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 Basic complex number 'magic' 
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Joined: 25 Feb 2008, 13:32
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Location: Cape Town, South Africa
Post Basic complex number 'magic'
Sameed posted a question about two months ago; I'd like to do the same, but this one is alot easier. Still interesting though.

Why does i=- \frac{1}{i} ?

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14 Oct 2008, 13:31

Joined: 12 Mar 2008, 10:57
Posts: 69
Location: India
Post Re: Basic complex number 'magic'
I see that Shaun most probably wants a geometric insight rather than mere algebraic manipulation.Best of luck to everyone!

P.S. You must be wondering why I haven't posted much lately.That's because I am busy reading Visual Complex Analysis by T.Needham.(Remember Prof. Penrose recommends the interested reader to Needham in Chapter 7)Its a beautiful,no an extremely beautiful book.(And it has the answer to Shaun's problem!)


16 Oct 2008, 06:39

Joined: 24 Dec 2008, 12:31
Posts: 2
Post Re: Basic complex number 'magic'
let's start with a basic equation:
-1=-\frac{1}{1}
now we'll take the root of either side:
\sqrt{-1}=i
and:
i=\sqrt{-\frac{1}{1}}=\frac{\sqrt{1}}{\sqrt{-1}}=\frac{1}{i}
but also:
i=\sqrt{-\frac{1}{1}}=\frac{-\sqrt{1}}{\sqrt{-1}}=\frac{-1}{i}=-\frac{1}{i}
qed


24 Dec 2008, 12:55
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Post Re: Basic complex number 'magic'
tulcod wrote:
now we'll take the root of either side:
\sqrt{-1}=i
and:
i=\sqrt{-\frac{1}{1}}=\frac{\sqrt{1}}{\sqrt{-1}}=\frac{1}{i}\ \ \ \ (2)
but also:
i=\sqrt{-\frac{1}{1}}=\frac{-\sqrt{1}}{\sqrt{-1}}=\frac{-1}{i}=-\frac{1}{i}
qed

Be careful what you are doing here!!
i^2=-1 by definition, but you can't then say \sqrt{-1}=i.
Ask yourself this question: Why did I take the positive square root of 1 and the positive square root of -1 in (2)?

It's like saying:
\frac{\sqrt{9}}{-3}=\frac{+3}{-3}=-1 choosing the positive square root of 9 and then saying:
But \sqrt{9}=-3 also, and so \frac{\sqrt{9}}{-3}=\frac{-3}{-3}=+1.
Therefore -1=+1!!!


You must say \sqrt{-1}=\pm i or \pm \sqrt{-1}=i. Otherwise you can end up saying i^2=1 just as you do in 2 above.

Look at it this way:
If you multiply both sides of Shaun's original equation by i then you get i^2=-1.
Another way of looking at it is that if you divide -1 by its square root (i) then by definition you must get i


24 Dec 2008, 20:18

Joined: 12 Mar 2008, 10:57
Posts: 69
Location: India
Post Re: Basic complex number 'magic'
Think of multiplication by "i" as a transformation of the complex plane i.e. rotation by a right angle in the anticlockwise direction. Multiplication by "1/i" is rotation by a right angle in the clockwise sense. Hence, the minus sign.

P.S. Back here after more than a year. This book did introduce me to the joy of mathematics after all!


03 Jul 2010, 13:31

Joined: 16 Jul 2010, 07:42
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Location: Florence, Italy
Post Re: Basic complex number 'magic'
One way of looking at the problem is to treat complex numbers as real number pairs: (a,b) with the usual rule for complex addition:
(a,b) + (c,d) = (a+b, c+d)
and a funny combination rule (complex multiplication) :
(a,b) . (c,d) = (ac - bd, bc + ad)
That way you don't have to worry about the 'square root of -1' at all.
Then, following the rule for division that I learnt in primary school, but extending it to complex division, your problem becomes:
'find the complex number (a,b) that when multiplied (using the above rule) by (0,1) becomes (-1, 0)'
and we find the answer is (0,1)


16 Jul 2010, 08:48
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Joined: 16 Jul 2010, 07:42
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Post Re: Basic complex number 'magic'
Sorry, I made a mistake in my last post (too much hurry). Addition rule is of course:
(a,b) + (c,d) = (a+c, b+d)
Tony


16 Jul 2010, 08:56
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Post P.S: Basic complex number 'magic'
Actually, I think that the section 4.1 of the book is misleading, in that we are not just extending the real numbers by adding a single element i. In fact we are creating a whole new set of abstract entities, each of which can be labelled with a real number pair (a,b), (Cartesian or polar as preferred) and with special rules for complex addition and complex multiplication. It is just lucky that this set of abstract entities, which incorporates the real numbers ((a,0) in Cartesian coordinates) turns out to be so useful.


16 Jul 2010, 10:41
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Post Re: Basic complex number 'magic'
tonychinnery wrote:
One way of looking at the problem is to treat complex numbers as real number pairs: (a,b) with the usual rule for complex addition:
(a,b) + (c,d) = (a+b, c+d)
and a funny combination rule (complex multiplication) :
(a,b) . (c,d) = (ac - bd, bc + ad)
That way you don't have to worry about the 'square root of -1' at all.
Then, following the rule for division that I learnt in primary school, but extending it to complex division, your problem becomes:
'find the complex number (a,b) that when multiplied (using the above rule) by (0,1) becomes (-1, 0)'
and we find the answer is (0,1)


Yes but this doesn't give you any insight into anything. One of the most interesting aspects of complex numbers is the amazing geometrical interpretations they have.
As you say "a funny combination rule (complex multiplication)" - but if you look at it geometrically, interpreting i as an anticlockwise rotation through 90, you see that complex multiplication is a combined magnification/reduction and a rotation.


09 Aug 2010, 14:01
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Post Re: P.S: Basic complex number 'magic'
tonychinnery wrote:
Actually, I think that the section 4.1 of the book is misleading, in that we are not just extending the real numbers by adding a single element i. In fact we are creating a whole new set of abstract entities, each of which can be labelled with a real number pair (a,b), (Cartesian or polar as preferred) and with special rules for complex addition and complex multiplication. It is just lucky that this set of abstract entities, which incorporates the real numbers ((a,0) in Cartesian coordinates) turns out to be so useful.

I don't think you are being fair to Roger Penrose. What he actually says is:
"introduce i and adjoin it to the system of reals to form expressions like:
a+ib".
Then we don't need any new/special rules for addition and multiplication.


09 Aug 2010, 14:36
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