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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

Joined: 22 May 2008, 19:08
Posts: 18
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

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: 235
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

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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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

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


05 Feb 2009, 03:35

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

Joined: 08 Dec 2010, 11:20
Posts: 5
Location: Durham, North Carolina, USA
Post Re: Exercise [04.01]
mexican wrote:
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.


Thanks mexican. Your approach is direct and simple. After struggling, I was amazed to see how easy you made the solution. Penrose should have suggested "multiply by (c-id)/(c-id)" as part of his assignment in the book, rather than saying "multiply by c-id," which threw me into the weeds.


08 Dec 2010, 12:27
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Post Re: Exercise [04.01]
Langing wrote:
mexican wrote:
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.


Thanks mexican. Your approach is direct and simple. After struggling, I was amazed to see how easy you made the solution. Penrose should have suggested "multiply by (c-id)/(c-id)" as part of his assignment in the book, rather than saying "multiply by c-id," which threw me into the weeds.


There is no real difference between multiplying throughout by c-id and multiplying the LHS by (c-id)/(c-id).


01 Nov 2013, 16:31
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