in reply to getting a cube root

All the replies posted so far finds a real root. Is there a method to find complex roots as well? Math::Complex doesn't seem to cut it.

Update: As pointed out below, c=r^(1/n)*e^(2*pi*i*j/n) (for j=0,1,...,n-1) will do the trick, but I was thinking someone should have written an easy access to the roots, as in: 

@roots=complex_roots($real, $n)

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to give all the n-th roots of $real. Hmmm.... This shouldn't be too difficult. I'll leave it as exercise to the reader. :)

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Re: Re: getting a cube root
by snax (Hermit) on Oct 08, 2003 at 20:06 UTC
    You can use trig and brute force a method -- I'm a little busy to give it a go, but the complex roots lie on a circle in the complex plane of the same radius as the real root. With a cube root, that means you need to find the coords of the points at 120 and 240 degrees (1/3 and 2/3 around the circle) on that circle -- hopefully that's clear. For the nth root you need to look at the (i * 360/n) degree points with i = 0..(n-1). i=0 corresponds to the real root.

    Update: 

    #!/usr/bin/perl -w
use strict;

my $twoPi = 4 * atan2(1, 0);

for (2..5) {
  my @roots = nthRoots(8, $_);
  print "$_ th roots of 8: $/";
  foreach my $root (@roots) {
    print "$root->{Real} $root->{Imag}$/";
  }
};

sub nthRoots {
  my $x = shift;
  my $n = shift;
  # Should check for integer powers
  my @roots;
  $roots[0] = {
                Real => $x ** (1/$n),
                Imag => 0
               };
  for (1..($n-1)) {
    push @roots, {
                  Real => $roots[0]->{Real} * cos( $_ * $twoPi/$n ),
                  Imag => $roots[0]->{Real} * sin( $_ * $twoPi/$n )
                  };
  };
    
  return @roots;
}

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