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 ???
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