Linear Algebra Chapter 9: Complex Scalars Section 9.1. Algebra of - - PowerPoint PPT Presentation

Linear Algebra

December 23, 2018 Chapter 9: Complex Scalars Section 9.1. Algebra of Complex Numbers—Proofs of Theorems

() Linear Algebra December 23, 2018 1 / 7

Table of contents

1

Page 463 Number 7(b)

2

Theorem 9.1(4). Properties of Conjugation in C

3

Page 463 Number 20

4

Page 464 Number 28(a)

() Linear Algebra December 23, 2018 2 / 7

Page 463 Number 7(b)

Page 463 Number 7(b)

Page 463 Number 7(b). Let z = 3 + i and w = 3 + 4i. Find z/w.

• Solution. We have

1 w = w |w|2 = 3 − 4i 32 + 42 = 3 25 − 4 25i, so z w = (3 + i)(3 − 4i) 25 = 9 − 12i + 3i + 4 25 = 13 − 9i 25 .

() Linear Algebra December 23, 2018 3 / 7

Page 463 Number 7(b)

Page 463 Number 7(b)

Page 463 Number 7(b). Let z = 3 + i and w = 3 + 4i. Find z/w.

• Solution. We have

1 w = w |w|2 = 3 − 4i 32 + 42 = 3 25 − 4 25i, so z w = (3 + i)(3 − 4i) 25 = 9 − 12i + 3i + 4 25 = 13 − 9i 25 .

() Linear Algebra December 23, 2018 3 / 7

Theorem 9.1(4). Properties of Conjugation in C

Theorem 9.1(4))

Theorem 9.1. Properties of Conjugation in C. Let z = a + bi and w = c + di be complex numbers. Then

• 4. z/w = z/w.
• Proof. We know a/w = (c − di)/(c2 + d2), so

z w = (a + bi)(c − di) c2 + d2 = ac − adi + bci + bd c2 + d2 = (ac + bd) + (bc − ad)i c2 + d2 ,

() Linear Algebra December 23, 2018 4 / 7

Theorem 9.1(4). Properties of Conjugation in C

Theorem 9.1(4))

Theorem 9.1. Properties of Conjugation in C. Let z = a + bi and w = c + di be complex numbers. Then

• 4. z/w = z/w.
• Proof. We know a/w = (c − di)/(c2 + d2), so

z w = (a + bi)(c − di) c2 + d2 = ac − adi + bci + bd c2 + d2 = (ac + bd) + (bc − ad)i c2 + d2 , so that z w

• = (ac + bd) + (bc − ad)i

c2 + d2 = ac + adi − bci + bd c2 + d2 = (a − bi)(c + di) c2 + d2 = z 1/w = z w .

() Linear Algebra December 23, 2018 4 / 7

Theorem 9.1(4). Properties of Conjugation in C

Theorem 9.1(4))

Theorem 9.1. Properties of Conjugation in C. Let z = a + bi and w = c + di be complex numbers. Then

• 4. z/w = z/w.
• Proof. We know a/w = (c − di)/(c2 + d2), so

z w = (a + bi)(c − di) c2 + d2 = ac − adi + bci + bd c2 + d2 = (ac + bd) + (bc − ad)i c2 + d2 , so that z w

• = (ac + bd) + (bc − ad)i

c2 + d2 = ac + adi − bci + bd c2 + d2 = (a − bi)(c + di) c2 + d2 = z 1/w = z w .

() Linear Algebra December 23, 2018 4 / 7

Page 463 Number 20

Page 463 Number 20

Page 463 Number 20. Find the three cube roots of −27.

• Solution. We have z = −27 = 27(cos π + i sin π). So the 3 cube roots are

(−27)1/3

• cos

π 3 + 2kπ 3

• + i sin

π 3 + 2kπ 3

• for k = 0, 1, 2.

() Linear Algebra December 23, 2018 5 / 7

Page 463 Number 20

Page 463 Number 20

Page 463 Number 20. Find the three cube roots of −27.

• Solution. We have z = −27 = 27(cos π + i sin π). So the 3 cube roots are

(−27)1/3

• cos

π 3 + 2kπ 3

• + i sin

π 3 + 2kπ 3

• for k = 0, 1, 2. That is, the roots are

3

• cos

π 3

• + i sin

π 3

• =

3

• 1

2 + i √ 3 3

• ,

3 (cos (π) + i sin (π)) = −3, 3

• cos

5π 3

• + i sin

5π 3

• =

3

• 1

2 − i √ 3 3

• .

() Linear Algebra December 23, 2018 5 / 7

Page 463 Number 20

Page 463 Number 20

Page 463 Number 20. Find the three cube roots of −27.

• Solution. We have z = −27 = 27(cos π + i sin π). So the 3 cube roots are

(−27)1/3

• cos

π 3 + 2kπ 3

• + i sin

π 3 + 2kπ 3

• for k = 0, 1, 2. That is, the roots are

3

• cos

π 3

• + i sin

π 3

• =

3

• 1

2 + i √ 3 3

• ,

3 (cos (π) + i sin (π)) = −3, 3

• cos

5π 3

• + i sin

5π 3

• =

3

• 1

2 − i √ 3 3

• .

() Linear Algebra December 23, 2018 5 / 7

Page 464 Number 28(a)

Page 464 Number 28(a)

Page 464 Number 28(a). In calculus it is shown that ex = 1 + x + x 2! + x3 3! + x4 4! + · · · sin x = x − x3 3! + x5 5! − x7 7! + x9 9! + · · · cos x = 1 − x2 2! + x4 4! − x6 6! + x8 8! + · · · . Proceed formally to show that eiθ − cos θ + i sin θ. This is Euler’s Formula. (The “formal” assumption you need is that each of the series converge

• absolutely. This allows you to rearrange the series without affecting their

limits.)

() Linear Algebra December 23, 2018 6 / 7

Page 464 Number 28(a)

Page 464 Number 28(a) (continued)

• Solution. From the series representations, we have

eiθ =

∞

• n=0

(iθ)n n! =

∞

• n=0

n≡0(mod 4)

(iθ)n n! +

∞

• n=0

n≡1(mod 4)

(iθ)n n! +

∞

• n=0

n≡2(mod 4)

(iθ)n n! +

∞

• n=0

n≡3(mod 4)

(iθ)n n! since the series converges absolutely =

∞

• n=0

n≡0(mod 4)

(θ)n n! +

∞

• n=0

n≡1(mod 4)

i θn n! +

∞

• n=0

n≡2(mod 4)

• −θn

n!

• +

∞

• n=0

n≡3(mod 4)

• −i θn

n!

• =

∞

• k=0

(−1)k θ2k (2k)! + i

∞

• k=0

(−1)k+1 θ2k+1 (2k + 1)! = cos θ + i sin θ.

• ()

Linear Algebra December 23, 2018 7 / 7