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Given is $f(z)=\sin(\exp(\frac{1}{z}))$. How do I find singularities and residue?

I know that singularity for my function is $z_0=0$. But how do I find residue?

Daniel Fischer
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Maica
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2 Answers2

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The residue of $f$ at $0$ is

$$\frac{1}{2\pi i} \int_{\lvert z\rvert = r} f(z)\,dz,\tag{1}$$

where since $f$ is holomorphic on $\mathbb{C}\setminus \{0\}$ the radius $r$ can be chosen arbitrarily. (Generally, it must be chosen small enough that the circle doesn't enclose nor pass through any other singularity.)

Making the substitution $w = 1/z$ in $(1)$ yields - note that the substitution leads to a negatively oriented circle; we then use the minus sign of the derivative to change the orientation -

$$\operatorname{Res} (f;0) = \frac{1}{2\pi i} \int_{\lvert w\rvert = R} \frac{\sin (\exp w)}{w^2}\,dw$$

(where $R = 1/r$), which is easy to evaluate if one remembers the integral formula for derivatives.

Daniel Fischer
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  • $Z=0$ is a singularity but it is a pole? – Emilio Novati Jun 15 '17 at 12:05
  • No, it's an essential singularity. – Daniel Fischer Jun 15 '17 at 12:05
  • But what is my 'r'? – Maica Jun 15 '17 at 12:08
  • @Majica Choose what you like. $r = 1$, or $r = 1/10$, it doesn't matter. Or don't choose at all and let it be an abstract value. The integrals don't depend on the radius of the respective circle. – Daniel Fischer Jun 15 '17 at 12:11
  • But we have R and that is 1/r it depends. – Maica Jun 15 '17 at 12:22
  • Not really @Majica. By Cauchy's integral theorem, the integral over $\lvert z\rvert = r$ is equal to the integral $\lvert z\rvert = s$ for all $r,s > 0$. It doesn't matter which radius one chooses. And you don't evaluate the last integral by parametrising the circle, but by appealing to the integral formula (or the residue theorem), so there is no need to choose a radius. – Daniel Fischer Jun 15 '17 at 12:32
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As an alternative to the way forward presented by Daniel, we can expand the function $f(z)$ in a Laurent series. Proceeding we write

$$\begin{align} f(z)&=\sin\left(e^{1/z}\right)\\\\ &=\sin\left(1+\frac1z+O\left(\frac{1}{z^2}\right)\right)\\\\ &=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\left(1+\frac1z+O\left(\frac{1}{z^2}\right)\right)^{2n+1}\\\\ &=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}+\left(\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}\right)\,\frac1z+O\left(\frac{1}{z^2}\right)\\\\ \end{align}$$

The residue is, therefore, $\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}=\cos(1)$.

Mark Viola
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