Understanding Fractional Reduction: Solving Complex Fractions with Ease

[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 can't 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

 

[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

 

[toneboy1]

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

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 can't see what you mean by grouping the positive and negative ones...

 

[tiny-tim]

hi toneboy1!

(just got up :zzz:)

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

 

[toneboy1]

Well hopefully you've had your coffee now.

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

Thanks

 

[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

 

[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.

 

[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? ​

 

[toneboy1]

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

Thanks

 

[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

 

[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