Imaginary unit/Introductory

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

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 $(-1)^{501} \times i^{2005} \times i^{2006} = -1 \times i \times -1 = i$ .

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Imaginary unit/Introductory

Contents

Solution 1

Solution 2