Re: Q - binomial theorem and complex numbers
- From: iandjmsmith@xxxxxxx
- Date: 13 Apr 2006 05:13:54 -0700
hanrahan398@xxxxxxxxxxx wrote:
Mitch wrote:
hanrahan398@xxxxxxxxxxx wrote:
Is there a neat expression, derived from the binomial theorem,
for the real and imaginary parts of (a+bi)^n,
where a and b are integers and n is a positive integer?
the others have given answers but I think what you finally wanted is
the more explicit:
(a+bi)^n = sum {n \choose 2k} (-1)^k a^{n-2k} b^{2k} +
i sum {n \choose 2k+1}(-1)^k a^{n-2k-1} b^{2k+1}
which, by DeMoivre's, also gives a combinatorial significance to sin
and cos.
I knew that (I understand the binomial theorem and De Moivre's :-) ),
but is there a neater expression for the sum, in the way that
n(n+1)/2 is a neat expression for the sum of x from 1 to n?
What I would like is a neat way to work out a power such as (3+8i)^101,
using integers only (so, not using De Moivre's theorem) and without
working out each term of the binomial expansion.
Chris
It sounds like all you want to do is calculate expressions such as
(3+8i)^101 using only integers. If so, you could use something simple
like
c^n = 1 IF n = 0,
= c IF n = 1,
= square(c^(n/2)) IF n even,
= c*square(c^((n-1)/2))) otherwise
where square(a+bi) is (a^2-b^2)+(2*a*b)i
Ian Smith
.
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