# Is order of poles of functions determined by Laurent series?

Suppose

$$f(z) = \frac{1}{(z-2)^5z}$$ is given.

By looking function, i will tell there is a $5$th-order pole at $z=2$ which is in fact true.

But on the other hand suppose

$$f(z) = \frac{\sin z}{z^5}$$ is given.

By looking just function i will again tell there is a $5$th-order pole at $z=0$, beep wrong answer says book.

If we look into Laurent series we see that highest negative power of $z-z_0$ where $f(z)$ is not defined is $4$.

At the end of day, what i want to ask is, while determining order of pole, do we need to look at laurent series?
When one can say intuitively order of pole just by looking to function?

• Regarding the second example recall that $\lim \limits_{z\to 0}\left(\dfrac{\sin(z)}{z}\right)$ exists, so you only need to account for $z\mapsto \dfrac 1{z^4}$. And yes, the order of a pole $z_0$ is determined by the laurent series around $z_0$. Commented Aug 22, 2015 at 11:11
• The order of the pole of $f$ at $a$ is precisely the $n \in \mathbb N$ such that $\lim_{z\to a} f(z)(z-a)^n$ exists and is nonzero.
– zhw.
Commented Aug 22, 2015 at 16:39
• Just think of the series for the sine function, and divide by $z^5$. The answer leaps out at you. Commented Aug 25, 2015 at 3:35