Euler's Formula
By Neil Morrison on Tuesday, November
23, 1999 - 11:03 am :
I already know the series proof of the more general de Moivre formula; I
merely used the version with p as an example. This formula has led me to
some interesting conclusions!
- ex is periodical (period 2pi)
hence ey = x has an infinite number of values of y for each x.
- lnx = ln|x| + i×arg(x)
eg: lni = 0 + i×p/2
ln(-1) = 0 ±i×p
- ii is a real number!
ii = (elni )i = eilni=ei i(p/2)=e-(p/2)=0.208¼
I would be grateful if you would elaborate on these conditions,
and point out any flaws, and also advise me what other areas I
can analyse using complex numbers.
Thanks
Yours,
Neil Morrison
By Matt Daws (Mdpd2) on Tuesday,
November 23, 1999 - 11:32 am :
You point out, correctly, that ex+2pn i = ex for
any integer n. Hence, if we define lnx to be the inverse of ex, we
have that lnx = ln|x| + i×arg(x) + 2pn i for any n. This is
just another way of pointing out that the argument function, arg(x), is not
well defined: we can add 2np to it and it will still behave the same.
Usually you just define arg(x) to be in a certain range, say
0 £ arg(z) < 2p, and have done with it. However, this can lead to problems:
consider a loop going about the origin, and what the value of arg(z) will
be on that loop. Using my definition, it'll jump between 0 and 2p at some
point, even if the loop is very well behaved (e.g. look at the circle about
the origin, moving in an anti-clockwise direction: arg(z) will increase from
0 to 2p but will then jump back to 0).
This sort of behaviour also means that any definition of lnz will jump
like this, which is not something we want. You can get around it by only
defining lnz on certain sections of the complex plane: i.e. only for
certain values of z, so that we can't get this jumping problem.
I don't know if you are at school or university or whatever: here at Cambridge
the courses Complex Analysis and Further Analysis deal with these issues: a
much more complex way of dealing with it is to use Riemann surfaces, which in
some manner allow you to define arg(x) to be all possible values.
By Neil Morrison (P1462) on Tuesday,
November 23, 1999 - 06:20 pm :
Thanks for your prompt reply! Today, I also worked out
(3+4i)2+2i as a test. As I am doing CSYS Maths (Papers
1&2), this was just some personal research. I found an
internet page (forgotten where) that gave Euler's formula, and
some properties to investigate.
By Matt Daws (Mdpd2) on Wednesday,
November 24, 1999 - 12:18 am :
Yeah, it's an interesting area: Complex
numbers behave very differently to Real numbers in some respects.
For example, it is approximately true to say that any integral in
the complex numbers doesn't depend on the path taken.
Path? In the real line, there is only one way to get between any
two numbers. In the complex plane, there are loads of ways, so to
integrate, you have to say what line you're integrating along for
it to make any sense. Luckily, it turns out that it actually
doesn't really matter! That's quite an odd idea, if you think
about it for a bit.
Anyway, I'm not sure I can do a good job of saying much more
without confusing you utterly. Maybe someone else has some ideas?
There are some good books around, but most probably assume more
than you know (e.g. about formal real analysis/calculus), so
probably won't be very useful.
By Andrew Wyld (Acew2) on Friday,
November 26, 1999 - 01:36 am :
I'd like to add a comment to the above.
Normally it doesn't matter what path you choose for integration, no, but
suppose you are taking the definite integral of some function between two
points. Let's say its indefinite integral is arg(z) plus some constant
(this works ok, and saves me having to define the actual function we're
integrating, which is tedious and irrelevant). The definite integral is
well-defined, we hope, since the constant drops out. So what you end up with
in your integral is the difference between the arguments of two points.
HOWEVER.
Let's choose the points 1 and -1 , for ease. If you go above the origin with
your path of integration, arg(z) increases along the path and so arg(-1)
is p plus some constant, arg(1) is zero plus the same constant, and
hence the difference between the two points - and hence the value of the
integral - is p. Take a path below the origin, however, and the value of
arg(z) decreases. Then arg(-1) is -p plus some constant, arg(1)
is still zero plus the same constant, but your integral now has the value
-p. Weird.
This is because the origin (which is the centre of all this trouble) is a
singularity for arg(z) - which is to say that arg(z) is not well-defined
there (worse things can happen, but that's all you need, really)! This is
obvious when you think about it. Whatever angle you choose, the point will
be the same.
So although you can choose any path of integration for complex numbers, if
you pass through a singularity (where the function is ill-defined) expect the
value of your integral to change.
Interesting thought - I haven't covered all the possibilities for arg(z) -
what about multi-spiral paths? This is where Riemann surfaces come in - you
think of yourself as being on a "different incarnation" of the complex plane,
so to speak, every time you sprial round, like a corkscrew shape. There's a
lot more to it than that. None of which I know.
I'll shut up now.
Laters,
Andrew Wyld
By Theo Zapata on Saturday, November
27, 1999 - 11:59 pm :
If you wanna see a good problem using the Moivre's formula....
calculate this sum.
[cos(p/4)]/2+[cos(2p/4)]/22+¼+[cos(np/4)]/2n