SOLVE EQUATIONS IN COMPLEX NUMBERS WITH CUBE ROOTS OF UNITY

Question 1 :

Solve the equation z³ + 27 = 0.

Solution :

z³ + 27 = 0

z³   =  -27

z³   =  (-1 ⋅ 3)³

z   =  [(-1 ⋅ (3)³] ^1/3

  = 3 (-1)^1/3

Polar form of -1 :

-1  =  3[cos π + i sin π]

  =  [cos(2kπ + π) + i sin (2kπ + π)]

  =  [cos π(2k + 1)) + i sin π(2k + 1)]

 (-1)^1/3  =  [cos π(2k + 1)) + i sin π(2k + 1)]^1/3

 (-1)^1/3  =  [cos (π/3)(2k + 1)) + i sin (π/3)(2k + 1)]

k = 0, 1, 2

If k = 0

  =  [cos (π/3)(2k + 1)) + i sin (π/3)(2k + 1)]

  =  3 cis (π/3)

If k = 1

  =  3 [cos π + i sin π]

  =  -3

If k = 2

  =  [cos (5π/3) + i sin (5π/3)]

  =  3 cis (5π/3)

Let us look into the next problem on "Solve Equations in Complex Numbers With Cube Roots of Unity"

Question 2 :

If ω ≠  1 is a cube root of unity, show that the roots of the equation (z −1)³ + 8 = 0 are −1, 1− 2ω, 1− 2ω²

Solution :

(z −1)³ + 8 = 0

(z −1)³ = -8

(z −1) = (-8)^1/3 

(z −1) = -2 ⋅ (1) ^1/3 

z  =  -2 ⋅ (1) ^1/3  + 1

z  =  1 - 2 ⋅ (1) ^1/3 

Cube root of 1 are 1, ω, ω²

z  =  1 - 2 ⋅ 1 

z  =  1 - 2  =  -1

z  =  1 - 2 ⋅ ω

 z  =  1 - 2ω

z  =  1 - 2 ⋅ ω²

 z  =  1 - 2ω²

Let us look into the next problem on "Solve Equations in Complex Numbers With Cube Roots of Unity"

Question 3 :

Find the value of

Solution :

If k = 1,

 =  cos 2π/9 + i sin 2π/9 ----(1)

If k = 2,

 =  cos 4π/9 + i sin 4π/9 ----(2)

If k = 3,

 =  cos 6π/9 + i sin 6π/9 ----(3)

If k = 4,

 =  cos 8π/9 + i sin 8π/9 ----(4)

...................

By adding all these, we get

  =  cis (π/9) (2 + 4 + 6 + 8 + 10 + 12 + 14 + 16)

  =  cis (π/9) 2(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8)

  =  cis (72π/9)

  =  cis 8π

  =  cos 8π + i sin 8π

  =  1 + i(0)

  =  1

Question 4 :

If ω ≠  1 is a cube root of unity, show that

(i) (1 − ω + ω²)⁶ + (1 + ω − ω²)⁶  =  128.

(ii) (1 − ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸).............(1 + ω^{2^11})  =  1

Solution :

(i) (1 − ω + ω²)⁶ + (1 + ω − ω²)⁶  =  128.

L.H.S: 

  =  (1 + ω² − ω)⁶ + (1 + ω − ω²)⁶ 

  =  (- ω − ω)⁶ + (-ω² − ω²)⁶ 

  =  (- 2ω) ⁶  + (- 2ω²) ⁶

  =  64 ω⁶  + 64ω¹²

  =  64 (ω³)² + 64 (ω³)⁴

  =  64 + 64

  =  128

R.H.S

Hence proved.

(ii) (1 − ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸).............(1 + ω^{2^11})  =  1

L.H.S

(1 − ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸)(1 + ω¹⁶) (1 + ω³²) (1 + ω⁶⁴)

(1 + ω¹²⁸)(1 + ω²⁵⁶)(1 + ω⁵¹²)(1 + ω¹⁰²⁴)(1 + ω²⁰⁴⁸)

First 2 terms are   =  (1 − ω)(1 + ω²)

3rd and 4th terms :

(1 + ω⁴)(1 + ω⁸)  =  (1 + ω)(1 + ω²)

5^th and 6^th terms :

(1 + ω¹⁶)(1 + ω³²)  =  (1 + ω)(1 + ω²)

Similarly by grouping these terms, we get 

=  [(1 + ω)(1 + ω²)]⁶

=  [1 + ω² + ω + ω³ ]⁶

=  [0 + ω³ ]⁶

=  1

Hence proved.

Question 5 :

If z = 2 - 2i, find rotation of z by θ radians in the counter clock wise direction about the origin when

(i)  θ  =  π/3    (ii) θ  =  2π/3    (iii) θ  =  3π/2

Solution :

z = 2 - 2i 

α  = tan^-1|(-2)/2|

α  = tan^-1|(-2)/2|

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