In my course "Algemene Natuurkunde Volume A: Trillingen" and also also in my course "Analyse 1: 1e kan Burgerlijk Ingenieur" (sorry it's flemish but it means a course physics and a course math) we encounter the following formula: cos(x) + i*sin(x)= exp(i*x) with i the complex number: sqrt(i) = -1 So in fact we are actually raising e to the power of a complex number???????? I'm not capapable anymore to understand how it's possible to raise a number to a complex number, I only know the definition x^10 = x*x*x*x*x*x*x*x*x*x (10 times) but x^i then would be x*x*x*... (i times) but "i times" doesn't exist??? Also, why is i^i = .2078795764 which is a real number? Almost any other complex power I tried (with Maple) has a complex result, and then just this one hasn't? Well now I know that e^i*x = cos(x) + i*sin(x) but how to put 2^i*x or i^i*x goniometrically? Can anyone please give an understandable explanation about the "i times" thing, how this is related to cos(x) and how to raise other number than e to a complex number? Thanks a lot. [EDIT] whoops, small mistake in a formula [edited by - Lode on June 18, 2002 7:15:29 PM]

It has more to do with how the exp function is defined for complex numbers, in fact exp(-1) or exp(0.5) also do not make any sense. You should not take the power things to literary.

Take a complex number: z1=x1 +i*y1
This can be regarded as a point in a 2D space. The problem is that the product behaves differently then the usual 2D vectors, because of the weird (-i)*(-i)=1.

You can also write the complex number using the distance to the origin R1 (is called the norm) and the rotation phi1 (called the argument). This results in:
z1=R1*(cos(phi)+i*sin(phi))

If you have two complex number z1 and z2 and you want to compute the product z1*z2 then you can just do:
z3=z1*z2=(x1+i*y1)*(x2+i*y2)=(x1*x2-y1*y2)+i*(x1*y2+x2*y1)
You can do this also with the sin and cos variant. The problem is that it requires lots of gonio to figure out what the norm and argumant of z3 would be.

A smart guy, i believe it was Euler, noticed that the norm was just the product, but the argument became an addition. So he defined:
z1=R1*exp(i*phi)
Then you could do:
z3=z1*z2=R1*exp(i*phi1)*R2*exp(i*phi2)=R1*R2*exp(i*(phi1+phi2))
So thanks to this definition we can see immediatly that R3=R1*R2 and phi3=phi1+phi2. So the power should not be regarded as a number of product. It is that the rules of power (like multiplication becomes addition and power become multiplication) hold for the argument of the complext number. It turns out to be very convenient.

Note also that if you take phi=pi then you make a rotation of 180 degrees and are facing the negative x-axis. So exp(i*pi)=-1, so that: exp(i*pi)-1=0. These are the five (e, i, pi, 1 and 0) most important math constants in one equation....

quote:

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Original post by Lode
with i the complex number: sqrt(i) = -1

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Is this what you meant? Or did you mean i squared = -1. Because if sqrt(i) = -1, then i isn''t complex (is it?), it is just 1.

Ah. Bummer, misread posts. Just ignore...

-Neophyte

[edited by - Neophyte on June 19, 2002 1:59:13 AM]

Neosmyle, I think he meant sqrt(-1) = i.

Lode, actually .2078... ( exp(-pi/2) ) is the principle branch of i^i, meaning that it is only a single possibility of infinitely many possibilities. It has a lot to do with angles and the fact that there are so many ways to describe an angle (Pi is the same as Pi + 2Pi or as Pi + 4Pi or as Pi + 2nPi).

After I found the equation to raise complex numbers to complex powers I went online and found these two articles as confirmation I did something right:

Using i as an Exponent (article)
Complex Exponents (from a text book)

I find that individually each article is lacking but together they explain a lot.

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*EDIT fixed broken link
...then fixed the < > symbols

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I will not make a list of links... I will not make a list of links... I will not make a list of links...
Invader's Realm

[edited by - Invader X on June 19, 2002 2:07:55 AM]

[edited by - Invader X on June 19, 2002 2:09:06 AM]

If you''ve taken calculus, you can express e^x as a taylor polynomial expansion

summation(x^n/n!) with n going from 0 to infinity

if you take every even term of this series when you put ix in for x, you get that their sum is equal to the taylor expansion of cos(x) and if you take every odd term, you get that their sum is equal to the taylor expansion of i*sin(x).

so you can say that e^(i*x) = cos(x) + i*sin(x)

OK. First off, sqrt(-1) = i, not hte way you put it. Now I have a reall easy way to raise a number to the i, get a ti-89 or derive 5. Or don''t bother. If it is of any use, here are a couple of neat formulae with i

e^ (pi * i) + 1 = 0
i ^ i = e ^ (-pi / 2)

quote:

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Original post by Puzzler183
Now I have a reall easy way to raise a number to the i, get a ti-89 or derive 5. Or don''t bother.

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Some people just don''t know how to have fun!

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I will not make a list of links... I will not make a list of links... I will not make a list of links...
Invader''s Realm

I''ll allow this homework-related thread to remain open, since Lode was asking for a theoretical discussion rather than specific help on homework. But please try to keep homework-related posts to a minimum . This site, this bandwidth, is to remain focused on game development activities.

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

"It is obvious to any mathematician that e^pi =-1"

(I forget who said that...)

[edited by - Magmai Kai Holmlor on June 26, 2002 2:43:08 AM]