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I am trying to break the function into partial fractions:

$$ r(z) = \frac{z^2+i}{z^2(z+1)} $$

There is some hint suggesting to divide the numerator by the denominator to get a polynomial & a rational function whose numerator degree < denominator degree.

Can someone please explain how this is done? I can't find any examples.

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So, in your example this is already done - $z^2 + i$ is of degree $2$, which is lower than the degree of $z^2(z+1)$ (which is degree $3$). So this is just a partial fractions problem; write $\frac{z^2 + i}{z^2(z+1)} = \frac{A}{z} + \frac{B}{z^2} + \frac{C}{z+1}$ and solve for $A$, $B$, and $C$.

But to address your question: the technique you're asking about would apply if we were looking at, for example, $\frac{z^3 + i}{z^2(z+1)}$. For this, partial fractions won't work right away; we have to do polynomial division first.

$z^3 + i$ divided by $z^2(z+1) = z^3+z^2$ gives the quotient $1$ with remainder $-z^2 + i$ (I'm omitting the steps of the long division; searching "polynomial long division" on Google will give plenty of excellent examples). What this means is that $\frac{z^3 + i}{z^2(z+1)} = 1 + \frac{-z^2+i}{z^2(z+1)}$. Now the fraction we have left is of the correct form to break into partial fractions.

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  • $\begingroup$ Cant thank you enough. Very helpful. $\endgroup$ – Navy Seal Feb 15 '17 at 3:57
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Here is an alternative way to proceed.

We write the term of interest as

$$\begin{align} \frac{z^2+i}{z^2(z+1)}&=\frac{1}{z+1}+i\left(\frac{1}{z^2(z+1)}\right)\\\\ &=\frac{1}{z+1}+i\left(\frac{(z+1)-z}{z^2(z+1)}\right)\\\\ &=\frac{1}{z+1}+i\left(\frac1{z^2}-\frac{1}{z(z+1)}\right)\\\\ &=\frac{1}{z+1}+i\left(\frac1{z^2}-\frac{1}{z}+\frac{1}{z+1}\right)\\\\ &=\frac{i}{z^2}-\frac{i}{z}+\frac{1+i}{z+1} \end{align}$$

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  • $\begingroup$ It would really help if you could please explain the logic behind the solution. 1) I am not sure how it was broken up like in the first step, 2) Not sure why (z+1) - z is being multiplied. $\endgroup$ – Navy Seal Feb 15 '17 at 3:53
  • $\begingroup$ Sure. Note that $1=(z+1)-z$. $\endgroup$ – Mark Viola Feb 15 '17 at 3:57

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