Difference between revisions of "Imaginary unit"

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The '''imaginary unit''', <math>i=\sqrt{-1}</math>, 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 <math>1 \mathrm{cis} \left(\frac{\pi}{2}\right)</math>.  
 
The '''imaginary unit''', <math>i=\sqrt{-1}</math>, 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 <math>1 \mathrm{cis} \left(\frac{\pi}{2}\right)</math>.  
  
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* [[Complex numbers]]
 
* [[Complex numbers]]
 
* [[Geometry]]
 
* [[Geometry]]
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[[Category:Constants]]

Revision as of 14:36, 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

  1. Find the sum of $i^1+i^2+\ldots+i^{2006}$


Solutions

Introductory

  1. Let's begin by computing powers of $i$.
    $i^1=\sqrt{-1}$
    $i^2=\sqrt{-1}\cdot\sqrt{-1}=-1$
    $i^3=-1\cdot i=-i$
    $i^4=-i\cdot i=-i^2=-(-1)=1$
    $i^5=1\cdot i=i$
    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 $-1+i$.

See also