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I really don't understand the concept behind finding poles in Complex Analysis and I can't find anything on the internet or in books that helps me grasp the concept...

The following are past exam questions that I'm looking at but don't know where to go with with them in order to find their poles, indicate their order to then compute their residues.. Any help would be greatly appreciated...

$(i) f(z)=\dfrac{\sin z }{(z-1)\sinh z}$

$(ii)g(z)= \dfrac{\sin z}{z(z^2+4)}$

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2 Answers 2

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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.
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Thank you so much, your explanation is so much clearer than anything else i've found!!! I fully understand the simpler example but im still a little confused on the harder example, like the examples I have given.. For (ii) would there be a pole of z=2i with order 1, z=-2i with order 1, then another at z=0 with order 1.. But im not really sure if that is all. Then the zero for the numerator isn't also a zero in the denominator, so that isnt a pole? –  Mathsstudent147 May 3 '13 at 20:30
    
Correct. There are simple poles at $z=\pm 2i$, and since $\sin 0=0$ $z=0$ will be a removable singularity. –  mscook May 3 '13 at 20:34
    
I got my z=0 with order 1 from the z part of the denomiator, as z=0 would make the denomiator zero.. is this right? –  Mathsstudent147 May 3 '13 at 20:37
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Yes, but at $z=0$ the numerator is 0 also ($\sin 0=0$), and they both have the same order. So the zero in the numerator "cancels" with the zero in the denominator, and it will be a removable singularity and not a pole. –  mscook May 3 '13 at 20:57

(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} $

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why does a different formula for the residue have to be used to $z=n\pi i$? –  Mathsstudent147 May 3 '13 at 20:39
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Either formula would work for any of the poles (they hold for all poles of order 1), however I think using them this way is far easier. For $z=1$ the $(z-1)$ cancels out with the first method, so all you have to do is plug in 1. For the $n\pi i$ poles, the limit method would be pretty untractable, but the derivative is easy. –  mscook May 3 '13 at 20:49
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Really they are actually just 2 different ways of approaching the same exact method, use the first approach if it's easier to cancel the $(z-c)$ term and use the second if it's easier to take the derivative of the denominator. –  mscook May 3 '13 at 20:55

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