Art of Problem Solving

Imaginary unit: Difference between revisions

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==Problems==
==Problems==
=== Introductory ===
=== Introductory ===
#Find the sum of <math>i^1+i^2+\ldots+i^{2006}</math>
*Find the sum of <math>i^1+i^2+\ldots+i^{2006}</math> ([[Imaginary unit/Introductory|Source]])


==Solutions ==
=== Introductory ===
#Let's begin by computing powers of <math>i</math>.
#:<math>i^1=\sqrt{-1}</math>
#:<math>i^2=\sqrt{-1}\cdot\sqrt{-1}=-1</math>
#:<math>i^3=-1\cdot i=-i</math>
#:<math>i^4=-i\cdot i=-i^2=-(-1)=1</math>
#:<math>i^5=1\cdot i=i</math>
#: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 <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>.


== See also ==
== See also ==
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* [[Complex numbers]]
* [[Complex numbers]]
* [[Geometry]]
* [[Geometry]]
 
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[[Category:Constants]]
[[Category:Constants]]

Revision as of 13:37, 26 October 2007

The imaginary unit, $i=\sqrt{-1}$, is the fundamental component of all complex numbers. In fact, it is a complex number itself. It has a magnitude of 1, and can be written as $1 \mathrm{cis} \left(\frac{\pi}{2}\right)$.

Problems

Introductory


See also

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