Square root branch cut Consider the following expression:
\begin{equation}
\phi(\delta)=i\,\sqrt{-3+i\,\delta},
\end{equation}
where $\delta$ is infinitesimal.
If we choose a branch cut along the negative real axes, it becomes $-\sqrt{3}$ if $\delta>0$ and $\sqrt{3}$ if opposite.
Now consider 
\begin{equation}
\psi(\delta)=\sqrt{3-i\,\delta}
\end{equation} 
which is just $\sqrt{3}$ with our choice of the branch cut.
Obviously $\phi\neq\psi$.
Is there a way to relate them? I am trying to do calculations with moving $i$ all around saving analytic structure of my expressions.
 A: You cannot relate them in any elegant way that does not make it a complete mess. The reason is that by multiplying by $-1$ inside the square root, you are doing a rotation before taking the square root, so it is essentially moving the branch cut. That means that the result is either $i$ or $-i$ times the original, and it depends on the input. In general you have no choice but to try not to manipulate things within a square root defined using branch cuts if you do not want things to become ugly.
There is another possible way, but you have to work with multi-valued functions instead of ordinary functions. In this setting every complex function takes a collection of complex numbers as input and produces a collection of complex numbers as output. Everything has to be built from scratch, but after you are done you get an exponential function then can be inverted. The multi-valued logarithm will take any collection $S$ and return $\{ z : e^z \in S \}$. If $S$ is a singleton, then the resulting collection will have members differing by a multiple of $2πi$. You can then define $S^T$ to mean $\exp(T\ln(S))$ for any collections $S,T$ of complex numbers. Then for any complex number $z$ it would always be the case that $\sqrt{\{z\}}$ is a collection with two members (the two square roots of $z$) except when $z = 0$. Note that using multi-valued arithmetic and exponentiation we preserve the fact $(S \cdot T)^U = S^U \cdot T^U$ for any collections $S,T \subseteq \mathbb{C}_{\ne 0}$ and singleton $U \subseteq \mathbb{C}$. For example:
  $\{1,-1\} = \{1\}^{\{0.5\}} = \{-1\}^{\{0.5\}} \cdot \{-1\}^{\{0.5\}} = \{i,-i\} \cdot \{i,-i\}$. [No contradiction!]
