Difference between revisions of "Imaginary unit/Introductory"
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| − | + | #Find the sum of <math>i^1+i^2+\ldots+i^{2006}</math>. ([[#Solution 1|solution]]) | |
| − | Find the sum of <math>i^1+i^2+\ldots+i^{2006}</math> | + | #Find the product of <math>i^1 \times i^2 \times \cdots \times i^{2006}</math>. ([[#Solution 1|solution]]) |
| − | == Solution == | + | |
| + | __TOC__ | ||
| + | |||
| + | == Solution 1 == | ||
Since <math>i</math> 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 <math>i^1+i^2+\ldots+i^{4k}</math> have a sum of zero (k is a natural number). Since <math>2006=4\cdot501+2</math>, the original series sums to the first two terms of the powers of i, which equals <math>-1+i</math>. | Since <math>i</math> 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 <math>i^1+i^2+\ldots+i^{4k}</math> have a sum of zero (k is a natural number). Since <math>2006=4\cdot501+2</math>, the original series sums to the first two terms of the powers of i, which equals <math>-1+i</math>. | ||
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| + | == Solution 2 == | ||
| + | <math>i \cdot -1 \cdot -i \cdot 1 = -1</math>, so the product is equal to <math>(-1)^{501} \times i^{2005} \times i^{2006} = -1 \times i \times -1 = i</math>. | ||
Revision as of 20:34, 26 October 2007
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Solution 1
Since 
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 
have a sum of zero (k is a natural number). Since 
, the original series sums to the first two terms of the powers of i, which equals 
.
Solution 2
, so the product is equal to 
.
