Imaginary unit/Introductory

Revision as of 14:38, 26 October 2007 by Temperal (talk | contribs) (creation)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

Find the sum of $i^1+i^2+\ldots+i^{2006} == Solution == Let's begin by computing powers of$i$.

  1. $ (Error compiling LaTeX. Unknown error_msg)i^1=\sqrt{-1}$#$ (Error compiling LaTeX. Unknown error_msg)i^2=\sqrt{-1}\cdot\sqrt{-1}=-1$#$ (Error compiling LaTeX. Unknown error_msg)i^3=-1\cdot i=-i$#$ (Error compiling LaTeX. Unknown error_msg)i^4=-i\cdot i=-i^2=-(-1)=1$#$ (Error compiling LaTeX. Unknown error_msg)i^5=1\cdot i=i$We can now stop because we have come back to our original term. This means that the sequence i, -1, -i, 1 repeats. Note that this sums to 0. That means that all sequences$i^1+i^2+\ldots+i^{4k}$have a sum of zero (k is a natural number). Since$2006=4\cdot501+2$, the original series sums to the first two terms of the powers of i, which equals$-1+i$.