# Analysis of singularities and taylor representation of $f(z)=\frac{z^2-1}{\sin \pi z}$

Let $$f(z)=\frac{z^2-1}{\sin \pi z}$$

A) Find all singulartities of $f$ in $\mathbb{C}$ and classify each as a pole (specifying the order), essential, removable, or other.

B) Explain why $f(z)$ has a series expansion of the form $\sum_{k=- \infty}^{\infty}c_kz^k$ valid for $z$ near 0.

C)Which, if any, of the coefficients $c_k$ for $k<0$ are not equal to 0?

D) Find $c_{-1},c_0,$ and $c_1$

E) What is the region of validity for the expansion discussed in part B?

F) Find $\int_{\gamma} f(z) \ dz$ where $\gamma$ is the circle of radius $1$ centered at the origin and traveled once counterclockwise.

My Approaches

A)

All the singularities of $f$ are at $z=k$ for $k \in \mathbb{Z}$. I've seperated into cases:

For $k=0$, we have a simple pole. Let $g(z_0)=z^2-1$, and $h(z_0)=\sin \pi z$. Then there is a residue such that $\operatorname{Res}(f,0)=\frac{g(z_0)}{h'(z_0)}=\frac{-1}{\pi}$. This implies the singularity is simple.

For $k=1$, we have a removable singularity? My claims are that $f(1)=\frac{0}{0}$, therefore by L'Hopital's rule, $f(1)= \lim_{z \to 1} \frac{2z}{\pi \cos \pi z}=\frac{-2}{\pi}$ and therefore the singularity is removable. This is the residue also.

For $k>1$, we have simple poles.

For $k=-1$, we have another removable singularity.

For $k<1$, we have simple poles.

Is my logic correct here?

B) In my complex analysis book we have a statement that says "there's a taylor series for $f(z)$ centered at $c$ which converges to $f(z)$ at least in the largest open disk centered at $c$ and $|z-c|<r$

Given this statement, I want to answer this question by saying that $f$ is analytic on $\mathbb{C} \setminus \{k, k \in \mathbb{C} \}$, which is analytic on a disk centered at the origin of radius less than $\pi$. Thus there's a Taylor expansion around $z=0$

C)

This question I do not know how to answer.

D)

How do I find $c_{-1}$? I'm using $$c_0=\frac{f(0)}{0!}=\lim_{z \to 0}\frac{z^2-1}{\sin \pi z}=0$$

$$c_1=\frac{f'(0)}{1!}=\lim_{z \to 0} \frac{\pi (z^2-1)\sin(\pi z)}{(\cos(\pi z))^2}+\frac{2z}{\cos (\pi z)}=0$$

E)

$$|z-z_0|<\pi$$

F)

$$\int_{\gamma} f(z) dz=2 \pi i\operatorname{Res}(f,0)=2 \pi i \cdot \frac{-1}{\pi}=-2i$$

Can we have a negative imaginary number?

• For C: $f(-z) = -f(z)$ so $f$ is an odd function and therefore $c_{2n} = 0$. $c_{-1}$ cannot be zero since the function has a pole at $z=0$. Commented May 11, 2014 at 1:32
• @Winther Great what about for part d. How would I find $c_{-1}$ Commented May 12, 2014 at 20:00

A: The logic is correct.

B: Since $f$ has a simple pole at $z=0$ the function $zf(z)$ is analytic here and there fore has a Taylor series

$$zf(z) = \sum_{i=0}a_i z^i$$ which gives $$f(z) = \sum_{i=-1}c_i z^i$$ with $c_i = a_{i+1}$

C/D: From $B$ we have that

$$c_{-1} = \lim_{z\to 0} zf(z) = -\frac{1}{\pi}$$

Your $c_0$ is wrong. As $f(z)$ is even we have $c_{2i} = 0$ and therefore $c_0 = 0$.

To find the smallest $c_i$ we can use $\sin \pi z = \pi z - (\pi z)^3/6 + ...$ to find

$$f(z) \approx \frac{z^2-1}{\pi z(1 - (\pi z)^2/6 + ...)} = \frac{(z^2-1)}{\pi z}(1 + (\pi z)^2/6 + ...)$$

Expand the product and read off $c_{-1},c_0,c_1$

E:

Hint: The series cannot be valid for $z=\pm 2$ as the original series has a pole here.

F: This seems correct. The integral is equal to $2\pi i c_{-1}$ which is correct. Yes you can have negative imaginary numbers: A imaginary number is a number on the from $a+bi$ where $a,b$ real so you can have $b<0$.