Trigonometry - Revision history

Milllo: /* Root of Unity */

2020-04-01T20:24:15Z

Root of Unity

2020-04-01T20:20:37Z

Root of Unity

2020-04-01T20:20:06Z

Root of Unity

2020-04-01T20:18:40Z

Root of Unity

2020-04-01T20:18:15Z

Root of Unity

2020-04-01T20:15:09Z

2020-04-01T20:14:35Z

Root of Unity

2020-04-01T19:56:07Z

Root of Unity

2020-04-01T19:53:24Z

2020-04-01T19:48:10Z

Root of Unity

@@ Line 47: / Line 47: @@
 These roots are:
-e<sup>''2kπi/n''</sup>, where k goes from 0 to n-1 (so there are n roots)
+e<sup>''2kπi/n''</sup>, where k goes from 1 to n (so there are n roots)
 using Euler's formula gives:

@@ Line 41: / Line 41: @@
 ==Root of Unity==
-A great way to get into complex numbers (a+b''i'') is via a root of unity, which is a number that satisfies the equation:
+A great way to get into complex numbers (''a''+''bi'') is via a root of unity, which is a number that satisfies the equation:
 z<sup>n</sup> = 1, where n is a positive integer

@@ Line 41: / Line 41: @@
 ==Root of Unity==
-A great way to get into complex numbers is via a root of unity, which is a number that satisfies the equation:
+A great way to get into complex numbers (a+b''i'') is via a root of unity, which is a number that satisfies the equation:
 z<sup>n</sup> = 1, where n is a positive integer

@@ Line 58: / Line 58: @@
 which can be converted to complex numbers:
-x = cos(2π/3) + isin(2π/3) = -1/2+''i''Sqrt(3)/2
+x = cos(2π/3) + isin(2π/3) = -1/2 + ''i''Sqrt(3)/2
-x = cos(2π/3) + isin(4π/3) = -1/2-''i''Sqrt(3)/2
+x = cos(2π/3) + isin(4π/3) = -1/2 - ''i''Sqrt(3)/2
 x = cos(2π/3) + isin(6π/3) = 1

← Older revision		Revision as of 20:18, 1 April 2020
Line 55:		Line 55:
	example:		example:
	roots of x<sup>3</sup> are ''e''<sup>''2iπ/3''</sup>, ''e''<sup>''4iπ/3''</sup>,''e''<sup>''6iπ/3''</sup>		roots of x<sup>3</sup> are ''e''<sup>''2iπ/3''</sup>, ''e''<sup>''4iπ/3''</sup>,''e''<sup>''6iπ/3''</sup>
		+
		+	which can be converted to complex numbers:
		+
		+	x = cos(2π/3) + isin(2π/3) = -1/2+''i''Sqrt(3)/2
		+
		+	x = cos(2π/3) + isin(4π/3) = -1/2-''i''Sqrt(3)/2
		+
		+	x = cos(2π/3) + isin(6π/3) = 1

← Older revision		Revision as of 20:15, 1 April 2020
Line 54:		Line 54:

	example:		example:
−	roots of x<sup>3</sup> are ''e''<sup>''~~2ip~~/3''</sup>, ''e''<sup>''~~4ip~~/3''</sup>,''e''<sup>''~~6ip~~/3''</sup>	+	roots of x<sup>3</sup> are ''e''<sup>''2iπ/3''</sup>, ''e''<sup>''4iπ/3''</sup>,''e''<sup>''6iπ/3''</sup>

← Older revision		Revision as of 19:56, 1 April 2020
Line 47:		Line 47:
	These roots are:		These roots are:

−	e<sup>''2kπi/n''</sup>	+	e<sup>''2kπi/n''</sup>, where k is an integer
		+
		+	using Euler's formula gives:
		+
		+	e<sup>''2kπi/n''</sup> = ''cos''(''2kπ/n'') + ''isin''(''2kπ/n'')