Continuity of the argument of a complex number.

Question

The argument of the complex number $x-i$ (real $x$) can be found using angle($x$,-1) in Wolfram Alpha. I asked WA if angle($x$,-1) is a continuous at $x=0$ it replied YES and drew a graph showing a discontinuity at zero. Yet when asked to find the left/right limits of angle($x-i$) at zero it returns $-\pi$ and $\pi$ (as does my ti-nspire CAS). Can anyone throw some light on this discrepancy in WA ?

Answer 1

The complex number $x-i\in{\mathbb C}$, where $x\in{\mathbb R}$, can be viewed as a vector emanating at $i$and having its tip at $x$. The signed angle from the negative $y$-axis to this vector is $=\arctan x$, and goes from $-{\pi\over2}$ to ${\pi\over2}$ as $x$ goes from $-\infty$ to $\infty$ It follows that the argument of this vector with respect to the positive $x$-direction is given by $${\rm Arg}(x-i)=\arctan x-{\pi\over2}\qquad(-\infty<x<\infty)\ ,$$ and is continuous on all of ${\mathbb R}$. I cannot help you with the idiosyncrasies of WA.

Answer 2

A single-valued function is continuous at a point ONLY if the negative and positive limits and the function output at that point all agree: No ifs, no buts!!

WA throws up many whacky answers, in fact, there is a whole (big) section devoted to WA on this site.

Ask yourself this buddy, if online-WA is so good why is it free but they ask you to sign up for a premium version. WA is based on MATHEMATICA which costs hundreds of green-backs(there ain't no such thing as a free lunch). If you want better solutions buy MATHEMATICA; online WA is there to draw you in so you buy the premium version or go the full hog with MATHEMATICA.