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Find the partial fractions of $\frac{z^4}{z^3-1}$ and $\frac{1}{z(z+1)^2(z+2)^3}$ where z is complex number

I am not sure about what form satisfies so called partial fraction $\frac{z^4}{z^3-1}$ can be decomposed as $\frac{1}{z(z+1)^2(z+2)^3}$. Is it partial fraction?

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  • $\begingroup$ Partial fractions --- each denominator should be linear, or a quadratic with no real roots, or a power of wither of those two. $\endgroup$ – Gerry Myerson Feb 6 '14 at 11:09
  • $\begingroup$ And the starting fraction must be proper (deg num. < deg den.). $\endgroup$ – Martín-Blas Pérez Pinilla Feb 6 '14 at 11:14
  • $\begingroup$ @GerryMyerson so 1/(z^2+z+1) is OK? $\endgroup$ – user98619 Feb 6 '14 at 12:57
  • $\begingroup$ Here z is a complex number.So quadratic should be decomposed also I guess.Is it right? $\endgroup$ – user98619 Feb 6 '14 at 13:12
  • $\begingroup$ You may be right; it's hard to read the intentions of the person asking the question from the bare statement of the question. If it's intended to do partial fractions over the complex numbers, then, yes, each denominator should be (a power of) a linear polynomial. $\endgroup$ – Gerry Myerson Feb 6 '14 at 23:47
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$ \frac{z^4-z+z}{(z^3-1)}= z+\frac{z}{(z^3-1)}$ and $$\frac{z}{(z^3-1)}=\frac{z}{(z-1)(z^2+z+1)}=\frac{A}{(z-1)}+\frac{Bz+C}{(z^2+z+1)}=$$ and $$\frac{1}{z(z+1)^2(z+2)^3}=\frac{A}{z}+\frac{B}{(z+1)}+\frac{C}{(z+1)^2}+\frac{D}{(z+2)}+\frac{E}{(z+2)^2}+\frac{F}{(z+2)^3}$$

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  • $\begingroup$ It is right when z is real.Consise answer $\endgroup$ – user98619 Feb 8 '14 at 19:31

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