Complex numbers?

Question

If "n" is a positive integer and W a complex number such that (W^n)=1, but W≠1.
What is it possible to say about S=1 + W + W^2 +.....+ W^(n-1)

Answer 1

S = 0
________

W^n = 1 has roots w = e^(i2k&pi/n) (k = 0, 1, 2, ..., n-1), i.e. &alpha^k (k = 0, 1, 2, ..., n-1), where &alpha = e^(i2&pi/n)

Thus
W^n − 1 = (W − 1)(W − &alpha)(W − &alpha²)...(W − &alpha^(n-1)).
The coefficient of W^(n-1) on RHS =
1 + &alpha + &alpha² + &alpha³ + ... + &alpha^(n-1) = coefficient of W^(n-1) on LHS = 0, i.e.
1 + &alpha + &alpha² + &alpha³ + ... + &alpha^(n-1) = 0.

________

ALTERNATIVELY:

(1 − &alpha)(1 + &alpha + &alpha² + &alpha³ + ... + &alpha^(n-1)) = 1 - &alpha^n
= 0.
Since &alpha is not 1, it follows that
1 + &alpha + &alpha² + &alpha³ + ... + &alpha^(n-1) = 0

_________

Hope this helps.

Answer 2

The square root of an imaginary number is another imaginary number. First note that (a+bi)^2 = a^2 +2abi -b^2 = a^2-b^2 + (2ab)i So for example, (2+i)^2 = 4 +4i -1) =3+4i Therefore the sqrt of 3+4i is 2+i and 2-1 since complex numbers always come in conjugate pairs. So let's see how to show that the sqrt of 3+4i is indeed 2 + i and 2 - i Remember (a+bi)^2= a^2 -b^2 + (2ab)i So we must show that a^2 - b^2 = 3 and 2ab = 4 Put b= 2/.a into first equation getting a^2-4/a^4=3 a^4 -3a^2 -4 =0This is a quadratic in a^2 which can be solve by factoring giving a^2 = 4 or -1. Reject the -1 as we want a real solution so x^2 = 4 and a = 2 or -2 b=2/a so b= 1 or -1 Hence the sqrt of 3+=4i is 2+i or 2-i Hope this helped.

Answer 3

S = 0 always, because:

(1) If w is a primitive root of unity of order n, then it generates ALL the roots of unity of order n and thus 1, w, w^2,...,w^(n-1) are all the roots of unity of order 1 , so S = 1 + w +...+ w^(n-1), being the sum of ALL the complex roots of the polynomial equation z^n - 1 = 0 , is the coefficient of z in the polynomial z^n - 1, i.e. it is zero, so S = 0 in this case.

2) w is NOT a primitive root of unity or order n ==> w^m =1 for some m < n. If we choose a minimal such m, then w is actually a primitive root of unity of order m, and in fact m divides n, so n = dm, for some integer d, so in S the value
1 + w +...w^(m-1) is repeated d times. But every time this value is zero (see explanation in (1) above for primitive unity root) . Pay attention! M is NOT 1 since we're assuming w is not 1.
So S = 0 again in this case.

Regards
Tonio

Answer 4

The sum of W^k, k=0 to k=infinity is 1/(1-W). So the sum of W^k, k=0 to n-1, is S = (1 - W^n)/(1 - W). This is true for all W, real or complex. (The proof of the sums is a simple but elegant one.)

We have a problem if W = 1 (no imaginary part). However, that is trivially shown to be S = n, from the original posted definition or through a tiny bit of work on this form.

Answer 5

In general, for ANY complex number W, we have the following result (it's called a geometric series):

1 + W + W^2 +.....+ W^(n-1)=(1-W^n)/(1-W)

So, automatically in your case, you get S=0.

Answer 6

If "n" is a positive integer that makes "nnnnnnnnnnnn" a very positive experience but it sounds like constipation to me!

And W is a complex number, well all those www web addresses are really deep.

S = 1 well ssssssssssssss is a singulalry impressive snake sound.


Ask me a history question

Answer 7

S = 0,,, as
1, w, w square, w cube.........so on
are the nth roots of unity



remember tht w = 1+root3i/2 iff when n = 3......
in tht case 1 + w + w2 = 0


* sum of the all nth roots of unity is ZERO

Answer 8

They are the n points, equally spaced, on a unit circle in the complex plane. This is also called the n n'th roots of unity. They always sum to 0.

Doug

Answer 9

that is N=1
W=1 IMPLIES NW=1+1

S=1+1+1+1*2+(1*)(1-1)


NOW, i hope you can deduce this by this simple formular.
hope i have tried. is so long to get the final item.

Answer 10

W=a+bi
W^n=1,W≠1
1,W,W^2,.....W^(n-1)
S=sum of the geometric series=1(1-W^(n-1+1))/(1-W)=0