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From: greg@jiff.berkeley.edu (Greg)
Newsgroups: sci.math,comp.graphics
Subject: Re: Solution of quartic equation
Message-ID: <7805@agate.BERKELEY.EDU>
Date: 18 Mar 88 23:55:39 GMT
References: <1656@thorin.cs.unc.edu> <7662@agate.BERKELEY.EDU> <2381@bsu-cs.UUCP>
Sender: usenet@agate.BERKELEY.EDU
Reply-To: greg@jiff.UUCP (Greg)
Organization: U. C. Berkeley
Lines: 18
Summary: Taking roots of complex numbers.

In article <2381@bsu-cs.UUCP> cfchiesa@bsu-cs.UUCP (Christopher Chiesa) writes:
>In article <7662@agate.BERKELEY.EDU>, greg@jiff.berkeley.edu (Greg) writes:
>> The direct solution is prone to programmer's instability.  It's very
>> messy and doesn't generalize.  The most painful step is taking the
>> cube root of a complex number,...
...
>I recall taking cube and higher roots of complex numbers
>quite easily by converting from "cartesian" (A+Bi) representation to "polar"
>(r cos Theta), and performing a few straightforward real-number calculations.

You are correct; you can take roots of complex number by passing to polar
coordinates.  Although the calculation is not quite as messy as I suggested,
it is still painful in the sense that it requires several invocations of
transcendental functions.  It does not seem like the fastest solution.

Be that as it may, the formula for the roots of a quartic is complicated.
Its derivation is even worse.
--
Greg