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Re: nth roots of complex numbers

  • To: mathgroup at smc.vnet.net
  • Subject: [mg39070] Re: nth roots of complex numbers
  • From: "Allan Hayes" <hay at haystack.demon.co.uk>
  • Date: Sun, 26 Jan 2003 18:44:07 -0500 (EST)
  • References: <b10ddh$mou$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

"Zachary Turner" <_NOzturner0826SPAM_ at hotmail.com> wrote in message
news:b10ddh$mou$1 at smc.vnet.net...
> Apparently Mathematica randomly returns roots a root of a complex number.
> Is there a way I can write my own function that will return a set
consisting
> of all n roots of a given complex number.  For example, Root[z,n] = {a set
> consisting of n elements}
>
>
>

Zachary,
The choice is not random - it is the principle root:

     z^n = Exp[n (Log[Abs[z]+ I principal argument of x)]

where the principal argument of x is that in (-Pi,Pi).

For numerical values of x and  n

Solve[x^3 == 5+2I,{x}]

{{x -> (5 + 2*I)^(1/3)},
  {x -> (-(-1)^(1/3))*(5 + 2*I)^(1/3)},
  {x -> (-1)^(2/3)*(5 + 2*I)^(1/3)}}

x/.%

{(5 + 2*I)^(1/3), (-(-1)^(1/3))*(5 + 2*I)^(1/3),
  (-1)^(2/3)*(5 + 2*I)^(1/3)}


--
Allan

---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565






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