Difference between revisions of "Imaginary unit"

m (proofreading)
Line 1: Line 1:
 
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 are sums, i.e. problems such as "Find the sum of <math>i^1+i^2+\ldots+i^{2006}</math>."
+
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>."
 
Let's begin by computing powers of <math>i</math>.
 
Let's begin by computing powers of <math>i</math>.
  
Line 14: Line 14:
 
<math>\displaystyle i^5=1\cdot i=i</math>
 
<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 11:57, 23 June 2006

The imaginary unit, $i=\sqrt{-1}$, 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 $i^1+i^2+\ldots+i^{2006}$." 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.