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For $z\in\Bbb C$ let $$ f(z) = \frac{\sin z}{z} $$ Along the real line this is well behaved, and approaches $1$ as $z\to 0$.

But is $f(z)$ analytic at the origin ($z=0$)?

I tried explicitly checking the Cauchy conditions but that gets ugly (unless I am missing something).

The function I was originally interested in is $$ g(z) = z\,\sin\left( \frac{1}{z} \right) $$ which my gut tells me must not be analytic at the origin, but is the singularity essential or a pole, or the end of a branch cut or something even uglier?

Mark Fischler
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    look at the Taylor series to figure out the behaviour of $f$ – Thomas May 27 '16 at 17:36
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    Of course the behaviours at the origin of $f(z)$ and $g(z)$ have nothing to do with each other – guestDiego May 27 '16 at 17:40
  • $\sin z$ has a simple zero at $z=0$, $z$ too, hence $f(z) = \frac{\sin z}{z}$ is analytic at $z=0$ (with $f(0) = 1$...) – reuns May 27 '16 at 17:40
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    @guestDiego: Took the words right out of my mouth. :-) – Brian Tung May 27 '16 at 17:40
  • On the other hand $g(z)$ at $z=0$ and $f(z)$ at $z=\infty$ are quite closely related... – Lee Mosher May 27 '16 at 17:43
  • The function $z \sin(1/z)$ has an essential singulairty at $z$; look at the Laurent series the main part (the negative powers of $z$) is not finite. – quid May 27 '16 at 17:45
  • The following are not proofs, but they can perhaps be made rigorous. 1) $\frac{\sin z}z=\prod_{i=1}^\infty\left(1-\frac{x^2}{n^2}\right)$, so it's a product of analytic functions near zero. 2) $\frac{\sin z}{z}$ is the zeroth spherical Bessel function, and so is the solution to an ODE which has a regular singular point at 0. 3) $\frac{\sin z}{z}$ is the Fourier transform of the rectangular step function and thus is the FT of a function with compact support. – Semiclassical May 27 '16 at 18:29

2 Answers2

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As defined, no, as it isn't well-defined at $z=0$. It does, however, have an analytic continuation to the entire plane. What you can observe is that $\sin(z)$ has a zero of order $1$ at $z=0$, and therefore the singularity of $\sin(z)/z$ has order 0 at $z=0$, meaning it is removable and an analytic continuation at $z=0$ exists (and is given by the limit). Equivalently, $$\lim_{z\rightarrow 0} \sin(z)/z = 1$$ holding over complex numbers means that the singularity is removable.

As for your $g(z)$ function, we can write $g(z)=f(1/z)$ and $2\sin(z) = e^{iz}-e^{-iz}$, and so for $x\in\mathbb R$ and $n\in\mathbb N$ in the following we have $$\lim_{x\rightarrow 0}|x^n g(x)| = \lim_{x\rightarrow\pm\infty} |f(x)/x^n| = 0$$ whereas $$\lim_{x\rightarrow 0}|x^n g(ix)| = \lim_{x\rightarrow\pm\infty} |f(ix)/x^n| = \infty,$$ which means that $g$ has an essential singularity at $z=0$

zibadawa timmy
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  • $\sin(z)$ has a simple zero at $z=0$, i.e. $\sin(z) = z ,h(z)$ for some analytic function $h(z)$, hence $\frac{\sin(z)}{z} = h(z)$ is analytic – reuns May 27 '16 at 17:41
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    @user1952009 Technically, $\sin(z)/z$ isn't defined at $z=0$, whereas $h(z)$ is, so they cannot be the same function. Granted it's not uncommon in (complex) analysis to agree to the convention that equalities of functions are all with respect to limits unless otherwise indicated, in which case, yes, you're spot on. – zibadawa timmy May 27 '16 at 17:46
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    @zibadawatimmy Another way to put it is to say that we're considering division in the ring of entire functions instead of pointwise... – David C. Ullrich May 27 '16 at 17:57
  • in complex analysis we don't consider functions but analytic continuation of functions. so in general $f(z) = \frac{\sin z}{z}$ is defined at $z=0$. but if you are afraid, you can write that $\frac{\sin z}{z} = h(z)$ for every $z\ne 0$, with $h(z)$ being analytic... – reuns May 27 '16 at 19:22
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Since $$ \sin z=\sum_{k\ge0}\frac{(-1)^k}{(2k+1)!}\,z^{2k+1} $$ we have, for $z\ne0$, $$ \frac{\sin z}{z}=\sum_{k\ge0}\frac{(-1)^k}{(2k+1)!}\,z^{2k} $$ Since this power series has radius of convergence infinite, it defines an entire function, which is an analytic continuation of $(\sin z)/z$ at $0$, where it has the value $1$.

So, as written, $(\sin z)/z$ is not analytic at $0$, because it's not defined there, but $0$ is a removable singularity.

For $z\sin(1/z)$ you can use the same series: $$ z\sin\frac{1}{z}=\sum_{k\ge0}\frac{(-1)^k}{(2k+1)!}\,z^{-2k} $$ which holds for $z\ne0$. This is the Laurent series for the analytic function $z\sin(1/z)$ at $0$ and it shows the singularity is essential, because the Laurent series at a pole has only a finite number of terms with $z$ having negative exponent.

egreg
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