The Road to Reality
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Exercise [04.01]
http://www.roadtoreality.info/viewtopic.php?f=19&t=107
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Author:  chag-art [ 12 May 2008, 19:06 ]
Post subject:  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.

Author:  ZZR Puig [ 22 May 2008, 19:31 ]
Post subject:  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.

Author:  chag-art [ 23 May 2008, 09:40 ]
Post subject:  Re: Exercise [4.1]

Thank you ZZR i didn't noticed it.

Author:  vasco [ 17 Jun 2008, 10:20 ]
Post subject:  Re: Exercise [04.01]

ZZR
I think your (c-ib) on the left hand side should be (c-id)
vasco

Author:  38fr17h9 [ 03 Feb 2009, 03:14 ]
Post subject:  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 -

Author:  vasco [ 03 Feb 2009, 11:29 ]
Post subject:  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

Author:  38fr17h9 [ 05 Feb 2009, 03:35 ]
Post subject:  Re: Exercise [04.01]

Thanks for the help.

Author:  mexican [ 25 May 2009, 04:04 ]
Post subject:  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.

Author:  Langing [ 08 Dec 2010, 12:27 ]
Post subject:  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.

Author:  vasco [ 01 Nov 2013, 16:31 ]
Post subject:  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).

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