Square rooting -1
By Emily Hall on Thursday, June 28, 2001 -
10:11 pm:
If i= root(-1)
then i2 = root(-1) xroot(-1)
= root(-1 x-1)
= root(+1)
= + or - 1
Am I right? All my A-level text books say
that i squared only = -1.
Can you help? Thanks
By Tim Martin on Thursday, June 28, 2001 -
10:22 pm:
You have to be careful when you write
i=root(-1) because this has two solutions (i and -i). It is more
accurate to say that i is the number such that i2 =
-1.
In this case the problem is that in the line
i2 = root(-1) x root(-1)
you have allowed the possibility that the two roots are different
values. If you use the same root of -1 for both values (it
doesn't matter which) you get the expected value, but if you take
i for one and -i for the other then you get the wrong
solution.
By Emily Hall on Thursday, June 28, 2001 -
11:52 pm:
I can see what you mean by saying that if I've said root(-1)
xroot(-1) this doesn't necessarily mean the two roots are the
same values and so do not necessarily give i squared. But if I
take the positive root of i squared, i, and then square it I
still don't end up with just -1, because surd rules say that
+root(-1) x+root(-1) = root(-1 x-1) = root(+1). I don't see why
this doesn't happen when you are using the same roots of i
squared.
By Dan Goodman on Friday, June 29, 2001 -
12:12 am:
The surd rules you know don't always
apply with complex numbers, this is because there is no unique
way to specify one complex square root or the other which
satisfies the rule that sqrt(ab)=sqrt(a)sqrt(b) for all a and b
(complex numbers). You can do this for real numbers (just always
choose the positive square root), but there is no way to do it
for complex numbers. This leads on to a difficult and interesting
branch of maths called Riemann Surfaces which I won't describe
here. Basically, all you need to know is that the surd rules no
longer work with complex numbers.
By Emily Hall on Friday, June 29, 2001 -
05:30 pm:
Thanks, that sort of clears it up. Is there a way of
describing Riemann Surfaces briefly that I'll understand? It's an
area of maths I'm finding very interesting.
By Dan Goodman on Saturday, June 30,
2001 - 02:30 am:
I'm not sure I'd be able to explain Riemann Surfaces in an
easy way without covering quite a lot of ground first. However, I'll give it a
go.
Do you know about the geometric interpretation of complex numbers and square
roots? Have you heard about the r eiq or polar coordinate (r,q)
notation for complex numbers? Get back to me and I'll see if I can help.
Also, you might be interested to know that I'm writing an introduction to
complex numbers for a future NRICH article at the moment.
See here .
By Andrew Hodges on Monday, July 02, 2001
- 07:27 pm:
Do you think you could explain some Riemannian Geometry to me,
in order that I shall find the study of General Relativity less
difficult! I have done the reix form of a complex
number and I am familiar with results such as De Moivre's
Theorem.
By Dan Goodman on Monday, July 02, 2001
- 09:06 pm:
Wow, lots of interest in Riemann surfaces!
OK, here's a very quick, rough and ready summary:
I described above the problem that you can't specify complex square roots so
that the surd rules apply. Clearly, for each complex number you can specify a
unique square root (for example, you could always choose the complex number
whose imaginary part is positive if it's nonzero, or whose real part is
positive if the imaginary part is zero), however the surd rules won't apply.
In fact, there is another problem here.
If you think of taking the complex square root as a function f(z) (so that
f(z)2 = z for all z) then there is no way to make sure that f(z) is
continuous (no sudden jumps in the value of f(z)) and single valued. Here's
why. Suppose f(z)2 = z for all z. Consider g(t)=f(e2pi t) for
0 £ t £ 1. g(0)=f(1)=±1. Let's assume that f(1)=1 (the same
argument works if f(1)=-1). If the value of g(t) doesn't "jump" for any
values of t (i.e. the value of f(z) doesn't jump for any z on a unit
circle) then we must have that for 0 £ t £ 1/2 g(t)=eipt (since
if at any point g(t)=-eipt then the value of g(t) would jump at that
point). So g(1/2)=eip/2 = i. Continue this argument for 1/2 £ t £ 1
to show that g(t)=eipt and so g(1)=eip = -1. But
g(1)=f(e2pi)=f(1)=1. In other words, f(z) is not single valued.
That was quite complicated, I hope you followed it. Basically, as we go round
the unit circle anticlockwise starting at 1 (taking square roots along the
way), when we get back to 1 the square root has changed from 1 to -1. In other
words, we can't choose a unique, continuous square root function.
There are lots of other functions like this, for example the complex logarithm
function and the complex nth root function.
There are two major ways of dealing with this problem, one is mostly used by
applied mathematicians and one is mostly used by pure mathematicians. The
applied mathematicians tend to use "branch cuts" and the pure mathematicians
tend to use "Riemann surfaces". I'll quickly describe both.
Let's start off with branch cuts. Suppose we said, I want a function f(z)
such that f(z)2 = z for all z except for negative real values of z. In
this case, we can have a single valued, continuous value of the complex square
root. Here is one way of doing it: f(r eit )=r0.5 eit/2 where
-p < t < p. Can you see why the argument I used above doesn't apply here?
If we do this, we say that the we have made a "branch cut" from 0 to -infinity
along the negative real axis. As long as there is a line of points from 0 to
infinity (a branch cut from 0 to infinity) which we do not want the value of
sqrt(z) on, we can make sqrt(z) single valued and continuous.
For more complicated functions, you might have to make more than one branch
cut, but I won't go into that now.
The other way of solving the problem is Riemann surfaces, but this is even
more difficult. Imagine, if you will, a 4 dimensional space. In fact, it's a
2 dimensional space where each dimension is complex. In other words, it's the
space of points (x+i y, u+i v) for all real values of x, y, u, v .
Two complex dimensions, four real dimensions. The "concrete Riemann surface"
of a function f(z) is the set of points (z,f(z)) in this 4 dimensional
space (z and f(z) are both complex numbers). So, for example, if f(z)=z2
then the concrete Riemann surface of f(z)=z2 is the set of points (z,z2).
What does this look like? Well, it's a surface which means that it is a 2
dimensional object sitting in a 4 dimensional space. It turns out that we can
study the square root "function" by studying this surface in this 4 dimensional
space. It turns out that if you take this surface of points (z,z2) and you
twist it and deform it in the right way, you can make it look like the surface
of a doughtnut (the mathematical word is a torus) with one point removed.
I'm not sure if that will have made any sense to you, it's an extremely
difficult subject to understand (we weren't taught it until the 3rd year at
Cambridge). If there were any bits you didn't follow, just post a note here
and I'll see if I can explain a bit more fully.
I'm not sure what relevance this has to general relativity, perhaps Sean can
tell you more?
By Arun Iyer on Tuesday, July 03, 2001 -
07:50 pm:
I understood the concept of Riemann Surfaces.
Does it help in the study of electromagnetic forces emitted by a
current carrying toroid?
love arun
By Sean Hartnoll on Wednesday, July 04,
2001 - 12:40 am:
Andrew - just a note that Riemann
surfaces and Riemannanian geometry are in fact different things
(although related)! Riemann has his name all over the
palce...
Arun - you probably wouldn't use Riemann surfaces for this,
although these sorts of electromagnetic problems involving shapes
with holes in them can be helped by sophisticated geometry - a
little hard to explain briefly though...
Sean
By Arun Iyer on Wednesday, July 04, 2001 -
06:47 pm:
Sean (or anyone reading this message), can you tell me some
good sites where I can come in to contact with physics?
love arun
By Sean Hartnoll on Thursday, July 05,
2001 - 04:09 pm:
Try http://physicsweb.org/ . It has many
links (got the section web links) to all sorts of physics
places.
Sean