2
$\begingroup$

This is a follow up of my previous question.

Of course, it makes no sense to discuss the asymptotic behavior of trigonometric functions on the real line, as they oscillates forever and does not even have a limit at infinity.

However, it happens that the limits exist for trigonometric functions with arguments tending to some complex infinity which is not real, i.e. $$\lim_{r\to\infty}f(re^{i\theta})$$ exists for $\theta\ne n\pi$ where $f$ is some trigonometric functions.

For example $$\lim_{r\to\infty}\csc(re^{i\theta})=0$$

Now I become interested in the asymptotic behavior of trigonometric functions near complex infinity. I tried to analyze the functions like $\csc(\frac1z)$ but unfortunately it is not differentiable at $0$ and thus I cannot obtain a Taylor series/Laurent series at $0$.

My question is ($f$ is some trigonometric function)

If we write $$f(re^{i\theta})\sim l+\text{lower order terms}$$ for large $r$, what are the lower order terms exactly?($l$ is the limit value of $f$ when $r\to\infty$, e.g. for $f=\csc$, $l=0$; $f=\cot$, $l=\pm i$)

$\endgroup$
  • $\begingroup$ Wouldn’t they be “higher order” terms? (After the change $w=1/z$). $\endgroup$ – Leo Bianco Jun 9 '18 at 2:03
  • $\begingroup$ @LeoBianco Well, I mean the terms that will vanish in the limit. I am not sure which is right. $\endgroup$ – Szeto Jun 9 '18 at 2:23
  • 1
    $\begingroup$ The trigonometric functions have essential singularities at $\infty$, and therefore by the Picard theorem, the image under them of any neighborhood of $\infty$ is the entire complex plane, other than possibly two points. $\endgroup$ – Paul Sinclair Jun 9 '18 at 13:31
  • $\begingroup$ @PaulSinclair How about for some complex infinity like $\infty e^{i\pi/3}$(not rigorous)? Does the trigonometric function still has an essential singularity at some complex infinity? $\endgroup$ – Szeto Jun 9 '18 at 14:05
  • $\begingroup$ In complex analysis, the standard is to have one infinity, $\infty$, which is approached as one increases without bound in every direction. So it wraps up the complex plane into the Riemann sphere (think of curving the two ends of the real line around until they meet at a single point. The Riemann sphere is the 2D version of the same thing. It is possible to compactify the complex plane by other means such as you describe. But the Reimann sphere is more matural, and gives very nice results. Effectively $\infty$ behaves pretty much like every other point. Only multiplication distinguishes it. $\endgroup$ – Paul Sinclair Jun 10 '18 at 5:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.