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Let $f(z)=(e^{3iz}-3e^{iz}+2)/z^3$, this clearly has a singularity at $z=0$, how do you show that this is a simple pole? i.e. a pole of order one, every way it definity expands to something with a minimum power of $-3$, so I come to the conclusion there is a pole of order $3$?

And if this is the case is $f(z)+3/z$ holomorphic? Does this term cancel some part of the expansion I have missed that eliminates the pole?

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Although the denominator has a root at z=0 of multiplicity 3, the numerator also has a root there. What is the multiplicity of the root z=0 of the numerator? –  hardmath Nov 4 '11 at 22:57

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up vote 1 down vote accepted

Consider $g(z) = e^{3iz} - 3e^{iz} + 2$. Then $g(0) = 1 - 3 + 2 = 0$, $g'(0) = 3i - 3i = 0$, and $g''(0) = (3i)^2 - 3i^2 = -9 + 3 = -6 \neq 0$.

So $g(z)$ has a zero of order $2$ at $z = 0$, which means that $g(z)/z^3$ has a simple pole at $z = 0$.

Since $g(z) = g(0) + g'(0)z + {1 \over 2}g''(0)z^2 + ...$, the function $g(z)$ can be written as $-3z^2 + $ higher order terms. So ${\displaystyle f(z) = {g(z) \over z^3} = -{3 \over z}}$ plus an function that is analytic near $z = 0$. Hence ${\displaystyle f(z) + {3 \over z}}$ is analytic on a neighborhood of $z = 0$, and therefore on all of ${\mathbb C}$ since the function is explicitly defined in an analytic fashion except at $z = 0$.

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Thank you, this is very well explained! –  Freeman Nov 5 '11 at 14:44

For $|z|\ll 1$, we have $$ \begin{split} e^{3iz} &= 1 + 3 i z - \frac{9}{2} z^2 + O(z^3) \\ e^{iz} &= 1 + i z - \frac{1}{2} z^2 + O(z^3) \end{split} $$ so $$ \begin{split} \frac{1}{z^3} (e^{3iz} - 3 e^{iz} +2 ) &= \frac{1}{z^3} \left( 1 - 3 + 2 + (3 - 3) i z + (-\frac{9}{2} + \frac{3}{2}) z^2 + O(z^3) \right) \\ &= - \frac{3}{z} + O(1) \end{split} $$

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Thanks! this is made it much more clear.. probably should have got here on my own!! –  Freeman Nov 5 '11 at 14:44

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