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Topic: [HM] History of complex numbers
Replies: 12   Last Post: Aug 9, 2000 1:05 PM

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Carlos Cesar de Araujo

Posts: 3
Registered: 12/3/04
Re: [HM] History of complex numbers
Posted: Jul 12, 2000 9:24 PM
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However, it is possible to define Sqrt[z] for ALL complex numbers z in such
a way that
Sqrt[-1] = I.
Even more can be said. For every $n$ in IN and every $z$ in C-{0} one can
choose, in the set

{r in C : r^n = z},

a "principal value" root[z,n] in such a way that

root[-1,2] = Sqrt[-1] = I.

The definition is as follows.

(1) If Im[z]=0 and Re[z] >= 0, set
root[z,n] = THE non-negative real number $r$ such that r^n = z.

(2) If $z$ is in C-{0}, set
root[z,n] = root[Abs[z],n]*(Cos[Arg[z]/n]+I*Sin[Arg[z]/n])

where

Arg[z] = THE $a$ in [-Pi,Pi] such that Cos[a]+I*Sin[a] = z/Abs[z].

This particular choice of the argument can be defined explicitly in many
ways. For example,

ArcCos[Re[z]/Abs[z]] if Im[z]>=0
Arg[z] =
- ArcCos[Re[z]/Abs[z]] if Im[z]<0.

You can test these ideas with any good Computer Algebra System. Try, for
example, the following commands written in Mathematica language:

root[z_,n_]:=Module[{a,r,num},
a=Arg[z];r=Power[Abs[z],Power[n,-1]];
num=r*(Cos[a/n]+I*Sin[a/n]);
FullSimplify[num]
]

In Mathematica, Arg[z] is a built-in function, so you don't need to define
it as above. In fact, you can experiment with other built-in functions like
Root.
----------------------------------------------------------
Exercise: Try to locate the principal value in the list

Table[Root[(#^n - z)&,i],{i,1,n}]
----------------------------------------------------------
Many more interesting things could be added here (about different approaches
to "standard" choices in other areas of mathematics), but I don't have
time now.

Carlos C\'esar de Ara\'ujo





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