x2 t01 11 nth roots of unity (2012)

nth Roots Of Unity 1nz

nth Roots Of Unity 1nzunity of rootsnth theare ,1 form, theof equations of solutions The nz

nth Roots Of Unity 1nzunity of rootsnth theare ,1 form, theof equations of solutions The nz

When placed on an Argand Diagram, they form a regular n sided polygon, with vertices on the unit circle.

nth Roots Of Unity 1nzunity of rootsnth theare ,1 form, theof equations of solutions The nz

1 .. 5 zge

When placed on an Argand Diagram, they form a regular n sided polygon, with vertices on the unit circle.

nth Roots Of Unity 1nzunity of rootsnth theare ,1 form, theof equations of solutions The nz

1 .. 5 zge

When placed on an Argand Diagram, they form a regular n sided polygon, with vertices on the unit circle.

2 0 0, 1, 25kz cis k

nth Roots Of Unity 1nzunity of rootsnth theare ,1 form, theof equations of solutions The nz

1 .. 5 zge

When placed on an Argand Diagram, they form a regular n sided polygon, with vertices on the unit circle.

2 0 0, 1, 25kz cis k 2 2 4 40, , , ,5 5 5 5

z cis cis cis cis cis

nth Roots Of Unity 1nzunity of rootsnth theare ,1 form, theof equations of solutions The nz

1 .. 5 zge

When placed on an Argand Diagram, they form a regular n sided polygon, with vertices on the unit circle.

y

x0cis

2 0 0, 1, 25kz cis k 2 2 4 40, , , ,5 5 5 5

z cis cis cis cis cis

nth Roots Of Unity 1nzunity of rootsnth theare ,1 form, theof equations of solutions The nz

1 .. 5 zge

When placed on an Argand Diagram, they form a regular n sided polygon, with vertices on the unit circle.

y

x0cis

52cis

52

cis

2 0 0, 1, 25kz cis k 2 2 4 40, , , ,5 5 5 5

z cis cis cis cis cis

nth Roots Of Unity 1nzunity of rootsnth theare ,1 form, theof equations of solutions The nz

1 .. 5 zge

When placed on an Argand Diagram, they form a regular n sided polygon, with vertices on the unit circle.

y

x0cis

52cis

52

cis

54cis

54

cis

2 0 0, 1, 25kz cis k 2 2 4 40, , , ,5 5 5 5

z cis cis cis cis cis

nth Roots Of Unity 1nzunity of rootsnth theare ,1 form, theof equations of solutions The nz

1 .. 5 zge

When placed on an Argand Diagram, they form a regular n sided polygon, with vertices on the unit circle.

y

x0cis

52cis

52

cis

54cis

54

cis

2 0 0, 1, 25kz cis k 2 2 4 40, , , ,5 5 5 5

z cis cis cis cis cis

roots.other he t

are and ,, that show ,01 ofroot complex a is If b) 54325 zi

roots.other he t

are and ,, that show ,01 ofroot complex a is If b) 54325 zi

15 z 1hen solution tais If 5

roots.other he t

are and ,, that show ,01 ofroot complex a is If b) 54325 zi

15 z 1hen solution tais If 5

11555

solutionais5

roots.other he t

are and ,, that show ,01 ofroot complex a is If b) 54325 zi

15 z 1hen solution tais If 5

11555

solutionais5

2552

solutionais 2112

roots.other he t

are and ,, that show ,01 ofroot complex a is If b) 54325 zi

15 z 1hen solution tais If 5

11555

solutionais5

2552

solutionais 2112

3553

solutionais3113

roots.other he t

are and ,, that show ,01 ofroot complex a is If b) 54325 zi

15 z 1hen solution tais If 5

11555

solutionais5

2552

solutionais 2112

3553

solutionais3113

4554

solutionais4114

roots.other he t

are and ,, that show ,01 ofroot complex a is If b) 54325 zi

15 z 1hen solution tais If 5

11555

solutionais5

2552

solutionais 2112

3553

solutionais3113

4554

solutionais4114

roots also are and ,,root then a is if Thus 5432

0 1 that Prove 432 ii

0 1 that Prove 432 ii

ab

5432 rootstheof sum

01 432 15

0 1 that Prove 432 ii

ab

5432 rootstheof sum

01 432 15

OR 015 011 432

011

01:

12 n

n

NOTE

01 432 1

0 1 that Prove 432 ii

ab

5432 rootstheof sum

01 432 15

OR 015 011 432

011

01:

12 n

n

NOTE

01 432 1

OR2 3 41 GP: 1, , 5a r n 1

1

na rr

51 1

1

0 5 1 0

0 1 that Prove 432 ii

ab

5432 rootstheof sum

01 432 15

OR 015 011 432

011

01:

12 n

n

NOTE

01 432 1

324 and are roots hoseequation w quadratic theFind iii

OR2 3 41 GP: 1, , 5a r n 1

1

na rr

51 1

1

0 5 1 0

0 1 that Prove 432 ii

ab

5432 rootstheof sum

01 432 15

OR 015 011 432

011

01:

12 n

n

NOTE

01 432 1

324 and are roots hoseequation w quadratic theFind iii324

1 324

7643

1

243

OR2 3 41 GP: 1, , 5a r n 1

1

na rr

51 1

1

0 5 1 0

0 1 that Prove 432 ii

ab

5432 rootstheof sum

01 432 15

OR 015 011 432

011

01:

12 n

n

NOTE

01 432 1

324 and are roots hoseequation w quadratic theFind iii324

1 324

7643

1

243

01 isequation 2 xx

OR2 3 41 GP: 1, , 5a r n 1

1

na rr

51 1

1

0 5 1 0

32( ) Evaluate 1i c) If is a complex cube root of unity, use the fact that to; 01 2 ωω

32( ) Evaluate 1i c) If is a complex cube root of unity, use the fact that to; 01 2 ωω

3

32( ) Evaluate 1i c) If is a complex cube root of unity, use the fact that to; 01 2 ωω

3

1

3

32( ) Evaluate 1i c) If is a complex cube root of unity, use the fact that to; 01 2 ωω

3

1

3

2

1 1( ) Evaluate 1 1

ii

32( ) Evaluate 1i c) If is a complex cube root of unity, use the fact that to; 01 2 ωω

3

1

3

2

1 1( ) Evaluate 1 1

ii

11

2

21

32( ) Evaluate 1i c) If is a complex cube root of unity, use the fact that to; 01 2 ωω

3

1

3

2

1 1( ) Evaluate 1 1

ii

11

2

21

1

2

2

32( ) Evaluate 1i c) If is a complex cube root of unity, use the fact that to; 01 2 ωω

3

1

3

2

1 1( ) Evaluate 1 1

ii

11

2

21

1

2

2

(iii) Form the cubic equation with roots 21 ,1 ,1

32( ) Evaluate 1i c) If is a complex cube root of unity, use the fact that to; 01 2 ωω

3

1

3

2

1 1( ) Evaluate 1 1

ii

11

2

21

1

2

2

(iii) Form the cubic equation with roots 21 ,1 ,1 0121 3222 zzz

32( ) Evaluate 1i c) If is a complex cube root of unity, use the fact that to; 01 2 ωω

3

1

3

2

1 1( ) Evaluate 1 1

ii

11

2

21

1

2

2

(iii) Form the cubic equation with roots 21 ,1 ,1 0121 3222 zzz

011 2 zzz

32( ) Evaluate 1i c) If is a complex cube root of unity, use the fact that to; 01 2 ωω

3

1

3

2

1 1( ) Evaluate 1 1

ii

11

2

21

1

2

2

(iii) Form the cubic equation with roots 21 ,1 ,1 0121 3222 zzz

011 2 zzz01223 zzzzz0122 23 zzz

5 4 3 2d) Solve 1 0z z z z z

5 4 3 2d) Solve 1 0z z z z z

111 Now 23456 zzzzzzz

5 4 3 2d) Solve 1 0z z z z z

111 Now 23456 zzzzzzz6And 1 0 has solutions;

2 0, 1, 2,36

zkz cis k

5 4 3 2d) Solve 1 0z z z z z

111 Now 23456 zzzzzzz

012345 zzzz

6And 1 0 has solutions;2 0, 1, 2,3

6

zkz cis k

5 4 3 2d) Solve 1 0z z z z z

111 Now 23456 zzzzzzz

012345 zzzz

1,016 zz

01

11 2345

z

zzzzz

6And 1 0 has solutions;2 0, 1, 2,3

6

zkz cis k

5 4 3 2d) Solve 1 0z z z z z

111 Now 23456 zzzzzzz

012345 zzzz

1,016 zz

01

11 2345

z

zzzzz

6And 1 0 has solutions;2 0, 1, 2,3

6

zkz cis k

2 2, , , ,3 3 3 3

z cis cis cis cis cis

1 3 1 3 1 3 1 3, , , , 12 2 2 2 2 2 2 2

z i i i i

9

e) 1996 HSC2 2 Let cos sin9 9

(i) Show that is a solution of 1 0, where is an integerk

i

z k

9

e) 1996 HSC2 2 Let cos sin9 9

(i) Show that is a solution of 1 0, where is an integerk

i

z k

19 z

9

e) 1996 HSC2 2 Let cos sin9 9

(i) Show that is a solution of 1 0, where is an integerk

i

z k

19 z

8,7,6,5,4,3,2,1,0 9

2

kkcisz

k

cisz

92

9

e) 1996 HSC2 2 Let cos sin9 9

(i) Show that is a solution of 1 0, where is an integerk

i

z k

19 z

8,7,6,5,4,3,2,1,0 9

2

kkcisz

k

cisz

92

kz

9

e) 1996 HSC2 2 Let cos sin9 9

(i) Show that is a solution of 1 0, where is an integerk

i

z k

19 z

8,7,6,5,4,3,2,1,0 9

2

kkcisz

k

cisz

92

kz 2 3 4 5 6 7 8(ii) Prove that 1

019 z

9

e) 1996 HSC2 2 Let cos sin9 9

(i) Show that is a solution of 1 0, where is an integerk

i

z k

19 z

8,7,6,5,4,3,2,1,0 9

2

kkcisz

k

cisz

92

kz 2 3 4 5 6 7 8(ii) Prove that 1

019 z01 8765432 rootsofsum

9

e) 1996 HSC2 2 Let cos sin9 9

(i) Show that is a solution of 1 0, where is an integerk

i

z k

19 z

8,7,6,5,4,3,2,1,0 9

2

kkcisz

k

cisz

92

kz 2 3 4 5 6 7 8(ii) Prove that 1

019 z01 8765432 rootsofsum

18765432

2 4 1(iii) Hence show that cos cos cos9 9 9 8

2 4 1(iii) Hence show that cos cos cos9 9 9 8

9

8 2 7 3 6 4 5

1

1

z

z z z z z z z z z

2 4 1(iii) Hence show that cos cos cos9 9 9 8

9

8 2 7 3 6 4 5

1

1

z

z z z z z z z z z

2 2

2 2

2 41 2cos 1 2cos 19 9

6 8 2cos 1 2cos 19 9

z z z z z

z z z z

2 2

2 2

2 41 2cos 1 2cos 19 9

8 1 2cos 19

z z z z z

z z z z

2 4 1(iii) Hence show that cos cos cos9 9 9 8

9

8 2 7 3 6 4 5

1

1

z

z z z z z z z z z

2 2

2 2

2 41 2cos 1 2cos 19 9

6 8 2cos 1 2cos 19 9

z z z z z

z z z z

2 2

2 2

2 41 2cos 1 2cos 19 9

8 1 2cos 19

z z z z z

z z z z

Let z i

9 2 4 81 1 2cos 2cos 2cos9 9 9

i i i i i i

9 2 4 81 1 2cos 2cos 2cos9 9 9

i i i i i i

9 2 4 81 1 2cos 2cos 2cos9 9 9

i i i i i i

4 2 4 81 1 2cos 2cos 2cos9 9 9

i i i

9 2 4 81 1 2cos 2cos 2cos9 9 9

i i i i i i

4 2 4 81 1 2cos 2cos 2cos9 9 9

i i i

2 4 81 8cos cos cos9 9 9

9 2 4 81 1 2cos 2cos 2cos9 9 9

i i i i i i

4 2 4 81 1 2cos 2cos 2cos9 9 9

i i i

2 4 81 8cos cos cos9 9 9

2 41 8cos cos cos9 9 9

9 2 4 81 1 2cos 2cos 2cos9 9 9

i i i i i i

4 2 4 81 1 2cos 2cos 2cos9 9 9

i i i

2 4 81 8cos cos cos9 9 9

2 4 1cos cos cos9 9 9 8

2 41 8cos cos cos9 9 9

9 2 4 81 1 2cos 2cos 2cos9 9 9

i i i i i i

4 2 4 81 1 2cos 2cos 2cos9 9 9

i i i

2 4 81 8cos cos cos9 9 9

2 4 1cos cos cos9 9 9 8

Cambridge: Exercise 7C; 1 to 4, 5ac, 6, 7, 8, 9, 11

Patel: Exercise 4I; 1, 3, 5 (need 4b), 7

Patel: Exercise 4J; 1 to 4, 7ac

2 41 8cos cos cos9 9 9