# Homework Help: Complex analysis- Oscillation/vibration class

1. Sep 9, 2005

### SteveDB

complex analysis-- Oscillation/vibration class

Hello all,
I'm taking a wave/vibration/oscillation class, and we're delving into complex notation for these.
One of our assigments dealt with a complex function that we didn't get a whole lot of practice out of in math methods.
I've gone back through my Boas' Math Methods text, as well as the Gradshtein Tables, and Zwillinger CRC Math tables book, and do not get very much from them in regards to solving these types of problems.
So-- perhaps someone with more skill than I can 'splain it to me.....
I get
Cos (w*t) + i* sin(w*t) = e^(i*w*t)
But the real to complex domain conversion still confuses me.
E.g. A^3=-1.
Maple 10 states that A = -1, .5+/- sqrt(3)*i/2 as the three solutions for this.
In trying to do the real to imaginary domain graph, as in the complex section of Boas' text, we have the typical x,y graph. Where the y becomes the imaginary plane.
If I do -1 as my point, it leaves me on the x axis, and no where in the imaginary plane.
However, it would appear that I should be somewhere off axis, and at .5 in the real space, and +/- sqrt(3)*i/2 in the imaginary plane.
anyone have a really good, comprehensive explanation to help stop my head-banging?
your assistance would be deeply appreciated.
Best,
SteveB.

2. Sep 9, 2005

### Astronuc

Staff Emeritus
This has to do with de Moivre's formula -

if z = r (cos $\theta$ + i sin $\theta$) and n is a positive integer, then

zn = rn (cos n$\theta$ + i sin n$\theta$ )

and by Euler's identity, z = r ei$\theta$,

zn = rn $e^{i n \theta}$

Let w = r (cos $\theta$ + i sin $\theta$), and zn, then

z =$\sqrt[n]{r}\,(cos(\frac{\theta}{n}+\frac{k}{n}2\pi)\,+\,i\,sin(\frac{\theta}{n}+\frac{k}{n}2\pi)\,)$, where k = 0, 1, . . . , n-1

Last edited: Sep 9, 2005
3. Sep 9, 2005

### SteveDB

Those parts I understand.
I got these out of my CRC Math book, as well as Boas, and other more standard math texts. We've done identity assigments, and I've been using these long enough to not have a problem with that-- I graduate this winter.
What's confusing to me is the graphic representation of this; I.e., understanding how to visualize these in the real/imaginary space, and solve them accordingly.
Do you have a copy of the Boas- Mathematical Methods in the Physical Sciences text?
I'm using the 2nd ed. Please go to pages 60-67, chapter 2, sections 9, and 10.
We didn't get a whole lot of practice with this, and so it was never real clear to me, beyond what you stated above.
With something like 1+i, there is a graphic repsentation of this in the real/imaginary planes. It works out to some trig function, using the x+iy system.
I hope I'm making this clear enough to understand.
Thanks.
SteveB.

4. Sep 9, 2005

### Astronuc

Staff Emeritus
5. Sep 9, 2005

### SteveDB

Ok, the second one....
Argand diagram is the one I'm referring to; well, actually now that the complex pge has loaded, they both show the graph I'm referring to.
Can you more succinctly explain this to me-- how it works, etc...
I know that the point on the 'unit' circle is the relation of x+iy = z, and z = r*e^i*theta. However, the functions derived from that confuse me.
My understanding is that we're to treat the circle similar to the trig circle, where pi/3 equals sqrt(3)/2, and pi/4 = sqrt(2)/2, pi/2 = i, etc.... except that it now includes i in the equation, because we're in both real, and the imaginary space.
However, when we're on the x axis, i goes away, and we're left only in real space.
No one has been able to explain this to me yet.
So, I guess what it is that I'm looking for is the break down on how this works.
Thanks again for your willingness to help.
SteveB.

6. Sep 9, 2005

### SteveDB

ok, I just remembered something when I was looking at the Wolfram site. I actually have Abramowitz, and Stegun-- in PDF.
If anyone wants it, let me know.
Ok, after looking more closely to this, I see that my comments about it being similar to the trig. unit circle, are accurate.
Picking the point on the unit circle, applying cos to x, and sin to y, I get the radian value conversions.
I.e., using the DeMoivre relation, I can get z = cos((2k+1)*pi/n) + i * sin((2k+1)*pi/n), and x = cos....
y = sin....
So, if sin... = sqrt (3) / 2, and cos ... = 1/2, then z = 1/2 + i sqrt (3) / 2. Ok, that makes sense. Now, one last thing-- this is not a series expansion, correct? Why the differing values of k? This leads me to think we're dealing with a series expansion.

FWIW, I think this discussion is one worth keeping. I know I'm not the only one who has been confused by this. Perhaps even a bit of expansion on the idea would be beneficial.
Thanks.