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Imaginary unit/Introductory

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  1. Find the sum of i^1+i^2+\ldots+i^{2006}.
  2. Find the product of i^1 \times i^2 \times \cdots \times i^{2006}.

Contents


Solution 1

Since i repeats in a n exponential series at every fifth turn, 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.

Solution 2

i \cdot -1 \cdot -i \cdot 1 = -1, so the product is equal to (-1)^{501} \times i^{2005} \times i^{2006} = -1 \times i \times -1 = i.