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It is well-known that trigonometric functions oscillate on the real axis and the limit does not exist as the argument approaches infinity.

However, I suspect that a limiting value exist if the argument approaches any complex infinity that is not real, i.e. $$\lim_{r\to\infty} f(re^{i\theta})$$ is suspected to exist for $\theta \ne n\pi$, where $f$ is a trigonometric function.

I confirmed it is the case for $\tan (z)$ by decomposing it into real and imaginary parts.

For real part: $$\lim_{r\to\infty}\frac{\sin 2r\cos\theta}{\cos 2r\cos\theta+\cosh 2r\sin \theta}=0$$ which is straightforward.

For imaginary part: $$\lim_{r\to\infty}\frac{\sinh 2r\sin\theta}{\cos 2r\cos\theta+\cosh 2r\sin \theta}=\text{sgn}(\sin\theta)$$

Thus $$\lim_{r\to\infty}\tan(re^{i\theta})= \text{sgn}(\sin\theta)i$$

My questions are:

  1. Are the above calculations correct?

  2. Is there an easier way to compute the limit, other than decomposing it into real and imaginary parts?

  3. Indeed, the limits should be well known. Are there some reliable references that summarize the results for various trigonometric functions?

Thanks in advance.

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  • $\begingroup$ The Wolfram Functions site includes limits at $\pm i \infty$ (i.e., $\theta = \pi/2$ or $3 \pi/2$) for all of the trigonometric functions; look under "Specific Values > Values at Infinities". It does not, however, treat the case for general $\theta$. $\endgroup$ – Michael Seifert Jun 8 '18 at 14:18
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You can base all calculations on the fact that $e^{iz} \to \infty$ as $z \to \infty$ with $-\pi+\varepsilon<\arg{z}<\varepsilon$ and $e^{iz} \to 0$ as $z \to \infty$ with $\varepsilon<\arg{z}<\pi-\varepsilon$ (this follows because $\lvert e^{i(x+iy)} \rvert = e^{-y}$). For example, $$ \tan{z} = \frac{e^{iz}-e^{-iz}}{i(e^{iz}+e^{-iz})} = -i\frac{e^{2iz}-1}{e^{2iz}+1}, $$ which tends to $i$ for $\varepsilon<\arg{z}<\pi-\varepsilon$ and $-i$ for $-\pi+\varepsilon<\arg{z}<\varepsilon$.

Similarly, for the other functions, $\cos{z}$ and $\sin{z}$ both tend to $\infty$ for $z \to \infty$ with $\arg{z}$ bounded away from $0$ and $\pi$. Therefore their reciprocals $\csc{z}$ and $\sec{z}$ both tend to $0$ in the same region.

Lastly, $\cot{z}=1/\tan{z}$, so it tends to $-i$ for $\varepsilon<\arg{z}<\pi-\varepsilon$ and $i$ for $-\pi+\varepsilon<\arg{z}<\varepsilon$.

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  • $\begingroup$ $e^{iz}$ has its real part and imaginary part are both oscillating as $|z|\to\infty$ with fixed argument. How can you justify your first sentence? $\endgroup$ – Szeto Jun 9 '18 at 0:07
  • $\begingroup$ "$\to \infty$" in the complex plane means that the absolute value tends to infinity; the actual values of the real and imaginary parts are unimportant. In this case, so long as $y$ is increasing (or decreasing), what $x$ does is totally irrelevant for the limit. If you don't like that, you can show that $|\sin{(x+iy)}|^2 = \sin^2{x}+\sinh^2{y}$, and similarly $|\cos{(x+iy)}|^2 = \cos^2{x}+\sinh^2{y}$, which gives the results for sine, cosine and their reciprocals. $\endgroup$ – Chappers Jun 9 '18 at 0:43
  • $\begingroup$ I asked this question because I need something more precise than just the magnitude. That is why I didn’t mention magnitude in my question at all. $\endgroup$ – Szeto Jun 9 '18 at 0:49

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