Taylor series for square root of exp[i x]

Question

I need to find $$ \textrm{Lim}_{x\rightarrow 0} \sqrt{-\exp[i x]} $$ mathematica says it is $- i$ but how it is so. I would say it is $i$ because if I take limit then $\exp[i x]$ becomes 1 and $\sqrt{-1}$ is iota. Can anyone help in understanding this.

Answer 1

Non-integer powers of complex numbers are defined as $z^\alpha \equiv e^{\alpha\,\text{Log}(z)}$ where $\text{Log}$ is some complex logarithm. The complex logarithm is multivalued (since $e^{z} = e^{z + 2\pi i n}$ for any integer $n$) so there are infinitely many choices. The common choice is to use the so-called principal branch of the logarithm for which $z^\alpha \equiv e^{\alpha(\log|z| + i\arg(z))}$ where $\arg(z) \in (-\pi,\pi]$ is the complex argument. Any definition of a complex logarithm requires a branch-cut which is a curve from $0$ to $\infty$ for which the logarithm has a discontinuity.

For this particular problem we have $$\sqrt{-e^{ix}} = \sqrt{e^{ix + \pi i}} = e^{\frac{i}{2}\arg(e^{ix + \pi i})}$$ and in Mathematica the branch cut is along the negative real axis so for small positive $x$ we have $\arg(e^{ix + \pi i})\approx -\pi$ and for small negative $x$ we have $\arg(e^{ix + \pi i}) \approx \pi$. This means that $$\lim_{x\to 0+}\sqrt{-e^{ix}} = -i~~~~\text{and}~~~~\lim_{x\to 0-}\sqrt{-e^{ix}} = +i$$ so the limit does not exist. The problem here is that the discontinuity of the logarithm is along where you want to take the limit.

Answer 2

When I solved this limit using MATLAB, the result was i, and analytically you do get i, so I guess Mathematica is probably giving you wrong results.

Answer 3

As usual, Mathematica is right and accurate.

It says $i$ for the left limit and $-i$ for the right limit. The limit itself does not exist. This is because in the complex you have to choose between branches.