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I want to find the residue of $\displaystyle f=\frac{e^z+1}{\sin z}$ at $z=0$.

My solution was $2$, which I got using $\displaystyle \frac{e^z+1}{\sin z}=\frac{e^z}{\sin z}+\frac{1}{\sin z}$.

The residue for $\displaystyle Res_0\left( \frac{1}{\sin z} \right)=\lim_{z\rightarrow 0}z\frac{1}{\sin z}=\lim_{z\rightarrow 0}\frac{1}{\cos z}=1$

The residue for $\displaystyle Res_0\left( \frac{e^z}{\sin z} \right)=\lim_{z\rightarrow 0}z\frac{e^z}{\sin z}=\lim_{z\rightarrow 0}\frac{e^z+ze^z}{\cos z}=1$

Adding up the residues, the answer would be $2$.

Another question I have is regarding the order of residues. I know that for $\displaystyle \frac{a}{z^n}$ the order is $n$, but what happens with non-polynomials function like trigonometric in my case? I assumed the order of $0$ in $\displaystyle\frac{1}{\sin z}$ to be $1$, but I am not sure of what argument I could give here.

Cure
  • 4,051

2 Answers2

1

We can use the Taylor series expansion of $\exp$ and $\sin$ to determine the residue. It is convenient to use the Landau Big-O notation.

We obtain \begin{align*} \frac{e^z+1}{\sin z}&=\frac{2+O(z)}{z+O(z^3)}\tag{1}\\ &=\frac{2}{z}\left(\frac{1+O(z)}{1+O(z^2)}\right)\tag{2}\\ &=\frac{2}{z}(1+O(z))(1+O(z^2))\tag{3}\\ &=\frac{2}{z}(1+O(z)) \end{align*} and it follows \begin{align*} \color{blue}{\mathrm{Res}_0\left( \frac{e^z+1}{\sin z} \right)=2} \end{align*}

Comment:

  • In (1) we expand $e^z=1+z+\frac{z}{2}+\cdots=1+O(z)$ and $\sin z=z-\frac{z^3}{3!}+\cdots=z+O(z^3)$.

  • In (2) we factor out $\frac{2}{z}$.

  • In (3) we use the geometric series expansion and obtain $$\frac{1}{1+O(z^2)}=1+O(z^2)+O(z^4)+\cdots=1+O(z^2)$$

Markus Scheuer
  • 108,315
1

The function has a pole of order $1$ at $0$, because $$ \lim_{z\to 0}z\frac{e^z+1}{\sin z}=2 $$ Thus the residue is $2$.

If you don't believe it, write $$ \frac{e^z+1}{\sin z}=\frac{a_{-1}}{z}+a_0+a_1z+\dotsb $$ and compute this way the limit above.

egreg
  • 238,574