Gerhard Niklasch on Sun, 1 Jul 2001 11:31:57 +0200 (MEST) |
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Re: modulus a+b*I |
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