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 Exercise [04.01] 
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Joined: 11 May 2008, 20:07
Posts: 11
Location: france
Post Exercise [04.01]
Let's start from the begginning.
We have the equation :

\frac{(a+ib)}{(c+id)}=\frac{ac+bd}{c^2+d^2}+i\frac{bc-ad}{c^2+d^2}

As recommended by Mr Penrose, we first verify this equation multiplying on both sides with c+id :

a+ib=\frac{(ac+bd)(c+id)}{c^2+d^2}+i\frac{(bc-ad)(c+id)}{c^2+d^2}

Then we develop separating reals and complex :

a+ib=\frac{ac^2+bcd-bcd+ad^2}{c^2+d^2}+i\frac{acd+bd^2-acd+bc^2}{c^2+d^2}

And there we are, with :

a+ib=a+ib



Now the other part of the exercise is not as quick as this one but still interesting :

\frac{(a+ib)(c-id)}{c+id}=\frac{(ac+bd)(c-id)}{c^2+d^2}+i\frac{(bc-ad)(c-id)}{c^2+d^2}

But now we can use the relation :

(c+id)(c-id)=c^2+d^2

So we can rewrite the previous equality as :

\frac{(a+ib)(c-id)}{c+id}=\frac{(ac+bd)(c-id)}{(c+id)(c-id)}+i\frac{(bc-ad)(c-id)}{(c+id)(c-id)}

And we find :

\frac{(a+ib)(c-id)}{c+id}=\frac{(ac+bd)}{(c+id)}+i\frac{(bc-ad)}{(c+id)}

This is a final equality, because we can get it explaining the left hand side or simplifying the right hand side.


Last edited by chag-art on 23 May 2008, 09:41, edited 1 time in total.

12 May 2008, 19:06
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Joined: 22 May 2008, 19:08
Posts: 17
Post Re: Exercise [4.1]
I think you made a small typing mistake, changing a plus sign with a minus sign, so the last equation should be:

\frac{(a+ib)(c-ib)}{c+id}=\frac{(ac+bd)}{(c+id)}+i\frac{(bc-ad)}{(c+id)}

Or simply:

\frac{(a+ib)(c-ib)}{c+id}=\frac{(ac+bd)+i(bc-ad)}{c+id}

P.S.: Already corrected in the original post.


Last edited by ZZR Puig on 23 May 2008, 14:29, edited 1 time in total.

22 May 2008, 19:31
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Joined: 11 May 2008, 20:07
Posts: 11
Location: france
Post Re: Exercise [4.1]
Thank you ZZR i didn't noticed it.


23 May 2008, 09:40
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Joined: 07 Jun 2008, 08:21
Posts: 182
Post Re: Exercise [04.01]
ZZR
I think your (c-ib) on the left hand side should be (c-id)
vasco


17 Jun 2008, 10:20
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Joined: 03 Feb 2009, 03:09
Posts: 3
Post Re: Exercise [04.01]
It looks as if the reduction labelled "Then we develop separating reals and complex" includes an intermediate step. Could someone please break this one out?

Many thanks -


03 Feb 2009, 03:14
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Joined: 07 Jun 2008, 08:21
Posts: 182
Post Re: Exercise [04.01]
Quote:
It looks as if the reduction labelled "Then we develop separating reals and complex" includes an intermediate step. Could someone please break this one out?

Quote:
a+ib=\frac{(ac+bd)(c+id)}{c^2+d^2}+i\frac{(bc-ad)(c+id)}{c^2+d^2}

Here goes:
All you do is multiply out the brackets as you would in real algebra, as follows:

=\frac{ac^2+bcd+acdi+bd^2i}{c^2+d^2}+i\frac{bc^2+bcdi-acd-ad^2i}{c^2+d^2}

=\frac{ac^2+bcd+acdi+bd^2i+bc^2i+bcdi^2-acdi-ad^2i^2}{c^2+d^2}

=\frac{ac^2+bcd+acdi+bd^2i+bc^2i-bcd-acdi+ad^2}{c^2+d^2} since i^2=-1

=\frac{(ac^2+bcd-bcd+ad^2)+i(acd+bd^2-acd+bc^2)}{c^2+d^2}

=\frac{a(c^2+d^2)+ib(c^2+d^2)}{c^2+d^2}

=a+ib

Hope this helps.

Vasco


03 Feb 2009, 11:29
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Joined: 03 Feb 2009, 03:09
Posts: 3
Post Re: Exercise [04.01]
Thanks for the help.


05 Feb 2009, 03:35
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Joined: 25 May 2009, 03:16
Posts: 2
Post Re: Exercise [04.01]
There is an easier way to do this: Simply multiply the left hand side by (c-id) / (c-id), and after a shorter algebra you get the right hand side. You can multiply only the left hand side because you are multiplying by one.


25 May 2009, 04:04
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