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

Revision as of 16:09, 23 September 2007

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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

One common problem involving the imaginary unit is along the lines of:

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

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