Difference between revisions of "Imaginary unit"
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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>i^1=\sqrt{-1}</math> | + | <math>\displaystyle i^1=\sqrt{-1}</math> |
| − | <math>i^2=\sqrt{-1}\cdot\sqrt{-1}=-1</math> | + | <math>\displaystyle i^2=\sqrt{-1}\cdot\sqrt{-1}=-1</math> |
| − | <math>i^3=-1\ | + | <math>\displaystyle i^3=-1\cdot i=-i</math> |
| − | <math>i^4=-i\ | + | <math>\displaystyle i^4=-i\cdot i=-i^2=-(-1)=1</math> |
| − | <math>i^5=1\ | + | <math>\displaystyle 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, 22 June 2006
The imaginary unit, 
, 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 are sums, i.e. problems such as "Find the sum of 
."
Let's begin by computing powers of 
.
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 
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 -1+i.
