Order of the pole of $f(z)=\frac{e^{\sin{z}}-1}{z^3}.$

Question

Let

$$f(z)=\frac{e^{\sin{z}}-1}{z^3}.$$

a) Determine if $f$ has a pole at $0$ and determine its order.

b) If $f(z)=\sum_{n=-\infty}^{\infty}a_n$ denote a Laurent serie of $f$ valid in the region $0<|z|<2,$ determine the coefficients $a_{-3},a_{-2},a_{-1}$ and $a_0.$

For a), it seems easy. the multiplicity is 3, thus the order of the pole should be 3. But the answer is 2. How come?

For b), I know the individual taylor series for the functions $\sin{z}$ and $e^{z}$. But I need their respective Laurent series in order to combine them. I'm not sure how I should approach this part.

Answer 1

(a) Near $z = 0$, $\sin z \approx z$, so $e^{\sin z} \approx 1 + z$, so $\frac{e^{\sin z} - 1}{z^3} \approx \frac{1}{z^2}$.

(b) The Taylor series of $e^{\sin z}$ up to order $3$ will give you the corresponding Laurent series for $f$. ($e^z$ and $\sin z$ are entire functions, so their "Laurent series" are just Taylor series). Approximating $e^z = 1 + z + \frac{z^2}{2} + \frac{z^3}{6}$ and $\sin z = z - \frac{z^3}{6}$ is enough.

Answer 2

Let $f(z)=e^{\sin z}$. Then:

• $f(0)=1$;
• $f'(0)=1$;
• $f''(0)=1$;
• $f'''(0)=0$,

you have$$\frac{e^{\sin z}-1}{z^3}=\frac{1+z+\frac{z^2}2+\cdots-1}{z^3}=z^{-2}+\frac{z^{-1}}2+(\text{terms of order greater than $0$)}.$$