Continuity of Quotient of Complex Values

[thatboi]

Hey all,
I have a very simple question regarding the quotient of complex values. Consider the function:
$$f(a) = \sqrt{\frac{a-1i}{a+1i}}$$
where $i$ is the imaginary unit. When I evaluate f(0) in Mathematica, I get $f(0) = 1i$, as expected. But if I evaluate at a very small value of $a$ such as $a = 10^{-20}$, I get $f(10^{-20}) = 10^{-20} - 1i$. I naively thought that $f$ would be continuous in $a$ but it is clear that somehow the imaginary part of $f$ flips sign the moment I introduce some small value for $a$. How do I explain this behavior? I feel like there is something obvious here I am missing.
Thanks!

 

[FactChecker]

The problem is not in the quotient, it is in the square root function. The square root of any real number, $r \in \mathbb{R}, r \gt 0$ can be positive or negative. It is a matter of convention that the positive number is considered the "principle branch" and is the answer given. That means that the "principle branch" excludes the negative answers, which are in another (non-principle) branch. The negative real axis in the z domain is called a "branch cut". When your input crosses the branch cut there will be a discontinuity because the continuous answer would cross into a different branch answer. That is what happened in your example. (see this )

 

[andrewkirk]

The problem arises from the use of the square root sign as if it denoted a function - ie a map that can only give one value per input. Recall that every complex number has square roots, Each is -1 times the other. So to make the square root sign able to work as a function we need to adopt a convention as to which of the two roots we select, For square roots of positive real numbers we use the convention That we always select the positive root. For square roots of negative numbers that can't apply so we need a different convention. that typically means it selects the root that lies within a designated set of two consecutive quadrants of the complex plane. That makes such a sqrt function discontinuous at the places in the domain that map to the boundary of that region of the range.

It looks like Mathematica is using the 1st and 4th quadrants, ie requiring the argument to lie in the range $(-\pi/2, \pi/2]$. That makes its square root function discontinuous at the negative x axis, as that maps to the negative y axis that is the omitted boundary of the selected half plane.

For your case with small $a$, the item inside the sqrt sign approaches -1 as a goes to zero, so the discontinuous sqrt function gives a discontinuity there.

It's easily fixed by selecting the other square root: $10^{-20}+i$, which is very close to $i$.

EDIT: Just when I posted it, I saw that @FactChecker posted essentially the same thing just before me (only more concisely)!

EDIT2: If you instead adopt a square root function that always selects the square root lying in quadrant 1 or 2 (more specifically: the root's argument lying in $(0,\pi]$) , your function will be continuous, because that sqrt function's discontinuity lies on the positive x axis, which is not in the image of the expression inside the square root sign.

 

[WWGD]

Isn't there an issue too, of how calculators rounding of, accuracy breaking down at such small numbers?

 

[FactChecker]

Isn't there an issue too, of how calculators rounding of, accuracy breaking down at such small numbers?

Good point. Those types of small problems near the discontinuity of a branch cut would cause large discontinuities.

 

[pasmith]

Hey all,
I have a very simple question regarding the quotient of complex values. Consider the function:
$$f(a) = \sqrt{\frac{a-1i}{a+1i}}$$

For real $a$, in the principal branch this is $$f(a) = \sqrt{\frac{(a-i)^2}{1 + a^2}} = \begin{cases} \frac{-a + i}{\sqrt{1 + a^2}} & a \leq 0 \\ \frac{a - i}{\sqrt{1 + a^2}} & a > 0. \end{cases}$$ You can see that $$\lim_{a \to 0^{-}} f(a) = f(0) = i \neq -i = \lim_{a \to 0^{+}} f(a).$$

Isn't there an issue too, of how calculators rounding of, accuracy breaking down at such small numbers?

In general, yes. But here from the exact result above you can see that the only inaccuracy in the numerical result of $10^{-20} - i$ comes from taking $\sqrt{1 + 10^{-40}} \approx \sqrt{1 + 2^{-132.88}}$ to be exactly 1. In particular, the sign of the imaginary part is correct.