The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

Help with a partial fraction please.

Can somebody partial fractionate the following for me please?

z2z4+1\dfrac{z^2}{z^4+1}

I think I'm being really stupid here. I've got to this point A(z2−i)+B(z2+i)=z2 A(z^2 - i) + B(z^2 + i) = z^2 and my brain has gone a bit fuzzy because I can't see what to sub in for z to deduce A and B. Help.
Original post by Preeka
Can somebody partial fractionate the following for me please?

z2z4+1\dfrac{z^2}{z^4+1}

I think I'm being really stupid here. I've got to this point A(z2−i)+B(z2+i)=z2 A(z^2 - i) + B(z^2 + i) = z^2 and my brain has gone a bit fuzzy because I can't see what to sub in for z to deduce A and B. Help.

I wouldn't factorise the denominator that way. Note that z4+1=(z2−2z+1)(z2+2z+1)z^4+1 = (z^2 - \sqrt2 z + 1)(z^2 + \sqrt2 z +1). Then attempt partial fractions.

You've made a mistake in breaking your fraction into the general partial fractions. I'm not even sure if you can do partial fractions with complex denominators using the standard method? Correct me if I'm wrong.

If it is allowed then you should have found that:

z2x4+1≡Az+Bz2+i+Cz+Dz2−i\dfrac{z^2}{x^4+1} \equiv \dfrac{Az+B}{z^2+i} + \dfrac{Cz+D}{z^2-i}.

Rather than just numerators of A and B, as you seem to have used.
(edited 13 years ago)
Reply 2
Avatar for Preeka
Preeka
OP
15
Original post by Farhan.Hanif93
I wouldn't factorise the denominator that way. Note that z4+1=(z2−2z+1)(z2+2z+1)z^4+1 = (z^2 - \sqrt2 z + 1)(z^2 + \sqrt2 z +1). Then attempt partial fractions.

You've made a mistake in breaking your fraction into the general partial fractions. I'm not even sure if you can do partial fractions with complex denominators using the standard method? Correct me if I'm wrong.

If it is allowed then you should have found that:

z2x4+1≡Az+Bz2+i+Cz+Dz2−i\dfrac{z^2}{x^4+1} \equiv \dfrac{Az+B}{z^2+i} + \dfrac{Cz+D}{z^2-i}.

Rather than just numerators of A and B, as you seem to have used.


Thanks :smile:
Original post by Preeka
Thanks :smile:

I know that the factorisation I proposed works fine but I'm not too sure about having denominators of z2+iz^2+i and z2−iz^2-i. I would wait for someone more knowledgable to confirm that.
Reply 4
Avatar for noobynoo
noobynoo
6
If you're trying to find the partial fraction of that then you're probably doing something wrong. Partial fractions are only really useful when they can be done in the real domain.

What is the full question? Or did you just make up this question?
Reply 5
Avatar for Preeka
Preeka
OP
15
Original post by Farhan.Hanif93
I wouldn't factorise the denominator that way. Note that z4+1=(z2−2z+1)(z2+2z+1)z^4+1 = (z^2 - \sqrt2 z + 1)(z^2 + \sqrt2 z +1). Then attempt partial fractions.

You've made a mistake in breaking your fraction into the general partial fractions. I'm not even sure if you can do partial fractions with complex denominators using the standard method? Correct me if I'm wrong.

If it is allowed then you should have found that:

z2x4+1≡Az+Bz2+i+Cz+Dz2−i\dfrac{z^2}{x^4+1} \equiv \dfrac{Az+B}{z^2+i} + \dfrac{Cz+D}{z^2-i}.

Rather than just numerators of A and B, as you seem to have used.


I wouldn't have originally broken it down into complex denominators but that's how my lecturer seems to have done it so it is valid. She's genius so I trust her enough on that account. Maybe I did mess up the numerators parts though.
Reply 6
Avatar for Preeka
Preeka
OP
15
Original post by noobynoo
If you're trying to find the partial fraction of that then you're probably doing something wrong. Partial fractions are only really useful when they can be done in the real domain.

What is the full question? Or did you just make up this question?


The whole question is to do with integration around a contour so it makes use of Cauchy's Integral formula as at one point, partial fractioning is required because both points lie inside the contour in question and in her answer sheet, my lecturer has broken it down like that but obviously assumed I'd be able to deduce the middle steps by myself (which I'm just failing at for this question :colondollar:)
Reply 7
Avatar for noobynoo
noobynoo
6
Original post by Preeka
The whole question is to do with integration around a contour so it makes use of Cauchy's Integral formula as at one point, partial fractioning is required because both points lie inside the contour in question and in her answer sheet, my lecturer has broken it down like that but obviously assumed I'd be able to deduce the middle steps by myself (which I'm just failing at for this question :colondollar:)


fair enough. :colondollar:

In your question have you tried
[br]A=B=1/2[br][br]A=B= 1/2[br]
(edited 13 years ago)
Reply 8
Avatar for Preeka
Preeka
OP
15
Original post by noobynoo
fair enough. :colondollar:

In your question have you tried
[br]A=B=1/2[br][br]A=B= 1/2[br]


yeah A and B do equal 1/2 according to the answers. I just had a hard time seeing why and how. I asked a friend and she said, just substitute z2=iz^2 = i and it works out to give the right values for A and B which I suppose seems obvious now that she told me. I still don't know how mathematically sound that is but if it works, it works. I'm not in the mood to investigate anymore after the realisation that I can't even do partial fractions at times :frown:. Stupid partial fractions. Grr.
Reply 9
Avatar for noobynoo
noobynoo
6
Original post by Preeka
yeah A and B do equal 1/2 according to the answers. I just had a hard time seeing why and how. I asked a friend and she said, just substitute z2=iz^2 = i and it works out to give the right values for A and B which I suppose seems obvious now that she told me. I still don't know how mathematically sound that is but if it works, it works. I'm not in the mood to investigate anymore after the realisation that I can't even do partial fractions at times :frown:. Stupid partial fractions. Grr.


First see how many z^2 there are:

[br]A+B=1[br][br]A+B=1[br]

then do the same with the constants:

[br]Ai−Bi=0[br][br]Ai-Bi=0[br]

solve the simultaneous equations.

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you