Z = Z^1.5

mistabell

Forums Newbie

Posts: 7

« on: July 13, 2010, 04:16:26 AM »

I'm trying to draw a Glynn fractal.

I've successfully rendered the mandelbrot and various julia sets. But how the heck do you raise a complex number to the 1.5 power?

I have this so far:

_{function glynnPow(a,b) {
var tempa = a*a-b*b; // Z^2 = (ac - bd, bc + ad)
var tempb = b*a+a*b;

var tempaB = tempa*a-tempb*b;
var tempbB = tempb*a+tempa*b;

a = tempaB; // Z^3
b = tempbB;

var r = Math.sqrt(a*a + b*b); // Rectangular to Polar form
var q = Math.atan(b/a);

a = Math.sqrt(r)*Math.cos(q/2); // De Moivre's Formula
b = Math.sqrt(r)*Math.sin(q/2);

return a + "," + b;
}}
However, sometimes the signs are flipped or the real and imaginary parts are flipped when I compare the answers with wolframalpha. How do you compute these?


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David Makin

Global Moderator
Fractal Senior

Posts: 2286

Re: Z = Z^1.5 - 0.2

« Reply #1 on: July 13, 2010, 11:38:03 PM »

See:

http://spanky.triumf.ca/www/fractint/append_a_misc.html

Just find the section on trig identities.

To calculate a fractional or fully complex power e.g. a^b then use exp(b*log(a)).

This method is clear and straightforward for complex numbers and standard hypercomplex (bicomplex) numbers but is a little woolly when it comes to quaternions or other number forms that are not commutative since then: exp(b*log(a)) is not necessarily the same as exp(log(a)*b).

(Of course here I'm using "exp" so "log" here is the natural log)

Edit: Of course the above is the complete general method, in fact for a real power for (x+i*y)^p I think one can use:

m = (x*x + y*y)^(p/2)
a = p*atan(y/x)
x = m*cos(a)
y = m*sin(a)

which is essentially just a simplification of the method you posted.


« Last Edit: July 14, 2010, 12:04:36 AM by David Makin »	Logged

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Schlega

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Posts: 63

Re: Z = Z^1.5 - 0.2

« Reply #2 on: July 14, 2010, 10:04:11 AM »

I think atan(y/x) does not always give the correct angle. You should use atan2(y,x) instead.


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mistabell

Forums Newbie

Posts: 7

Re: Z = Z^1.5 - 0.2

« Reply #3 on: July 14, 2010, 11:52:20 AM »

Awesome. I had to switch the atan function out with atan2, and multiply by -1 (not sure why.)


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yv3

Conqueror

Posts: 149

Re: Z = Z^1.5 - 0.2

« Reply #4 on: December 20, 2010, 08:48:52 AM »

Thx to David and Schlega! Now my glynn fractal is rendered 4,5 x faster! I found out that the complex class of Microsoft is very slow, i was searching a long time to make my own pow function. In most descriptions i was missing how to calculate the real and imaginery part directly. I like this forum. Happy X-Mas.

Code:

void yComplex::powr(double l) 
{
	temp = pow(real*real + imag*imag, l/2.0);
	temp2 = l*atan2(imag, real); //*-1.0?
		
	real = temp*cos(temp2);
	imag = temp*sin(temp2);
}


« Last Edit: December 20, 2010, 08:59:03 PM by yv3 »	Logged

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