Difference between revisions of "Imaginary unit"

Revision as of 14:37, 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

Find the sum of $i^1+i^2+\ldots+i^{2006}$ (Source)

@@ Line 3: / Line 3: @@
 ==Problems==
 === Introductory ===
-#Find the sum of <math>i^1+i^2+\ldots+i^{2006}</math>
+*Find the sum of <math>i^1+i^2+\ldots+i^{2006}</math> ([[Imaginary unit/Introductory|Source]])
-==Solutions ==
-=== Introductory ===
-#Let's begin by computing powers of <math>i</math>.
-#:<math>i^1=\sqrt{-1}</math>
-#:<math>i^2=\sqrt{-1}\cdot\sqrt{-1}=-1</math>
-#:<math>i^3=-1\cdot i=-i</math>
-#:<math>i^4=-i\cdot i=-i^2=-(-1)=1</math>
-#:<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 <math>-1+i</math>.
 == See also ==
@@ Line 20: / Line 10: @@
 * [[Complex numbers]]
 * [[Geometry]]
+{{stub}}
 [[Category:Constants]]

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Difference between revisions of "Imaginary unit"

Revision as of 14:37, 26 October 2007

Problems

Introductory

See also