Perry S. Glenn on Sun, 1 Jul 2001 14:55:59 -0700 (PDT)

 Re Modulus a + b*I

```Hello,
thank you for clearing that up for me
I did mean
abs(z)=sqrt(a^2+b^2).

I see I was making assumptions about a and b, and I was not
condsidering the validity of this for the general case.

The PARI-GP is the best caculator I have come across and I
am using it to extend my knowledge in mathematics. I am new to
symbolic calculations but I'm interested in increasing
my understanding of this and Number Theory. PARI-GP is a very nice
tool for this end.

Again, thank you very much for your quick response, and please excuse
my
bumbling.

-PSGlenn

On Sun, Jul 01, 2001 at 11:31:57AM +0200, Gerhard Niklasch wrote:
> In response to:
> > Message-ID: <20010701080627.73568.qmail@web13607.mail.yahoo.com>
> > Date: Sun, 1 Jul 2001 01:06:27 -0700 (PDT)
> > From: "Perry S. Glenn" <psglenn@yahoo.com>
>
> > I would expect to get the result
>
> > z=a+b*I
> > abs(z)= a^2+b^2
>
> I hope not!  First, that would be the square of abs(z).  Second,
> even that only when it is known in advance that a and b stand
> for real numbers -- which gp has no reason to assume.
>
> > ? z=a+I*b
> > %1 = a + I*b
> > ? abs(z)
> > %2 = a + I*b
>
> As with many other functions, this got applied to each
> coefficient of the multivariate polynomial.  (The only
> surprise here is that it left I alone, whereas abs(I)
> returns 1.000000000000000000000000000 .)
>
> If you intend a and b to stand for real numbers, then
>
> (12:07) gp > norm(a+b*I)
> %1 = a^2 + b^2
>
> does what you seem to want:  compute the square of the
> absolute value.  If a and b can themselves be complex
> (or if they are indeterminates which can take values
> in any old commutative ring containing something which
> behaves like I),  there is no simple formula - the answer
> would depend both on the ring you're working in and on
> the precise shape of a and b written as elements of that
> ring.
>
> Enjoy, Gerhard

--f61LtHN51936.994024517/big.psf.his.org--

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