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 ???

--Dan

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Quoting Dan Asimov <dasimov@earthlink.net>:

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).

What's even more pleasant (for purposes of numerical analysis) is that it can be done only using square roots of positive real numbers. Even if a = -1 and b=0. ------------------------------------------------- www.correo.unam.mx UNAMonos Comunicándonos