# How do I find singularities and residue?

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?

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.

• $Z=0$ 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'? – Majica 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. – Majica Jun 15 '17 at 12:22

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

• Please let me know how I can improve my answer. I really want to give you the best answer I can. – Mark Viola Jun 30 '17 at 22:27