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I will start by writing the definition of a pole of order m:

Let $f(z) = \Sigma_{n=-\infty}^{\infty}a_k(z-c)^k$ converge in $D_{\rho}(c) $\ $\{c\}$

Then z is a pole of order m if $a_k = O, k<-m$ and $a_{-m} ≠ 0, m≥1$ (translated from french)

Now I will illustrate this definition by an example we saw in class:

$\dfrac{1}{1+z^4} = \dfrac{1}{(z-e^{i\pi/4})(z-e^{i5\pi/4})(z-e^{i3\pi/4})(z-e^{i7\pi/4})}$

We therefore have poles of order $1$ at $z = e^{i\pi/4}$, at $z= e^{i5\pi/4}$ at $z= e^{i3\pi/4}$ and at $z= e^{i7\pi/4}$

I don't understand why those points are poles of order m

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    Of order $\;m=1\;$ ... Each one of them. – 2017-01-08
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    @DonAntonio Yes but I have trouble understanding why they are poles of order 1 – 2017-01-08
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    Because you already have the denominator factored, and each factor is a *different* linear polynomial...! – 2017-01-08
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    @DonAntonio how did youderive that from the definition? – 2017-01-08
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    The answer below tells *exactly* what I intended to tell you. The most important thing: you don't necessarily need the Laurent series. – 2017-01-08

2 Answers 2

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$f(z)=\frac{1}{g(z)}$ has a pole at $z=a$ of order $m$ if $g(z)$ has a zero at $z=a$ of order $m$.

In your case $g(e^{iπ/4})=0$ but $g'(e^{iπ/4})\neq 0$ which implies $g(z)$ has a simple zero ($m=1$) at $z=e^{iπ/4}$, thus $f(z)$ has a simple zero thereat.

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Consider the function

$$g_k(z)=(z-c)^kf(z)$$

If infinitely many of the negative terms in $f(z)=\sum_{-\infty}^{\infty}a_n (z-c)^n$ are non-zero, then $g_k(z)\to\infty$ as $z\to c$, for any choice of $k\in\mathbb{Z}$.

On the other hand, if only finitely many negative terms are non-zero, say $a_m$ is the first non-zero term, then $g_k(z)$ has a limit at $c$ if and only if $k\ge m$ (otherwise $g_k(z)\to \infty$).

In other words, the order of the pole at $c$ is the smallest $k$ such that $(z-c)^kf(z)$ doesn't blow up at $c$. Applying this to your function shows that each pole has order $1$.