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