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

Linear Algebra Chapter 9: Complex Scalars Section 9.1. Algebra of Complex Numbers—Proofs of Theorems December 23, 2018 () Linear Algebra December 23, 2018 1 / 7

Table of contents Page 463 Number 7(b) 1 Theorem 9.1(4). Properties of Conjugation in C 2 Page 463 Number 20 3 Page 464 Number 28(a) 4 () 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 + 4 i . Find z / w . Solution. We have 1 | w | 2 = 3 − 4 i 3 2 + 4 2 = 3 25 − 4 w w = 25 i , so w = (3 + i )(3 − 4 i ) z = 9 − 12 i + 3 i + 4 13 − 9 i = . 25 25 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 + 4 i . Find z / w . Solution. We have 1 | w | 2 = 3 − 4 i 3 2 + 4 2 = 3 25 − 4 w w = 25 i , so w = (3 + i )(3 − 4 i ) z = 9 − 12 i + 3 i + 4 13 − 9 i = . 25 25 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 ) / ( c 2 + d 2 ), so w = ( a + bi )( c − di ) = ac − adi + bci + bd = ( ac + bd ) + ( bc − ad ) i z , c 2 + d 2 c 2 + d 2 c 2 + d 2 () 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 ) / ( c 2 + d 2 ), so w = ( a + bi )( c − di ) = ac − adi + bci + bd = ( ac + bd ) + ( bc − ad ) i z , c 2 + d 2 c 2 + d 2 c 2 + d 2 so that � z = ( ac + bd ) + ( bc − ad ) i = ac + adi − bci + bd � c 2 + d 2 c 2 + d 2 w = ( a − bi )( c + di ) 1 / w = z z = w . c 2 + d 2 () 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 ) / ( c 2 + d 2 ), so w = ( a + bi )( c − di ) = ac − adi + bci + bd = ( ac + bd ) + ( bc − ad ) i z , c 2 + d 2 c 2 + d 2 c 2 + d 2 so that � z = ( ac + bd ) + ( bc − ad ) i = ac + adi − bci + bd � c 2 + d 2 c 2 + d 2 w = ( a − bi )( c + di ) 1 / w = z z = w . c 2 + d 2 () 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 � � π 3 + 2 k π � � π 3 + 2 k π �� ( − 27) 1 / 3 cos + i sin 3 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 � � π 3 + 2 k π � � π 3 + 2 k π �� ( − 27) 1 / 3 cos + i sin 3 3 for k = 0 , 1 , 2. That is, the roots are √ � � � π � π 1 3 � � �� 3 cos + i sin = 3 2 + i , 3 3 3 3 (cos ( π ) + i sin ( π )) = − 3 , √ � � � � 5 π � � 5 π �� 1 3 3 cos + i sin = 3 2 − i . 3 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 � � π 3 + 2 k π � � π 3 + 2 k π �� ( − 27) 1 / 3 cos + i sin 3 3 for k = 0 , 1 , 2. That is, the roots are √ � � � π � π 1 3 � � �� 3 cos + i sin = 3 2 + i , 3 3 3 3 (cos ( π ) + i sin ( π )) = − 3 , √ � � � � 5 π � � 5 π �� 1 3 3 cos + i sin = 3 2 − i . 3 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 2! + x 3 3! + x 4 1 + x + x e x = 4! + · · · x − x 3 3! + x 5 5! − x 7 7! + x 9 sin x = 9! + · · · 1 − x 2 2! + x 4 4! − x 6 6! + x 8 cos x = 8! + · · · . Proceed formally to show that e i θ − 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 ∞ ( i θ ) n e i θ = � n ! n =0 ∞ ( i θ ) n ∞ ( i θ ) n ∞ ( i θ ) n ∞ ( i θ ) n � � � � = + + + n ! n ! n ! n ! n =0 n =0 n =0 n =0 n ≡ 0( mod 4) n ≡ 1( mod 4) n ≡ 2( mod 4) n ≡ 3( mod 4) since the series converges absolutely ∞ ( θ ) n ∞ i θ n ∞ − θ n ∞ − i θ n � � � � � � � � = n ! + n ! + + n ! n ! n =0 n =0 n =0 n =0 n ≡ 0( mod 4) n ≡ 1( mod 4) n ≡ 2( mod 4) n ≡ 3( mod 4) ∞ ( − 1) k θ 2 k ∞ θ 2 k +1 � � ( − 1) k +1 = (2 k )! + i (2 k + 1)! = cos θ + i sin θ. k =0 k =0 � () Linear Algebra December 23, 2018 7 / 7