Complex Number Arithmetic

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.