Math 211 Math 211 Lecture #18 October 31, 2000 2 Complex Numbers - PDF document

1 Math 211 Math 211 Lecture #18 October 31, 2000 2 Complex Numbers Complex Numbers A complex number is one of the form z = x + iy , where x and y are real numbers. • i 2 = − 1 . • x is the real part of z ; x = Re z . • y is the imaginary part of z ; y = Im z . ⋄ The imaginary part is a real number. • Addition and multiplication. 3 Complex Conjugate Complex Conjugate The conjugate of z = x + iy is z = x − iy. • z = z ⇔ z is a real number. • x = Re z = z + z 2 • y = Im z = z − z 2 i • z + w = z + w ; z − w = z − w � z = z � • zw = z · w ; w w return 1 John C. Polking

4 Absolute Value Absolute Value The absolute value of z = x + iy is the real x 2 + y 2 . � number | z | = • z · z = | z | 2 = x 2 + y 2 . • | zw | = | z || w | � z � = | z | � � • � � w | w | 5 Quotients Quotients • Reciprocal of z = x + iy 1 z = 1 z · z z = z z zz = | z | 2 . x + iy = x − iy 1 x 2 + y 2 • Quotient z/w w = z · 1 z w = zw | w | 2 6 Geometric Representation Geometric Representation • Complex plane. • Addition • Polar representation z = r [cos θ + i sin θ ] . ⋄ θ is the argument of z : tan θ = y/x. ⋄ r = | z | . • Euler’s formula e iθ = cos θ + i sin θ. ⋄ z = | z | e iθ • Multiplication 2 John C. Polking

7 Complex Exponential Complex Exponential For z = x + iy define e z = e x + iy = e x · e iy = e x [cos y + i sin y ] . Properties: • e z + w = e z · e w ; e z − w = e z · e − w = e z /e w • e z = e z • | e z | = e Re z • If λ is a complex number, then d dte λt = λe λt 8 Complex Matrices Complex Matrices Matrices (or vectors) with complex entries inherit many of the properties of complex numbers. • M = A + iB where A = Re M and B = Im M are real matrices. • M = A − iB ; M = M ⇔ M is real. • Re M = 1 Im M = 1 2 ( M + M ) ; 2 i ( M − M ) • M + N = M + N • M z = M z Conjugate return 9 Consequences Consequences A a real matrix; complex eigenvalue λ ; associated eigenvector w , so A w = λ w • A w = A w = A w . • λ w = λ w . • ⇒ A w = λ w ⇒ λ is an eigenvalue of A with associated eigenvector w • λ � = λ ⇒ w and w are linearly independent. Complex Matrices return 3 John C. Polking

10 Solutions Solutions z ( t ) = e λt w z ( t ) = e λt w . and • z and z are linearly independent complex solutions to x ′ = A x . • z ( t ) = x ( t ) + i y ( t ) & z ( t ) = x ( t ) − i y ( t ) • x ( t ) = 1 2 ( z ( t )+ z ( t )) & y ( t ) = 1 2 i ( z ( t ) − z ( t )) • x ( t ) and y ( t ) are real valued solutions. • x ( t ) and y ( t ) are linearly independent. Complex Matrices Consequences 11 Example Example � − 5 20 � x ′ = A x where A = − 2 7 • p ( λ ) = λ 2 − 2 λ + 5 . • Eigenvalue: λ = 1 + 2 i � 3 − i � • Eigenvector: w = 1 return 12 Complex Solutions Complex Solutions � 3 − i � z ( t ) = e λt w = e (1+2 i ) t 1 � 3 + i � z ( t ) = e λt w = e (1 − 2 i ) t 1 System return 4 John C. Polking

13 Real Solutions Real Solutions � 3 cos 2 t + sin 2 t � x ( t ) = Re( z ( t )) = e t cos 2 t � 3 sin 2 t − cos 2 t � y ( t ) = Im( z ( t )) = e t sin 2 t return Complex 14 Initial Value Problem Initial Value Problem Solve � − 5 20 � x ′ = A x where A = − 2 7 with the intial condition � 5 � x (0) = . 3 15 Initial Value Problem Initial Value Problem Solution is � 3 cos 2 t + sin 2 t � u ( t ) = 3 e t cos 2 t � 3 sin 2 t − cos 2 t � + 4 e t sin 2 t � 5 cos 2 t + 15 sin 2 t � = e t 3 cos 2 t + 4 sin 2 t Real solutions 5 John C. Polking

16 Summary Summary Suppose A is a real 2 × 2 matrix with • complex conjugate eigenvalues λ and λ , and • associated nonzero eigenvectors w and w . Then • z ( t ) = e λt w and z ( t ) = e λt w form a complex valued fundamental set of solutions, and • x ( t ) = Re( z ( t )) and y ( t ) = Im( z ( t )) form a real valued fundamental set of solutions. return 17 Double Real Root Double Real Root √ T 2 − 4 D λ = T ± = T 2 . 2 • T 2 − 4 D = 0 • Eigenspace has dimension 2: ⇒ A = λI . • Every vector is an eigenvector. Every solution has the form x ( t ) = e λt v . return 18 Double Real Root Double Real Root • Eigenspace has dimension 1. • Standard procedure gives only one solution. If v 1 � = 0 is an eigenvector, then x 1 ( t ) = e λt v 1 is a solution. • The solution to the initial value problem x ′ = A x with x (0) = x 0 is x ( t ) = e λt [ I + t ( A − λI )] x 0 Degenerate return 6 John C. Polking

19 Example Example � 1 9 � x ′ = A x where A = − 1 − 5 • p ( λ ) = λ 2 + 4 λ + 4 = ( λ + 2) 2 ; λ = − 2 � 3 � − 3 9 � � • A − λI = ; v 1 = − 1 − 3 1 • One solution � − 3 � x 1 ( t ) = e λt v 1 = e − 2 t . 1 return 20 Example (cont.) Example (cont.) � 1 + 3 t 9 t � I + t ( A − λI ) = − t 1 − 3 t • Solution with initial value x 0 is � 1 + 3 t 9 t � x ( t ) = e − 2 t x 0 . − t 1 − 3 t • Easy fundamental set of solutions � 1 + 3 t � � 9 t � y 1 ( t ) = e − 2 t & y 2 ( t ) = e − 2 t . − t 1 − 3 t Solution Example 21 Example (cont.) Example (cont.) • Fundamental set including x 1 ( t ) ⋄ Find v 2 with ( A − λI ) v 2 = v 1 . � − 1 � v 2 = 0 ⋄ x 2 ( t ) = e λt [ v 2 + t v 1 ] �� − 1 � − 3 � �� = e − 2 t + t 0 1 Solution Example 7 John C. Polking

22 Summary Summary Suppose A is a real 2 × 2 matrix with • a double real eigenvalue λ , and • the dimension of the eigenspace is one. If v 1 is an eigenvector, and if v 2 satisfies ( A − λI ) v 2 = v 1 , then • x 1 ( t ) = e λt v 1 and x 2 ( t ) = e λt [ v 2 + t v 1 ] form a fundamental set of solutions. • the general solution is x ( t ) = e λt [( C 1 + C 2 t ) v 1 + C 2 v 2 ] . 8 John C. Polking