[darsham]

Maybe someone can help me understand this. On slide 39 of the first slideshow, it says :

"For any irrational power p, there are an infinite number of solutions to z^p=c, all lying on a circle."

This means that most of the solutions have an angle larger than a full circle, right ? But if complex numbers can be represented as the sum of the real and complex parts, how can their angle be superior to 360 degrees ?

[darkmighty]

Taylor series explanation: (simpler)

p irrational:

z^p = a1 z +a2 z^2+... =

    = a1 k1+a2 k2 +...

but there is 1 extra z'!=z which has the same value k1 for z'^2 as a solution z of z^2, 2 extra z' which have the same value k2 for z'^3 and so on. As long as mdc(a,b) is 1, for distinct a and b, z^b and z^b won't share all z' alternatives. You can then choose prime numbers s.t. you get infinitely amount of distinct solutions.

Exponential expanation: (more intuitive)

So we're taking roots of complex numbers: all z s.t. z^p=exp(a+jb). Without loss of generality, take z^p=exp(jb) instead, since exp(a) is just a positive constant. A trivial solution for that is simply exp(jb/p). But remember that for any x, exp(jx) = exp(jx+2 k pi) for integer k. For example, exp(jpi/6) can be written either as that or exp(j 13 pi/6) -- that's because exp(jx) is a sum of periodic components cos(x) and j*sin(x). So we can add new solutions of the form exp(j(b+2 k pi)/p), for any k, which may or may not overlap with previous solutions. But if 1/p=m/n (rational), you can set k=n to get back to exp(jb/p+2 pi)=exp(jb/p); otherwise, you get infinitely many solutions, since there is no k/p=integer. However, you can arbitrarily close by an approximation p~m'/n'; in fact, you get arbitrarily close to any other rational exponent exp(jq), so that the solutions are scattered everywhere.

[zygy]

This is the 'collapsing' that he talks about. If one of the complex solutions had a magnitude of 1 and an angle of 395 degrees, e.g., it would be equal to a number with a magnitude of 1 and an angle of 35 degrees.

[darsham]

Thanks (and thanks to the other replies !). So a complex number has a single representation when using real and imaginary parts, but an infinity of representations if you use angle and magnitude (just add or subtract 2π radians). I guess that should have been clear from the article, I just needed to sleep on it.

[NAFV_P]

  z = x + i*y = e^(a + i*b) = e^a*(e^(i*b)) = e^a(cos(b)+i*sin(b))

Both sin and cos are many to one functions. In the equation above, replacing

  b

with

  b + 2*pi*t

where t is any positive or negative integer, would result in the same complex number.

[NaNaN]

We've used the same number for those different rotations. That does not do anything directly about irrational power p.

If p is a rational number represented by m/n, then

    360 / (m/n) * m = 360n

So there are multiples of 360 that are divisible by m/n, and we can get a unique complex number from z^p.

Now p is irrational, what happens? According to what you said, there are so many different graphs that can not be merged into just one.

[NaNaN]

EDIT: So ... we can get FINITE complex numbers (or graphs) from z^p.

Now p is irrational, what happens? According to what you said, there are infinite different graphs that can not be merged into just SOME.

[NAFV_P]

I am replying to the last paragraph of darsham's comment.