StudentAidUnit 1: Complex Numbers — Exercise 1 4
11th Class Mathematics · Unit 1: Complex Numbers
1.Find the three cube roots of: (i) 8, (ii) -8, (iii) -27, (iv) 64, (v) -125
Given
(i) Cube roots of 8: Let x be the cube root
x3 = 8
x3 - 8 = 0
(x-2)(x2 + 2x + 4) = 0
x = 2 or x2 + 2x + 4 = 0
Working
Using quadratic formula
x = frac{-2 ± sqrt{4-16}{2} = frac{-2 ± 2isqrt{3}{2}
Where ω = (-1+i√3)/2 is a cube root of unity
x = 2, 2omega, 2omega2
Result
(i) Answer
2, 2omega, 2omega2
Given
(ii) Cube roots of -8: Let x be the cube root
x3 = -8
x3 + 8 = 0
(x+2)(x2 - 2x + 4) = 0
Result
(ii) Answer
-2, -2omega, -2omega2
Given
(iii) Cube roots of -27: Let x be the cube root
x3 = -27
x3 + 27 = 0
(x+3)(x2 - 3x + 9) = 0
Result
(iii) Answer
-3, -3omega, -3omega2
Given
(iv) Cube roots of 64: Let x be the cube root
x3 = 64
Result
(iv) Answer
4, 4omega, 4omega2
Given
(v) Cube roots of -125: Let x be the cube root
x3 = -125
Result
(v) Answer
-5, -5omega, -5omega2
2.Find the four fourth roots of: (i) 16, (ii) 81, (iii) 625. Also show that their sum is zero in each case.
Given
(i) Fourth roots of 16: Let x be the fourth root
x4 = 16
x4 - 16 = 0
(x2)2 - 42 = 0
(x2 - 4)(x2 + 4) = 0
x2 = 4 or x2 = -4
x = ± 2 or x = ± 2i
Result
(i) Answer
± 2, ± 2i
Given
(ii) Fourth roots of 81: Let x be the fourth root
x4 = 81
x2 = ± 9
x = ± 3 or x = ± 3i
Result
(ii) Answer
± 3, ± 3i
Given
(iii) Fourth roots of 625: Let x be the fourth root
x4 = 625
x2 = ± 25
x = ± 5 or x = ± 5i
Result
(iii) Answer
± 5, ± 5i
4.Evaluate: (i) ((−1+√3i)/2)⁷ + ((−1−√3i)/2)⁷, (ii) (−1+√3i)⁵ + (−1−√3i)⁵
Given
(i) Let ω = (−1+√3i)/2, a primitive cube root of unity
omega3 = 1, omega2 = frac{-1-sqrt{3}i}{2}
Rewrite using ω
= omega7 + (omega2)7
Reduce exponents using ω³ = 1
= omega6 · omega + omega12 · omega2 = (omega3)2 · omega + (omega3)4 · omega2
= omega + omega2
Since 1 + ω + ω² = 0 for primitive cube root of unity
= -1
Result
(i) Answer
-1
Given
(ii) Note that −1+√3i = 2ω
2omega = -1 + sqrt{3}i, quad 2omega2 = -1 - sqrt{3}i
Rewrite using ω
= (2omega)5 + (2omega2)5
= 32omega5 + 32(omega2)5 = 32omega5 + 32omega10
Reduce exponents using ω³ = 1
= 32omega3 · omega2 + 32(omega3)3 · omega = 32omega2 + 32omega
= 32(omega2 + omega) = 32(omega + omega2)
= 32(-1) = -32
Result
(ii) Answer
-32
7.Evaluate ∑(k=0 to 5) ω^(2k), where ω is an imaginary cube root of unity
Given
Sum ω² to the powers 0, 2, 4, 6, 8, 10
sumk=05 omega2k = omega0 + omega2 + omega4 + omega6 + omega8 + omega10
Note that ω³ = 1
= 1 + omega2 + omega4 + omega6 + omega8 + omega10
Reduce exponents modulo 3
= 1 + omega2 + omega + 1 + omega2 + omega
Group and simplify
= 2(1 + omega + omega2)
Since 1 + ω + ω² = 0 for any primitive cube root of unity
= 2(0) = 0
Result
Answer