm behaves. Here is the description (this task is not homework!):
Assume I have a (complex) function $f:\mathbb{R}\rightarrow\mathbb{C}$, and I want to find a real, smooth function $\theta$ such that $\tilde{f}:=e^{i\theta}f$ is real, i.e. I just rotate $f$.
Now, my question is: which properties do I have to impose on $f$ and $\tilde{f}$ such that such smooth function $\theta$ exists?
Attempt at the solution:
$\tilde{f}=e^{i\theta}f$ 
$\log(\tilde{f})=i\theta+i2\pi n+\log(f)$, $n\in\mathbb{Z}$
$i\theta=\log(\tilde{f})-i2\pi n-\log(f)$
$\theta=-2\pi n-i(\log(\tilde{f})-\log(f))$
Now I'm not sure how to proceed from here. $\log(\tilde{f})$ is real but $\log(f)$ is complex. If I want $\theta$ to be smooth then the right hand side also has to be smooth. Which properties does it then imply for $f,\tilde{f}$?
Any hint will be appreciated.
 A: The process of constructing an angle function is rather lengthy but the key point is to keep track of how many complete turns we have made around $0$.
If $F$ is smooth then you can obtain a smooth $\theta$.
But note that you can consider discontinuous functions and still have a smooth $\theta$. For instance $G(t) = F^{it} + \delta_{(0,\infty)}(F(t))$
To construct a smooth angle function from a smooth $F$ It is sufficient to decompose your function $F(t) = r(F(t)) e^{i g(F(t))}= r(F(t)) e^{i h(F(t))}$ where $g(z)$ is the angle measured in counter-clockwise manner beginning at the positive real axis ($g(z) \in[0,2 \pi)$) and $h(z)$ is the angle measured counter-clockwise manner beginning at the negative real  axis ($h(z) \in[-\pi,\pi)$).Note that $$g(z) = h(z) + 2 k(z)\pi $$ and  that $g\circ F,h \circ F$ are not  smooth functions in general ( look at $F(t) = e^{it}$ - Note that $\theta(t) = t$ is smooth)
Note also that $g(z)$ is smooth as long as $z \in \Bbb{C} \setminus \{z  \mid\Re(z) <0, \Im(z)=0 \} = A_g$ and that $h(z)$ is smooth as long as $z \in \Bbb{C} \setminus \{z  \mid\Re(z) >0, \Im(z)=0 \} = A_h$.
Assume withou loss that $F(0) \in A_g$ So you can consider $\theta(0) = g(F(0))$ as long as $F(0)\in A_g$ but once you "approach" $\{z  \mid\Re(z) <0, \Im(z)=0 \}$ you need to consider $\theta(t) = h(F(t)) + 2k(z)\pi $

Hope this helps
