Special topics in Algebra Volume 1 Roots of Unity Section 1: square roots The square roots of 1 are 1 and -1. 1 * 1 = 1 and (-1) * (-1) = 1. We say the principle square root of 1 is 1. And we write sqrt(1) = 1. We say the primitive square root of 1 is -1. This means that -1 squared is equal to 1, and no smaller power of -1 is equal to 1. Section 2: cube roots To find the 3 cube roots of 1, we solve the equation z^3 - 1 = 0. z^3 means z raised to the third power. z^3 - 1 = 0 (z-1)(z^2 + z + 1) = 0 z-1 = 0 z^2 + z + 1 = 0 From z^2 + z + 1 = 0, we get z1 = -1/2 + i [ sqrt(3) ] / 2 z2 = -1/2 - i [ sqrt(3) ] / 2 Where i means sqrt(-1). From z-1 = 0, we get z3 = 1. The three cube roots of 1 are z1 = -1/2 + i [ sqrt(3) ] / 2 z2 = -1/2 - i [ sqrt(3) ] / 2 z3 = 1. To confirm that z1 is a cube root, we calculate z1^3. z1^2 = [ (-1/2)^2 - ( [ sqrt(3) ]/2 )^2 ] + i 2 (-1/2) ([ sqrt(3) ]/2 ) = [ 1/4 - 3/4 ] - i [ sqrt(3) ]/2 = -1/2 - i [ sqrt(3) ]/2 = z2 z1^3 = z1 * z2 = (-1/2)^2 + ( [ sqrt(3) ]/2 )^2 = 1/4 + 3/4 = 1 To confirm that z2 ^3 = 1, we proceed as follows: z2 ^3 = (z1 ^2) ^3 = z1 ^6 = ( z1 ^3 ) ^2 = 1 ^2 = 1. z1 and z2 are the primitive cube roots of 1, and 1 is the principle cube root of 1. Section 3: 4th roots z^4 - 1 = 0 (z^2 - 1)(z^2 + 1) = 0 (z-1)(z+1)(z^2+1) = 0 z-1 = 0 z+1 = 0 z^2 + 1 = 0 z1 = sqrt(-1) = i z2 = -1 z3 = -sqrt(-1) = -i z4 = 1 z1 and z3 are the primitive 4th roots of 1. Section 4: 5th roots z^5 - 1 = 0 (z-1)(z^4 + z^3 + z^2 + z + 1) It is not obvious how we may factor (z^4 + z^3 + z^2 + z + 1). If w is a 5th root of 1, then also 1/w is a 5th root of 1. Let w1 and w2 be two 5th roots of 1 where w2 is not equal to 1/w1. z^4 + z^3 + z^2 + z + 1 = (z - w1)(z - 1/w1)(z - w2)(z - 1/w2) = (z - [w1 + 1/w1] z + 1)(z - [w2 + 1/w2] z + 1) Let a = w1 + 1/w1 and b = w2 + 1/w2 (z^4 + z^3 + z^2 + z + 1) = (z^2 - a z + 1)(z^2 - b z + 1) (z^2 - a z + 1)(z^2 - b z + 1) = z^4 - (a+b) z^3 + (2 + ab) z^2 - (a+b) z + 1 This gives us the equations a + b = -1 2 + ab = 1 a + b = -1 ab = -1 (x - a)(x-b) = x^2 + (a+b) x + ab = x^2 + x -1 if x^2 + x - 1 = 0 Then (x-a)(x-b) = 0 and thus x = a or x = b. Let a = -1/2 + [ sqrt(5) ]/2 and b = -1/2 - [ sqrt(5) ] /2 We confirm that a+b = -1/2 + [ sqrt(5) ]/2 - 1/2 - [ sqrt(5) ] /2 = -1 and that ab = (-1/2 + [ sqrt(5) ]/2) *(-1/2 - [ sqrt(5) ] /2) = 1/4 - 5/4 = -4/4 = -1 we defined a = w1 + 1/w1 and b = w2 + 1/w2 Solve w1 + 1/w1 = a and w2 + 1/w2 = b w1^2 + 1 = a w1 and w2^2 + 1 = b w2 This leads to z^2 - a z + 1 = 0 z = a/2 + [ sqrt(a^2 - 4) ] / 2 or z = a/2 - [ sqrt(a^2 - 4) ] / 2 z^2 - b z + 1 = 0 z = b/2 + [ sqrt(b^2 - 4) ] / 2 or z = b/2 - [ sqrt(b^2 - 4) ] / 2 Since we calculated a = -1/2 + [ sqrt(5) ] /2 and b = -1/2 - [ sqrt(5) ] /2 we have a^2 = 3/2 - [ sqrt(5) ] /2 b^2 = 3/2 + [ sqrt(5) ] /2 a^2 - 4 = -5/2 - [ sqrt(5) ] /2 b^2 - 4 = -5/2 + [ sqrt(5) ] /2 z1 = b/2 + [ sqrt(b^2 - 4) ] / 2 = -[ 1/2 - [ sqrt(5) ] /2]/2 + i [sqrt( 5/2 + [ sqrt(5) ]/2)]/2 = ( -1 + [ sqrt(5) ]) /4 + i (sqrt[ 10 + 2 sqrt(5) ])/4 z2 = a/2 + [ sqrt(a^2 - 4) ] / 2 = - [ 1/2 + [ sqrt(5) ] /2]/2 + i [sqrt( 5/2 - [ sqrt(5) ]/2)]/2 = ( -1 - [ sqrt(5) ]) /4 + i (sqrt[ 10 - 2 sqrt(5) ])/4 z3 = a/2 - [ sqrt(a^2 - 4) ] / 2 = - [ 1/2 - [ sqrt(5) ] /2]/2 + i [sqrt( 5/2 - [ sqrt(5) ]/2)]/2 = ( -1 - [ sqrt(5) ]) /4 - i (sqrt[ 10 - 2 sqrt(5) ])/4 z4 = b/2 - [ sqrt(b^2 - 4) ] / 2 = -[ 1/2 - [ sqrt(5) ] /2]/2 - i [sqrt( 5/2 + [ sqrt(5) ]/2)]/2 = ( -1 + [ sqrt(5) ]) /4 - i (sqrt[ 10 + 2 sqrt(5) ])/4 To confirm that z1^5 = 1, we proceed as follows. z1 = ( -1 + [ sqrt(5) ]) /4 + i (sqrt[ 10 + 2 sqrt(5) ])/4 z1^2 = [6 - 2 sqrt(5) ]/16 - [10 + 2 sqrt(5) ]/16 + i (2 [ ( -1 + [ sqrt(5) ] ) ] (sqrt[ 10 + 2 sqrt(5) ] )/16 ) = [-4 - 4 sqrt(5)]/16 + i sqrt[ (2 [ ( -1 + [ sqrt(5) ]) ] (sqrt[ 10 + 2 sqrt(5) ] )/16)^2] = [-1 - sqrt(5)]/4 + i sqrt[ (4 [ -1 + sqrt(5) ]^2 * [ 10 + 2 sqrt(5) ]/256 )] = [-1 - sqrt(5)]/4 + i sqrt[ ( [ -1 + sqrt(5) ]^2 * [ 10 + 2 sqrt(5) ]/64 )] = [-1 - sqrt(5)]/4 + i sqrt[ [ 6 - 2 sqrt(5) ] * [ 10 + 2 sqrt(5) ]/64 ] = [-1 - sqrt(5)]/4 + i sqrt[ [ (60 - 20) +( 12 - 20) sqrt(5) ]/64 ] = [-1 - sqrt(5)]/4 + i sqrt[ [ 40 - 8 sqrt(5) ]/64 ] = [-1 - sqrt(5)]/4 + i sqrt[ [ 10 - 2 sqrt(5) ]/16 ] = [-1 - sqrt(5)]/4 + i sqrt[ 10 - 2 sqrt(5) ] / 4 z2 = ( -1 - [ sqrt(5) ]) /4 + i (sqrt[ 10 - 2 sqrt(5) ])/4 z1 ^ 2 = z2. z2^2 = [6 + 2 sqrt(5) ]/16 - [10 - 2 sqrt(5) ]/16 - iÿ (2 [ ( 1 + [ sqrt(5) ] ) ] (sqrt[ 10 - 2 sqrt(5) ] )/16 ) = [-4 + 4 sqrt(5)]/16 - iÿ sqrt[ (2 [ ( 1 + [ sqrt(5) ]) ] (sqrt[ 10 - 2 sqrt(5)])/16)^2] = [-1 + sqrt(5)]/4 - iÿ sqrt[ (4 [ 1 + sqrt(5) ]^2 * [ 10 - 2 sqrt(5) ]/256 )] = [-1 + sqrt(5)]/4 - iÿ sqrt[ ( [ 1 + sqrt(5) ]^2 * [ 10 - 2 sqrt(5) ]/64 )] = [-1 + sqrt(5)]/4 - iÿ sqrt[ [ 6 + 2 sqrt(5) ] * [ 10 - 2 sqrt(5) ]/64 ] = [-1 + sqrt(5)]/4 - iÿ sqrt[ [ (60 - 20) +( 20 - 12) sqrt(5) ]/64 ] = [-1 + sqrt(5)]/4 - iÿ sqrt[ [ 40 + 8 sqrt(5) ]/64 ] = [-1 + sqrt(5)]/4 - iÿ sqrt[ [ 10 + 2 sqrt(5) ]/16 ] = [-1 - sqrt(5)]/4 - iÿ sqrt[ 10 + 2 sqrt(5) ] / 4 z4 = ( -1 + [ sqrt(5) ]) /4 - i (sqrt[ 10 + 2 sqrt(5) ])/4 z2 ^ 2 = z4 z1 = ( -1 + [ sqrt(5) ]) /4 + i (sqrt[ 10 + 2 sqrt(5) ])/4 z4 = ( -1 + [ sqrt(5) ]) /4 - i (sqrt[ 10 + 2 sqrt(5) ])/4 z1 * z4 = [ ( -1 + [ sqrt(5) ]) /4 ]^2 + [(sqrt[ 10 + 2 sqrt(5) ])/4 ]^2 = [ 6 - 2 sqrt(5) ]/16 + [ 10 + 2 sqrt(5) ]/16 = 16 /16 = 1 z1 ^ 5 = z1 * ( z1 ^ 2 ) ^ 2 = z1 * z2 ^ 2 = z1 * z4 = 1