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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?

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    $\begingroup$ 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 $z_0$. $\endgroup$
    – Git Gud
    Commented Aug 22, 2015 at 11:11
  • $\begingroup$ 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. $\endgroup$
    – zhw.
    Commented Aug 22, 2015 at 16:39
  • $\begingroup$ Just think of the series for the sine function, and divide by $z^5$. The answer leaps out at you. $\endgroup$
    – Lubin
    Commented Aug 25, 2015 at 3:35

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Hint: While finding the order of a pole at z=a for a rational function f(z)/g(z) , just keep in mind that at point z=a, f(z) should be analytic and non-zero there.

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