Re: complex numbers


"jem" <xxx@xxxxxxx> wrote in message news:p6Are.107571$sy6.98621@xxxxxxxxxxxxx
> N:dlzc D:aol T:com (dlzc) wrote:
> > Dear jem:
> >
> > "jem" <xxx@xxxxxxx> wrote in message
> > news:FVere.91014$sy6.84187@xxxxxxxxxxxxx
> >
> >>The TimeLord wrote:
> >
> > ...
> >
> >>>You can see that unless you keep straight just what
> >>>the square root is defined to be,
> >>
> >>Sqrt() is defined to be a function so e.g. Sqrt(1) = 1,
> >>not +-1, and of course i^2 = -1, not +-1.
> >
> >
> > sqrt() may be defined by programming languages to be "the
> > positive square root", allowing the programmer to assign whatever
> > sign(s) the programmer chooses. But the result of the square
> > root is bivalued (except for sqrt(0) ). sqrt(-1) is the
> > conundrum, i^2 is the solution. And note that +/-i *is* a
> > solution, and valid result.
>
> Dear Mr. Smith,
>
> The equation x^2 = a does indeed have 2 roots (even for a=0 according to
> the Fundamental Theorem of Arithmetic), but the *functional*
> relationship x = sqrt(a) has only 1 root.


Actually, "roots" are numbers that make a polynomial zero.
Equations don't have roots. They have "solutions".

So, in terms of polynomials and roots:
The polynomial
x^2 - a
has two roots, namely
x1 = sqrt(a) and x2 = -sqrt(a) if a > 0
or
x1 = i sqrt(-a) and x2 = -i sqrt(-a) if a < 0
or it has one root, namely
x1 = 0 if a = 0

Expressed in terms of equations and solutions:
The equation
x^2 - a = 0
has two solutions, namely
x1 = sqrt(a) and x2 = -sqrt(a) if a > 0
or
x1 = i sqrt(-a) and x2 = -i sqrt(-a) if a < 0
or it has one solution, namely
x1 = 0 if a = 0

Dirk Vdm


.

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