Re: [math-fun] Cube root of a complex number
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 ???
_____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
Yes but my first line should read: Let the cube root of a+bi be y^(1/3) + x^(1/3) i. (or else swap x,y in the two cubics). On Wed, Nov 11, 2009 at 9:38 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 ???
_____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
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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 ???
_____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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 ???
_____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Re: [math-fun] Cube root of a complex number somewhat tangential to the original question, Möbius transformations map three points into three points. So, why not map the three roots of the cubic into the three complex roots of unity? You only get to use the coefficients of the polynomial. I recall a paper in an ACM journal humorously dated March 32 some years ago where someond did that; I don't remember if the solution was relevant to the present inquiry. ------------------------------------------------- www.correo.unam.mx UNAMonos Comunicándonos
That was my paper! http://home.pipeline.com/~hbaker1/sigplannotices/sigcol07.pdf At 09:19 AM 11/12/2009, mcintosh@servidor.unam.mx wrote:
Re: [math-fun] Cube root of a complex number
somewhat tangential to the original question, Möbius transformations map three points into three points. So, why not map the three roots of the cubic into the three complex roots of unity? You only get to use the coefficients of the polynomial.
I recall a paper in an ACM journal humorously dated March 32 some years ago where someond did that; I don't remember if the solution was relevant to the present inquiry.
Figure three seems wrong in this rendering--shouldn't it be a fractal? On Thu, Nov 12, 2009 at 9:47 AM, Henry Baker <hbaker1@pipeline.com> wrote:
That was my paper!
http://home.pipeline.com/~hbaker1/sigplannotices/sigcol07.pdf
At 09:19 AM 11/12/2009, mcintosh@servidor.unam.mx wrote:
Re: [math-fun] Cube root of a complex number
somewhat tangential to the original question, Möbius transformations map three points into three points. So, why not map the three roots of the cubic into the three complex roots of unity? You only get to use the coefficients of the polynomial.
I recall a paper in an ACM journal humorously dated March 32 some years ago where someond did that; I don't remember if the solution was relevant to the present inquiry.
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-- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike http://reperiendi.wordpress.com
For those who can't see the ps version I recreated H. Baker's pdf from his postscript and it seems to render properly. I put it here: http://dl.dropbox.com/u/531485/Mobius_Madness-Baker.pdf But it is 1257 KB and I'm constantly running out of space so will probably delete it in a week or so. On Thu, Nov 12, 2009 at 12:06 PM, Mike Stay <metaweta@gmail.com> wrote:
Figure three seems wrong in this rendering--shouldn't it be a fractal?
On Thu, Nov 12, 2009 at 9:47 AM, Henry Baker <hbaker1@pipeline.com> wrote:
That was my paper!
http://home.pipeline.com/~hbaker1/sigplannotices/sigcol07.pdf
At 09:19 AM 11/12/2009, mcintosh@servidor.unam.mx wrote:
Re: [math-fun] Cube root of a complex number
somewhat tangential to the original question, Möbius transformations map three points into three points. So, why not map the three roots of the cubic into the three complex roots of unity? You only get to use the coefficients of the polynomial.
I recall a paper in an ACM journal humorously dated March 32 some years ago where someond did that; I don't remember if the solution was relevant to the present inquiry.
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-- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike http://reperiendi.wordpress.com
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participants (7)
-
Dan Asimov -
Henry Baker -
James Buddenhagen -
mcintosh＠servidor.unam.mx -
Mike Stay -
Schroeppel, Richard -
victor miller