Finding poles, indicating their order and then computing their residues

Solution 1:

The general, intuitive idea of poles is that they are points where evaluating your function would entail dividing by zero. The order of the pole is the exponent in the factor that is going to zero in the denominator. It's best to start with some simple examples, such as rational functions:

$$ f(z) = \frac{(z + 1)(z - 2)}{(z + 1)(z - 1)(z-3)^2}$$

Notice that the denominator goes to zero at $z = -1, 1, 3$. For $z = -1$, however, there's also a copy of $(z + 1)$ on the top, so this is a pole of order zero, or a removable singularity, so it normally doesn't count. At $z = 1$, we have one copy of $(z - 1)$ in the denominator, so it's a pole of order $1$. For $z = 3$, we have $(z - 3)$ with multiplicity $2$, so it's a pole of order $2$.

Now let's look at a slightly more interesting example:

$$f(z) = \frac{z}{ \sin{z}}$$

As we all know, $\sin(z) = 0$ when $z = n \pi$, where $n$ is an integer, and furthermore, all these zeros of $\sin(z)$ are single roots, so naturally we might think that $f(z)$ has poles of order $1$ at $z = n \pi$ for every $n$. However, the $z$ in the numerator cancels out the zero at $z = 0$ in the denominator, so in fact $f(z)$ in this case has a pole of order $1$ for $z = n \pi$, where $n$ is a nonzero integer.

So the general strategy can be described as this:

• Identify all the zeros in the denominator, along with their multiplicities.
• Identify any zeros in the numerator that are also zeros in the denominator, along with their multiplicities.
• Each zero in the denominator is a pole whose order is given by taking the multiplicity in the denominator and subtracting away the multiplicity in the numerator. If the order isn't positive, then it actually isn't a pole.

Solution 2:

(i) The poles of $f(z)=\dfrac{\sin z}{(z-1)\sinh z}$ will be the poles of $\sin z$ along with the zeroes of $(z-1)\sinh z$). Since $\sin(z)$ is analytic it will have no poles. Thus the poles will by $z=1$ and the zeros of $\sinh z$, which are $z=n\pi i$ for $z\in \mathbb{Z}$. and they will all be simple.

At $z=1$, we can use the formula $Res(f,c)=\lim_{z\rightarrow c}(z-c)f(z)$. Thus $Res(f,1)=\lim_{z\rightarrow 1}\dfrac{\sin z}{\sinh z}=\dfrac{\sin 1}{\sinh 1}$

For the $z=n\pi i$ poles, we want to use the formula $Res(\dfrac{g}{h},c)=\dfrac{g(c)}{h^{\prime}(c)}$, so $Res(f,n\pi i)=\dfrac{sin(n\pi i)}{(n\pi i-1)\cosh (n\pi i)}=\dfrac{\sinh(n\pi)}{n\pi i-1} $