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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.

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  • $\begingroup$ It seems to me like $\phi=\psi$, from factoring $-1$ out of the radical. $\endgroup$ Aug 7, 2015 at 5:32
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    $\begingroup$ @MoseWintner: You cannot factor things out of a radical unless you know what you are doing. Otherwise $1 = \sqrt{-1 \times -1} = \sqrt{-1} \times \sqrt{-1} = -1$. $\endgroup$
    – user21820
    Aug 7, 2015 at 6:02
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    $\begingroup$ If you choose a branch cut $C$ for the square root appearing in $\phi$, and you choose the branch cut $-C$ for the square root in $\psi$, then $\phi$ and $\psi$ have the same domain, and depending on the chosen branches on the domains of the square root, you have either $\phi \equiv \psi$ or $\phi \equiv -\psi$. If you choose the same branch cut for both square roots, then $\operatorname{dom} \psi = - \operatorname{dom} \phi$, the intersection of the domains consists of two components, and on one of the components you have $\phi\equiv\psi$, on the other $\phi\equiv -\psi$. $\endgroup$ Aug 7, 2015 at 6:26
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    $\begingroup$ @Axoren: You must know precisely what is the definition of square-root. If not, then factoring anything in or out is meaningless. You certainly cannot randomly take rules that apply to positive real numbers and assume they apply to anything else. And you should learn why they apply for reals. $\endgroup$
    – user21820
    Aug 7, 2015 at 7:00
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    $\begingroup$ @Axoren You are probably familiar with examples such as $\sqrt{-5}=i\sqrt5$. This works because $5$ is a positive real number. Factoring under a square root always works if one of the arguments is positive real. Presumably you wouldn't try the same thing for $\sqrt{-(-1)}=i\sqrt{-1}$, though. $\endgroup$ Feb 4, 2016 at 12:45

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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!]

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  • $\begingroup$ The "collection" approach to multivalued functions looks appealing at first, but it doesn't actually solve the problems it is invented to fix, and in particular $(S\cdot T)^U=S^U\cdot T^U$ does not hold. Let $S=T=\{2\}$ and $U=\{1/2,1\}$. Then $\{4\}^{\{1/2,1\}}=\{-2,2,4\}$ but $\{2\}^{\{1/2,1\}}=\{-\sqrt2,\sqrt2,2\}$ and $\{2\}^{\{1/2,1\}}\cdot\{2\}^{\{1/2,1\}}=\{-2,2,-2\sqrt2,2\sqrt2,4\}$. $\endgroup$ Feb 1, 2016 at 1:27
  • $\begingroup$ @MarioCarneiro: You are absolutely right. I was all the while thinking only of a singleton exponent, and in that case it's true because of the earlier part of my answer. $\endgroup$
    – user21820
    Feb 1, 2016 at 15:26

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