Fractional Reduction TROUBLE

1. Nov 17, 2012

toneboy1

Hi, this seems like the right section to post, I have a few questions, three fractions that I know the answer for but I just cant figure out how they were arrived at (the middle step).
The only way I can think of solving them is expanding it all out etc. which would result in a 100 term cubic polynomial, so I'm sure there is an easier way to get to the final fraction.
If anyone can help me solve either of these three that would be great.
Please see the pictures for the questions.

P.S the 'j' is 'i', that is the square root of -1.

Thanks

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2. Nov 17, 2012

tiny-tim

hi toneboy1!

for the first one, two hints:

i] the denominator of the solution is a2 + b2, which factors as … ?

ii] the first two denominators are almost identical, only some of the terms are multiplied by -1, so group together all the terms that are and all the terms that aren't

3. Nov 17, 2012

toneboy1

Thanks for the reply, I appreciate it.

Err...well I do remember a2+b2=(a+b)2-2ab
and I can see the solution of the first is squared like that but I don't see how it was originally (a+b)2-2ab.

Likewise I can see they're almost identical but I cant see what you mean by grouping the positive and negative ones....

4. Nov 18, 2012

tiny-tim

hi toneboy1!

(just got up :zzz:)

a2+b2 = a2 - (ib)2 = (a + ib)(a - ib) ?

5. Nov 18, 2012

toneboy1

Please, treat me like an idiot, I still don't see quite what your getting at.

Thanks

6. Nov 18, 2012

tiny-tim

try factoring the denominator of the RHS (the a2 + b2 one) …

then compare it with the LHS, and you may see what's going on

7. Nov 18, 2012

toneboy1

Ok, applying '(a + ib)(a - ib)' to the RHS I got: (-w2+4jw-1-4w+j8)(-w2+4jw-1+4w-8j)...which doesn't mean a hell of a lot to me.

But I only included the RHS answer to the question for varifications sake, it shouldn't be really necessary.

8. Nov 18, 2012

tiny-tim

ok, for the LHS, write out separately the terms in the denominators which are the same for both denominators, and the terms which are minus each other …

what do you get?

9. Nov 19, 2012

toneboy1

You're the boss.
Ok well than you'd get: 16j - 8jw - 8w on the denominator (?)

Thanks

10. Nov 19, 2012

tiny-tim

uhh?

if you expand each denonimator on the LHS, there are 9 terms which are the same apart possibly for a minus sign (5 are the same, 4 are minus)

write out the ones that are the same, and then the ones that are minus

11. Nov 19, 2012

toneboy1

what I previously posted was what I thought the result was, as the denominators I expanded to be:

[-w2+4w+4jw-8j-1]-[-w2-4w+8j+4jw-1] = 16j - 8jw - 8w ...or was it -8w + 16j...

Thanks