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.
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