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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
by darsham 12y ago
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 ?
- zygy 12y agoThis 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 12y agoThanks (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 12y agoz = 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 12y agoWe'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.
- darkmighty 12y agoTaylor 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.