Powers and Roots of Complex numbers

Operations on Complex Numbers

Mathematics 4

November 29, 2011

Mathematics 4 () Operations on Complex Numbers November 29, 2011 1 / 18

Review of Multiplication of Complex Numbers

Find the product of 4 + 4i and −2− 3i

1. Multiply Algebraically

(4 + 4i)(−2− 3i) = −8− 12i− 8i− 12i2

= −8− 20i+ 12

= 4− 20i



Find the product of 4 + 4i and −2− 3i

2. Multiply in their polar forms

(4 + 4i)(−2− 3i) = (4√2 cis 45o) · (

√13 cis 236.31o)

= (4√2 ·√13) cis(45 + 236.31)o

= 4√26 cis 281.31o

= 4− 20i



Rule for Multiplication of Complex Numbers in Polar Form

Given:

z1 = r1 cisα z2 = r2 cisβ

z1 · z2 = (r1 · r2) cis(α+ β)



Rule for Multiplication of Complex Numbers in Polar Form

Given:

z1 = r1 cisα z2 = r2 cisβ

z1 · z2 = (r1 · r2) cis(α+ β)


Raising Complex Numbers to a Power

Given: z = r cis θ

z0 =

1z1 = r cis θz2 = (r cis θ) · (r cis θ) = r2 cis 2θz3 = (r2 cis 2θ) · (r cis θ) = r3 cis 3θ



Given: z = r cis θ

z0 = 1

z1 = r cis θz2 = (r cis θ) · (r cis θ) = r2 cis 2θz3 = (r2 cis 2θ) · (r cis θ) = r3 cis 3θ



Given: z = r cis θ

z0 = 1z1 =

r cis θz2 = (r cis θ) · (r cis θ) = r2 cis 2θz3 = (r2 cis 2θ) · (r cis θ) = r3 cis 3θ



Given: z = r cis θ

z0 = 1z1 = r cis θ

z2 = (r cis θ) · (r cis θ) = r2 cis 2θz3 = (r2 cis 2θ) · (r cis θ) = r3 cis 3θ



Given: z = r cis θ

z0 = 1z1 = r cis θz2 =

(r cis θ) · (r cis θ) = r2 cis 2θz3 = (r2 cis 2θ) · (r cis θ) = r3 cis 3θ



Given: z = r cis θ

z0 = 1z1 = r cis θz2 = (r cis θ) · (r cis θ)

= r2 cis 2θz3 = (r2 cis 2θ) · (r cis θ) = r3 cis 3θ



Given: z = r cis θ

z0 = 1z1 = r cis θz2 = (r cis θ) · (r cis θ) = r2 cis 2θ

z3 = (r2 cis 2θ) · (r cis θ) = r3 cis 3θ



Given: z = r cis θ

z0 = 1z1 = r cis θz2 = (r cis θ) · (r cis θ) = r2 cis 2θz3 =

(r2 cis 2θ) · (r cis θ) = r3 cis 3θ



Given: z = r cis θ

z0 = 1z1 = r cis θz2 = (r cis θ) · (r cis θ) = r2 cis 2θz3 = (r2 cis 2θ) · (r cis θ)

= r3 cis 3θ



Given: z = r cis θ

z0 = 1z1 = r cis θz2 = (r cis θ) · (r cis θ) = r2 cis 2θz3 = (r2 cis 2θ) · (r cis θ) = r3 cis 3θ



Given: z = r cis θ

z0 = 1z1 = r cis θz2 = (r cis θ) · (r cis θ) = r2 cis 2θz3 = (r2 cis 2θ) · (r cis θ) = r3 cis 3θ



De Moivre’s Theorem

(r cis θ)n = rn cis(n · θ)



Example 1: Find (√2 cis 20o)10

(√2 cis 20o)10 = (

√2)10 cis(10 · 20)o

(√2 cis 20o)10 = 32 cis 200o



Example 1: Find (√2 cis 20o)10

(√2 cis 20o)10 = (

√2)10 cis(10 · 20)o

(√2 cis 20o)10 = 32 cis 200o


De Moivre’s TheoremExample 1: Find (

√2 cis 20o)10

(√2 cis 20o)0 = 1 cis 0



√2 cis 20o)10

(√2 cis 20o)1 =

√2 cis 20o



√2 cis 20o)10

(√2 cis 20o)2 = 2 cis 400



√2 cis 20o)10

(√2 cis 20o)3 = 2

√2 cis 60o



√2 cis 20o)10

(√2 cis 20o)4 = 4 cis 80o



√2 cis 20o)10

(√2 cis 20o)5 = 4

√2 cis 100o



√2 cis 20o)10

(√2 cis 20o)6 = 8 cis 120o



√2 cis 20o)10

(√2 cis 20o)7 = 8

√2 cis 140o



√2 cis 20o)10

(√2 cis 20o)8 = 16 cis 160o



√2 cis 20o)10

(√2 cis 20o)9 = 16

√2 cis 180o



√2 cis 20o)10

(√2 cis 20o)10 = 32 cis 200o



Example 2: Find (3 cis 120o)5

(3 cis 120o)5 = 35 cis(5 · 120)o

(3 cis 120o)5 = 243 cis 600o = 243 cis 240o



Example 2: Find (3 cis 120o)5

(3 cis 120o)5 = 35 cis(5 · 120)o(3 cis 120o)5 = 243 cis 600o = 243 cis 240o


De Moivre’s TheoremExample 2: Find (3 cis 120o)5

(3 cis 120o)0 = 1 cis 0



(3 cis 120o)1 = 3 cis 120o



(3 cis 120o)2 = 9 cis 240o



(3 cis 120o)3 = 27 cis 360o



(3 cis 120o)4 = 81 cis 480o



(3 cis 120o)5 = 243 cis 600o = 243 cis 240o



Example 3: Find (1− i)8

(1− i)8 = (√2 cis 315o)8

(√2 cis 315o)8 = 16 cis 2520o = 16 cis 0 = 16



Example 3: Find (1− i)8

(1− i)8 = (√2 cis 315o)8

(√2 cis 315o)8 = 16 cis 2520o = 16 cis 0 = 16


De Moivre’s TheoremExample 3: Find (1− i)8

(1− i)0 = (√2 cis(−45)o)0 = 1 cis 0



(1− i)1 = (√2 cis(−45)o)1 =

√2 cis(−45)o



(1− i)2 = (√2 cis(−45)o)2 = 2 cis(−90)o



(1− i)3 = (√2 cis(−45)o)3 = 2

√2 cis(−135)o



(1− i)4 = (√2 cis(−45)o)4 = 4 cis(−180)o



(1− i)5 = (√2 cis(−45)o)5 = 4

√2 cis(−225)o



(1− i)6 = (√2 cis(−45)o)6 = 8 cis(−270)o



(1− i)7 = (√2 cis(−45)o)7 = 8

√2 cis(−315)o



(1− i)8 = (√2 cis(−45)o)8 = 16 cis(−360)o = 16 cis 0



(1− i)8 = (√2 cis(−45)o)8 = 16 cis(−360)o = 16 cis 0



De Moivre’s Theorem can be used to find the nth of a complex number:

Find the three cube roots of −2− i2√3.

We wish to find values of r and θ such that:

(r cis θ)3 = −2− i2√3

Using De Moivre’s Theorem and expressing the complex number in polarform:

r3 cis 3θ = 4 cis 240o

Therefore:

r3 = 4 and 3θ = 240o + k · 360o, k ∈ Zr = 3√4 and θ = 80o + k · 120o, k ∈ Z






(r cis θ)3 = −2− i2√3



Therefore:

r3 = 4 and 3θ = 240o + k · 360o, k ∈ Z

r = 3√4 and θ = 80o + k · 120o, k ∈ Z






(r cis θ)3 = −2− i2√3



Therefore:

r3 = 4 and 3θ = 240o + k · 360o, k ∈ Zr = 3√4 and θ = 80o + k · 120o, k ∈ Z


Finding the nth roots of complex numbers

For any complex number r cis θ and n ∈ Z+:

The nth roots of r cis θ is given by:

n√r cis θk

θk =θ + k360o

n, k = 0, 1, 2, ...(n− 1)



Example 1: Find the fourth roots of 16 cis 120o

r4 cis 4θ = 16 cis 120o

r4 = 16 and 4θ = 120o + k360o

r = 2 and θ = 30o + k90o

2 cis 30o

2 cis 120o

2 cis 210o

2 cis 300o




r4 cis 4θ = 16 cis 120o

r4 = 16 and 4θ = 120o + k360o

r = 2 and θ = 30o + k90o

2 cis 30o

2 cis 120o

2 cis 210o

2 cis 300o




r4 cis 4θ = 16 cis 120o

r4 = 16 and 4θ = 120o + k360o

r = 2 and θ = 30o + k90o

2 cis 30o

2 cis 120o

2 cis 210o

2 cis 300o




r4 cis 4θ = 16 cis 120o

r4 = 16 and 4θ = 120o + k360o

r = 2 and θ = 30o + k90o

2 cis 30o

2 cis 120o

2 cis 210o

2 cis 300o




r4 cis 4θ = 16 cis 120o

r4 = 16 and 4θ = 120o + k360o

r = 2 and θ = 30o + k90o

2 cis 30o

2 cis 120o

2 cis 210o

2 cis 300o


Finding the nth roots of complex numbersExample 1: Find the fourth roots of 16 cis 120o

(2 cis 30)0

(2 cis 120)0

(2 cis 210)0

(2 cis 300)0



(2 cis 30)1

(2 cis 120)1

(2 cis 210)1

(2 cis 300)1



(2 cis 30)2

(2 cis 120)2

(2 cis 210)2

(2 cis 300)2



(2 cis 30)3

(2 cis 120)3

(2 cis 210)3

(2 cis 300)3



(2 cis 30)4

(2 cis 120)4

(2 cis 210)4

(2 cis 300)4



(2 cis 30)4

(2 cis 120)4

(2 cis 210)4

(2 cis 300)4



Example 2: Find the cube roots of −8

r3 cis 3θ = −8 = 8 cis 180o

r3 = 8 and 3θ = 180o + k360o

r = 2 and θ = 60o + k120o

2 cis 60o

2 cis 180o = −22 cis 300o



Example 2: Find the cube roots of −8r3 cis 3θ = −8 = 8 cis 180o

r3 = 8 and 3θ = 180o + k360o

r = 2 and θ = 60o + k120o

2 cis 60o

2 cis 180o = −22 cis 300o




r3 = 8 and 3θ = 180o + k360o

r = 2 and θ = 60o + k120o

2 cis 60o

2 cis 180o = −22 cis 300o




r3 = 8 and 3θ = 180o + k360o

r = 2 and θ = 60o + k120o

2 cis 60o

2 cis 180o = −22 cis 300o




r3 = 8 and 3θ = 180o + k360o

r = 2 and θ = 60o + k120o

2 cis 60o

2 cis 180o = −22 cis 300o


Finding the nth roots of complex numbersExample 2: Find the cube roots of −8

(2 cis 60)0 (2 cis 180)0 (2 cis 300)0



(2 cis 60)1 (2 cis 180)1 (2 cis 300)1



(2 cis 60)2 (2 cis 180)2 (2 cis 300)2



(2 cis 60)3 (2 cis 180)3 (2 cis 300)3


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Powers and Roots of Complex numbers