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.
 
The '''imaginary unit''', <math>i=\sqrt{-1}</math>, is the fundamental component of all [[complex numbers]]. In fact, it is a complex number, itself.
  
The imaginary unit shows up frequently in contest problems. The most common type of problem involving it is sums, i.e. problems such as "Find the sum of <math>i^1+i^2+\ldots+i^{2006}</math>."
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==Problems==
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The imaginary unit shows up frequently in contest problems. One type of problem involving it is sums, i.e. problems such as "Find the sum of <math>i^1+i^2+\ldots+i^{2006}</math>."
 
Let's begin by computing powers of <math>i</math>.
 
Let's begin by computing powers of <math>i</math>.
  

Revision as of 12:41, 20 July 2006

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The imaginary unit, $i=\sqrt{-1}$, is the fundamental component of all complex numbers. In fact, it is a complex number, itself.



Problems

The imaginary unit shows up frequently in contest problems. One type of problem involving it is sums, i.e. problems such as "Find the sum of $i^1+i^2+\ldots+i^{2006}$." Let's begin by computing powers of $i$.

$\displaystyle i^1=\sqrt{-1}$

$\displaystyle i^2=\sqrt{-1}\cdot\sqrt{-1}=-1$

$\displaystyle i^3=-1\cdot i=-i$

$\displaystyle i^4=-i\cdot i=-i^2=-(-1)=1$

$\displaystyle i^5=1\cdot i=i$

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.