# SOLVE EQUATIONS IN COMPLEX NUMBERS WITH CUBE ROOTS OF UNITY

Solve Equations in Complex Numbers With Cube Roots of Unity :

Here we are going to see some example problems of solving equations in complex numbers with cube roots of unity.

## Solve Equations in Complex Numbers With Cube Roots of Unity  - Questions

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 + ω2^11)  =  1

Solution :

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

L.H.S:

=  (1 + ω− ω)6 + (1 + ω − ω2)6

=  (- ω − ω)6 + (-ω− 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 + ω2^11)  =  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)

5th and 6th 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| After having gone through the stuff given above, we hope that the students would have understood, "Solve Equations in Complex Numbers With Cube Roots of Unity".

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