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Can someone explain to me how trig functions work in the complex plane? I'm trying to show that $f(z) = \cos(1-\frac{1}{z})$ has an essential singularity at $z=0$, and part of doing that requires I find a sequence $z_n$ so that when $z = z_n$, $\lim\limits_{z \to 0} f(z)=\infty$. Originally I assumed $\cos(z)$ would never approach $\infty$, as it is bounded between 1 and -1, however my thinking was clearly with real variables, as I was informed that $coz(z)$ can in fact approach infinity in the complex plane. I'm trying to understand when this will happen, any information to help my understanding is greatly appreciated!

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You should approach this using the following formula, which is valid for all complex numbers $z$

$$\cos(z)=\frac{e^{iz}+e^{-iz}}{2}$$

We also have the cooresponding formula for sine

$$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$$

These can both be proven by using $e^{iz}=\cos(z)+i\sin(x)$ and solving for the respective quantities.

As pointed out by Rob Israel, these two formulas are sometimes used to define the sine and claims functions on the complex plane. The other way that they are sometimes defined is via their Taylor polynomials.

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  • $\begingroup$ These formulas are often used as the definitions of $\cos(z)$ and $\sin(z)$ for complex $z$. $\endgroup$ – Robert Israel Apr 7 '17 at 21:37
  • $\begingroup$ @RobertIsrael Good point! I'll add that information to my answer. $\endgroup$ – Stella Biderman Apr 7 '17 at 21:41
  • $\begingroup$ @StellaBiderman: I tried to fix the misspelled "corresponding" but it won't let me change just two characters - so you'll have to do it :-) $\endgroup$ – NickD Apr 7 '17 at 22:08
  • $\begingroup$ @Nick thanks for the correction! $\endgroup$ – Stella Biderman Apr 7 '17 at 22:14
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    $\begingroup$ If $z = x + i y$ for real $x$ and $y$, $|\cos(z)|^2 = \cos(x)^2 + \sinh(y)^2$, so this goes to $\infty$ on any path where $|y| \to \infty$. $\endgroup$ – Robert Israel Apr 7 '17 at 23:16
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To explain the deeper answer to the "why" of why it will happen is first off to point out that one cannot and should not expect that the behavior of at least analytic complex functions should mirror that on the real number line.

In particular, one very important theorem of complex analysis is that any non-constant entire function is unbounded (more often-stated converse: a bounded entire function is constant), where "entire" means it is complex differentiable everywhere. The same is not true for real-valued functions, as $\cos$ and $\sin$ illustrate.

(There is another interesting theorem in the same vein which says that an injective entire function is linear, which means that almost all the "nicest" complex functions are not invertible!)

Thus if we are to extend $\cos$ and $\sin$ in any way to the complex plane that preserves their differentiability they must be unbounded and thus there must be some way to approach $\infty$ under them following some suitable path in the plane (which can be discretized to a sequence). It does not matter how we do it -- as long as we want to be able to take a derivative everywhere this must hold. It's even better than that -- if we want to take $\cos$ and $\sin$ as agreeing with their real number equivalents but are willing to admit only an isolated set of nondifferentiability points (i.e. just a few scattered points), we will actually get them to be differentiable everywhere and so they will still be unbounded.

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