0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ $2z^2+3z+1 = (2z+1)(z+1)$. $\endgroup$ – Math Lover Feb 18 '18 at 18:32
  • $\begingroup$ Yes, you need to do long division first to get down to something that's (at most) linear in the numerator. $\endgroup$ – Ted Shifrin Feb 18 '18 at 18:32
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 .

$\endgroup$
0
$\begingroup$

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$$\frac{\frac{111}8z+\frac{39}8}{(2z+1)(z+1)}=\frac a{2z+1}+\frac b{z+1}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.