Re: [math-fun] Cube root of a complex number
We can compute cos(t)+isin(t)=a'+b'i = (a+bi)/|a+bi| with a real sqrt. We can compute |a+bi|^(1/3) with a real cbrt. So cbrt(a+bi) = (cos(acos(a')/3)+isin(asin(b')/3))*|a+bi|^(1/3). The real part of cbrt(a'+b'i) is cos(acos(a')/3), which looks like Chebyshev 1/3. Are there Chebyshev functions with non-integral n? Does the explicit formula Chebyshev(n) = (1/2)[(x-sqrt(xx-1))^n+(x+sqrt(xx-1))^n] work for n=1/3 ? http://en.wikipedia.org/wiki/Chebyshev_polynomials At 08:48 AM 11/12/2009, Schroeppel, Richard wrote:
The cubic formula requires taking cube roots of the two solutions of a quadratic equation. When you try to use it to find the real (or imaginary) part of cbrt(a+bi), you wind up facing the same problem (or an equivalent) over again.
Numerically, your options for cbrt(a+bi) are trigonometric (choice of sin/cos/tan), logarithmic & exponentials of complex numbers (coming back to trig), or Newton's method (& siblings), or the HP-calculator method, or a power series solution based on reverting the series for x-x^3 (it's cute).
It looks like the schoolbook method for cbrt could be adapted to work, by rotating a+bi so that Realpart>=|Imagpart|, and going alternately after a bit/digit of the real & imaginary roots.
Rich
PS: Suppose you have a subroutine for cbrt(a+bi) on the unit circle, but nothing for cbrt(real). Can you somehow get cbrt(2)? --R
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of victor miller Sent: Wednesday, November 11, 2009 9:57 PM To: Dan Asimov; math-fun Subject: Re: [math-fun] Cube root of a complex number
Another way of seeing this is (despite your admonishment not to mention trig functions) is that we really only have to do it when |a+bi|=1. Using the fact that
cos(3 theta) = 4 cos^3 theta - 3 cos theta, so we see that we need to solve the cubic:
4 t^3 - 3 t - a = 0. This is the casus irreducibilis of solving the cubic. Even though all the roots are real, when using radicals, one needs to take the cube root of a non-real quantity:
http://en.wikipedia.org/wiki/Casus_irreducibilis
Victor
On Wed, Nov 11, 2009 at 10:38 PM, Dan Asimov <dasimov@earthlink.net> wrote:
No, that's perfect, since the equations for x and y can be solved by the solution to the general cubic . . . and will *hopefully* give real answers -- leading to expressions for the real and imaginary parts of (a + bi)^(1/3) in terms of radicals, which is what I was looking for.
--Dan
James wrote:
<< Oops, I seem to have x and y reversed below.
On Wed, Nov 11, 2009 at 9:01 PM, James Buddenhagen <jbuddenh@gmail.com> wrote: Not as pretty, and not really a formula, but sort of interesting:
Let the cube root of a+bi be x^(1/3) + y^(1/3) i. Then x is a root of 64x^3 - 48bx^2 - (27a^2+15b^2)x - b^3 = 0, and y is a root of 64y^3 + 48ay^2 - (27b^2+15a^2)y + a^3 = 0.
On Wed, Nov 11, 2009 at 7:13 PM, Dan Asimov <dasimov@earthlink.net> wrote: << It's a pleasant exercise to calculate the square root of the complex number a + bi directly (without making use of polar form or trig functions).
(Be sure to consider all cases.)
But is there a similar formula for the cube root of a + bi ???
* Henry Baker <hbaker1@pipeline.com> [Nov 13. 2009 08:29]:
[...]
The real part of cbrt(a'+b'i) is cos(acos(a')/3), which looks like Chebyshev 1/3. Are there Chebyshev functions with non-integral n?
yes, take T_n(x) = cos( n*acos(x) ) or (hypergeometric 2F1) T_n(x) = F( -n, +n; 1/2; x ) Note that T_n( T_{1/n}( x ) ) == x (because T_n( T_m( x ) ) == T_{n*m}(x) )
[...]
* Joerg Arndt <arndt@jjj.de> [Nov 13. 2009 12:21]:
* Henry Baker <hbaker1@pipeline.com> [Nov 13. 2009 08:29]:
[...]
The real part of cbrt(a'+b'i) is cos(acos(a')/3), which looks like Chebyshev 1/3. Are there Chebyshev functions with non-integral n?
yes, take T_n(x) = cos( n*acos(x) ) or (hypergeometric 2F1) T_n(x) = F( -n, +n; 1/2; x )
correction: T_n(x) = F( -n, +n; 1/2; (1-x)/2 )
Note that T_n( T_{1/n}( x ) ) == x (because T_n( T_m( x ) ) == T_{n*m}(x) )
[...]
Ok, so if we are given c+is, c,s real, cc+ss=1, then we want to compute cos(acos(c)/3) for the real part of one of the cube roots of c+is. cos(acos(c)/3) = T_(1/3)(c) = d, so cos(3acos(d)) = T_3(d) = c, i.e., 4d^3-3d=c So we have reduced the problem to one of finding a real root of a real cubic 4d^3-3d-c=0. Once we have d, we can easily compute the imaginary part, since we know that the absolute value is 1. Question: Since we know that the real parts of all three of the complex cube roots of c+is are real (duh!), this would seem to imply that the equation 4d^3-3d-c=0 always has 3 real roots. Is this easy to see? Also, all three roots should lie within the range [-1,1]. Is this easy to see? At 05:27 PM 11/12/2009, Joerg Arndt wrote:
* Joerg Arndt <arndt@jjj.de> [Nov 13. 2009 12:21]:
* Henry Baker <hbaker1@pipeline.com> [Nov 13. 2009 08:29]:
[...]
The real part of cbrt(a'+b'i) is cos(acos(a')/3), which looks like Chebyshev 1/3. Are there Chebyshev functions with non-integral n?
yes, take T_n(x) = cos( n*acos(x) ) or (hypergeometric 2F1) T_n(x) = F( -n, +n; 1/2; x )
correction: T_n(x) = F( -n, +n; 1/2; (1-x)/2 )
Note that T_n( T_{1/n}( x ) ) == x (because T_n( T_m( x ) ) == T_{n*m}(x) )
[...]
* Henry Baker <hbaker1@pipeline.com> [Nov 14. 2009 18:50]:
Ok, so if we are given c+is, c,s real, cc+ss=1, then we want to compute cos(acos(c)/3) for the real part of one of the cube roots of c+is.
cos(acos(c)/3) = T_(1/3)(c) = d, so
cos(3acos(d)) = T_3(d) = c, i.e.,
4d^3-3d=c
So we have reduced the problem to one of finding a real root of a real cubic 4d^3-3d-c=0. Once we have d, we can easily compute the imaginary part, since we know that the absolute value is 1.
Question:
Since we know that the real parts of all three of the complex cube roots of c+is are real (duh!), this would seem to imply that the equation 4d^3-3d-c=0 always has 3 real roots. Is this easy to see?
Also, all three roots should lie within the range [-1,1]. Is this easy to see?
[...]
Taking "see" literally: (using pari/gp) ploth(x=-1,1,poltchebi(3)) Plot also at http://mathworld.wolfram.com/ChebyshevPolynomialoftheFirstKind.html
participants (2)
-
Henry Baker -
Joerg Arndt