Compute \(i^{-100}+i^{-99}+i^{-98}+...+i^{-1}+i^{0}+i^{1}+...+i^{99}+i^{100}\)

I got 0 but it was incorrect. I think I might be doing something wrong. Thanks!

jilin73 Mar 30, 2024

#1**-1 **

Let's break it down step-by-step:

Grouping Terms:

We notice that the series alternates between terms with a negative imaginary unit exponent (i^-n) and terms with a positive imaginary unit exponent (i^n). This allows us to group the terms together:

(i^-100 + i^-99 + ... + i^-1) + (i^0 + i^1 + ... + i^99 + i^100)

Simplifying Each Group:

First Group:

In this group, all the terms except the first (i^-100) and the last (i^-1) will cancel out because i raised to any negative power is the same as its reciprocal multiplied by i^-1.

Therefore, the first group simplifies to: i^-100 + i^-1

Second Group:

Similarly, in the second group, all the terms except the first (i^0) and the last (i^100) will cancel out because for any non-zero integer n, i^n * i^-n = i^(n-n) = i^0 = 1.

Therefore, the second group simplifies to: 1 + i^100

Recombining and Simplifying:

Now, we have the simplified version of the series:

(i^-100 + i^-1) + (1 + i^100)

i^-100 can be simplified as 1/i^100 (since i raised to any negative power is the reciprocal).

i^-1 = -i (because i raised to the odd power -1 is -i).

Putting it all together:

(1/i^100 - i) + (1 + i^100)

Analyzing i^100:

The key here is to recognize that i raised to any multiple of 4 results in 1 (i^4 = (i^2)^2 = (-1)^2 = 1). Since 100 is a multiple of 4, i^100 = 1.

Final Simplification:

(1/i^100 - i) + (1 + i^100) becomes:

(1/1 - i) + (1 + 1)

Simplifying further: (1 - i) + 2

Combining like terms: 1 - i + 2

Final result: 3 - i

The sum is 3 - i.

ChiIIBill Mar 30, 2024

#2**+1 **

i^1 + i^2 + i^3 + i^4 =

i + -1 - i + 1 = 0

And this pattern will repeat

So....sum of i^1 tp i^100 inclusive = 0

i^(-1) = 1 / i = i / i^2 = -i

i^(-2) = 1/i^2 = 1/ -1 = -1

i^(-3) = 1/^2 * 1/i = -1 * 1/i = -1 *-i = i

i^(-4) = 1/I^2 * 1/i^2 = -1 * -1 = 1

And this pattern will repeat

So....sum of i^(-100) to i^(-1) inclusive = 0

The only term not evaluated is i^0 = 1

So....the sum will be 1

CPhill Mar 31, 2024