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Topic: Binomial theorem (Read 1660 times) |
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Steve_B.
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Binomial theorem
« on: Jan 16th, 2012, 3:35pm » |
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Binomial theorem to complex numbers;
(a + i*b)^n
(i*b)^1 = i*b, (i*b)^3 = -i*b^3, (i*b)^5 = i*b^5, (i*b)^7 = -i*b^7, (i*b)^9 = i*b^9, (i*b)^11 = -i*b^11, (i*b)^13 = i*b^13, (i*b)^15 = -i*b^15, (i*b)^17 = i*b^17, (i*b)^19 = -i*b^19,...
So, we get a negative sign for n of the form 4*n + 3
For even numbers of n:
(i*b)^2 = -b^2, (i*b)^4 = b^4, (i*b)^6 = -b^6, (i*b)^8 = b^8
The signs alternate, we get a negative sign before a real number for numbers congruent to 2 mod 4: a(n) = 4n+2, and a positive sign for multiples of 4.
(a + i*b)^n = a^n + C(1,n) a^(n-1) (i*b) + C(2,n) a^(a-n) * (-b^2) + C(3,n) a^(n-3) * (-b^3) + C(4,n) a^(n-4) * (b^4) + ... + (i*b)^n
how can the last term be expressed?
Is z^n = (a + i*b)^n of the form x^2 + i * y^2 ?
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