Partial fraction decomposition of $\frac{5z^4 + 3z^2 + 1}{2z^2 + 3z + 1}$

Question

So I'm working on this problem:

$$\frac{5z^4 + 3z^2 + 1}{2z^2 + 3z + 1}$$ where $z = x + yi$

It might be that my brain is just blanking right now but, how would one find the partial fraction decomposition of this expression? I tried long division but that didn't simplify out too nicely.

Answer 1

$$\frac{5z^4+3z^2+1}{2z^2+3z+1}=\frac{5}{2}z^2-\frac{15}{4}z+\frac{47}{8}-\frac{9}{z+1}+\frac{\frac{33}{16}}{z+\frac{1}{2}}.$$

Answer 2

hint

the denominator is $$(2z+1)(z+1) $$

the decomposition will take the form

$$az^2+bz+c+\frac {d}{2z+1}+\frac {e}{z+1} $$

you will find $a,b,c $ by Euclidian division and $d,e $ after multiplying by $(2z+1) $ and $(z+1) $ and making $z=-1/2$ to get $d $ then $z=-1$ to get $e $.

If you don't find that $e=-9$, it means you made a mistake .

Answer 3

First of all, you should divide $5z^4+3z^2+1$ by $2z^2+3z+1$, getting$$5z^4+3z^2+1=\left(\frac52z^2-\frac{15}4z+\frac{47}8\right)\left(2z^2+3z+1\right)-\frac{111}8z-\frac{39}8.$$So$$\frac{5z^4+3z^2+1}{2z^2+3z+1}=\frac52z^2-\frac{15}4z+\frac{47}8-\frac{\frac{111}8z+\frac{39}8}{(2z+1)(z+1)}.$$Now, find numbers $a$ and $b$ such that$$\frac{\frac{111}8z+\frac{39}8}{(2z+1)(z+1)}=\frac a{2z+1}+\frac b{z+1}.$$