Re: Q - binomial theorem and complex numbers

hanrahan398@xxxxxxxxxxx wrote:
Mitch wrote:

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?

Do you mean, is there a simplification of the summation I gave (or some
other formula) that doesn't involve summation or recurrence and
additionally is integer based? As William alluded, probably not (that
is, there is no currently known 'closed form' solution to such a sum,
but that doesn't mean that there might be one). Or rather, the left
hand side -is- the simplification, and quite a good one at that.

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.

Well, as the other respondent did, the power of a complex number (or
any similar domain) can be computed easily by just by... computing
recursively. There's the simple linear algorithm (a+bi)^n =
(a+bi)*(a+bi)^(n-1). They gave the much cleverer efficient 'repeated
squaring' algorithm.

Are you trying to compute the value or simplify the form algebraically
or something else?

Mitch

.

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