# Square root of a complex number with infinitesimal imaginary part

I am going through lecture in Quantum Field Theory and I am encountering the square root of a complex number with infinitesimally small imaginary part: $$\lim_{y\rightarrow0^+}\sqrt{-x+iy}=-i\sqrt{x}$$ and $$\lim_{y\rightarrow0^+}\sqrt{-x-iy}=i\sqrt{x}$$ But I don't understand how these equalities are established. The standard way of taking the square root is by writing the complex number in polar form and then the result is just the square root of the magnitude times the exponential of the imaginary unit times the argument of the complex number divided by two in addition to multiples of $$\pi$$. But when I take the limit of the imaginary part going to zero I just get a vanishing argument and I don't get the minus or plus sign in both the square roots. What am I not getting right here?

• Well if we naively substitute $y=0$ we get $\pm i \sqrt{x}$. Perhaps the polar form can illuminate which root is chosen for each case. Commented Jun 25, 2022 at 14:29
• Could it also have something to do with a complex logarithm? Commented Jun 25, 2022 at 14:32
• For positive real $x$, the argument of $-x+iy$ goes to $\pi$, not $0$. Commented Jun 25, 2022 at 14:35
• The roots here are all real except for $-1$ so I'm not sure the logarithm is required. Commented Jun 25, 2022 at 14:36
• @peterwhy yes you are right. that was a stupid mistake. Commented Jun 25, 2022 at 14:38

$$\lim_{y\rightarrow 0+} \sqrt{-x\pm y}=\lim_{y\rightarrow 0+} \sqrt{x^2+y^2}e^{i\,arctan2(\pm y/-x)/2}$$ For the plus case: $$\lim_{y\rightarrow 0+} arctan2(y/-x)=\pi$$ and for the minus case: $$\lim_{y\rightarrow 0+} arctan2(-y/-x)=-\pi$$ Therefore $$\lim_{y\rightarrow 0+} \sqrt{-x\pm iy}=\mp i \sqrt{x}$$
• About your notation, for positive $x$, $$\lim_{y=0^+}\arctan[\pm y/(-x)] = 0$$ The argument function you are looking for may be atan2. Commented Jun 25, 2022 at 14:57
• In the final expression I believe you're missing an $i$ in the square root on the left and that it should be $\mp$ not $\pm$ on the right hand side of the equal sign. Commented Jun 25, 2022 at 15:49