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

The isolated singularities are I and $1\pm 2i$

How do I determine if they are poles or not? and specificly how do I determine their order?. Also what are their connection to the residues of f?

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Hint

The singularities are $\pm i$ and $1\pm 2i$. For the order, if $a$ is an isolated singularity of $f$, then the smallest $n$ s.t. $f(z)(z-a)^n$ is holomorphic gives you the order.

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    I know that. but how do I determine if the limit $f(z)(z-a)^n$ exists? – 2017-01-25
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    In general, if $f$ and $g$ are derivable and $g(x),g'(x)\neq 0$ in $A$, then $\frac{f}{g}$ is well defined and derivable on $A$. – 2017-01-25
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And $-i$ are they poles...

Is $f(z)$ defined at these values? i.e. are you trying to divide by 0? If you are then you have a pole.

Poles of order 2 would be a root of multiplicity in the denominator. Do you have any of those?

The residuals. You can break $f(z)$ into partial fractions.

Or for each pole $(z = a), \lim_\limits {z\to a} = f(z) (z-a)$