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2025lecture_11 - Chapter 11 Complex Numbers And Complex...

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Unformatted text preview: Chapter 11 Complex Numbers And Complex Vector Spaces Complex Vector Spaces Integral Transforms - p. 54/78 Introduction Complex Spaces • Introduction • Complex numbers • Operations on Up to now we have only considered vector spaces over the field of real numbers. For the Fourier transform we need vector spaces where we can multiply vectors by complex numbers. C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Complex Vector Spaces Integral Transforms - p. 55/78 Introduction Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector Up to now we have only considered vector spaces over the field of real numbers. For the Fourier transform we need vector spaces where we can multiply vectors by complex numbers. If x ∈ R, then know that x2 ≥ 0 and x2 = 0 only if x = 0. Therefore there is no real number such that x2 = −1, or: There is no real solution to the equation x2 + 1 = 0. spaces • Complex inner product • Complex norm Complex Vector Spaces Integral Transforms - p. 55/78 Introduction Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Up to now we have only considered vector spaces over the field of real numbers. For the Fourier transform we need vector spaces where we can multiply vectors by complex numbers. If x ∈ R, then know that x2 ≥ 0 and x2 = 0 only if x = 0. Therefore there is no real number such that x2 = −1, or: There is no real solution to the equation x2 + 1 = 0. More generally let us look at the equation x2 + bx + c = 0 By completing the square we get b1 x=− ± 22 Complex Vector Spaces b 2 − 4c Integral Transforms - p. 55/78 Introduction There are now 3 possibilities: Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Complex Vector Spaces Integral Transforms - p. 56/78 Introduction Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector There are now 3 possibilities: 1. b2 − 4c > 0. Then the quadratic equation above gives two solutions b1 x=− + 22 and b 2 − 4c b1 x=− − 22 b 2 − 4c spaces • Complex inner product • Complex norm Complex Vector Spaces Integral Transforms - p. 56/78 Introduction Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Complex Vector Spaces There are now 3 possibilities: 1. b2 − 4c > 0. Then the quadratic equation above gives two solutions b1 x=− + 22 and b 2 − 4c b1 x=− − 22 b 2 − 4c 2. b2 − 4c = 0. Then we have one solution x=− b 2 Integral Transforms - p. 56/78 Introduction Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm There are now 3 possibilities: 1. b2 − 4c > 0. Then the quadratic equation above gives two solutions b1 x=− + 22 and b 2 − 4c b1 x=− − 22 b 2 − 4c 2. b2 − 4c = 0. Then we have one solution x=− b 2 3. b2 − 4c < 0. Then there is no real solution to quadratic equation x2 + bx + c = 0. Complex Vector Spaces Integral Transforms - p. 56/78 Complex numbers We now introduce a new number i = Complex Spaces • Introduction • Complex numbers • Operations on √ −1 such that i2 = −1 C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Complex Vector Spaces Integral Transforms - p. 57/78 Complex numbers We now introduce a new number i = Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Complex Vector Spaces √ −1 such that i2 = −1 The complex numbers are all expressions of the form z = x + iy , x, y ∈ R. The set of complex numbers is denoted by C. We say that x = ℜz is the real part of z and y = ℑz is the imaginary part of z . Integral Transforms - p. 57/78 Complex numbers We now introduce a new number i = Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Complex Vector Spaces √ −1 such that i2 = −1 The complex numbers are all expressions of the form z = x + iy , x, y ∈ R. The set of complex numbers is denoted by C. We say that x = ℜz is the real part of z and y = ℑz is the imaginary part of z . Recall that the set of real numbers can be thought of as a line, the real line. To picture the set of complex numbers we use the plane. A vector v = (x, y ) corresponds to the complex number z = x + iy . Integral Transforms - p. 57/78 Addition and multiplication on C Complex Spaces • Introduction • Complex numbers • Operations on C The addition of two complex numbers z = x + iy and w = s + it then corresponds to the addition of the corresponding vectors. Thus (x + iy ) + (s + it) = (x + s) + i(y + t). • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Complex Vector Spaces Integral Transforms - p. 58/78 Addition and multiplication on C Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm The addition of two complex numbers z = x + iy and w = s + it then corresponds to the addition of the corresponding vectors. Thus (x + iy ) + (s + it) = (x + s) + i(y + t). To find out what the product of z and w is, we use the familiar rules along with i2 = −1. Thus (x + iy ) · (s + it) = xs + xit + iys + iyit = xs + i2 yt + i(xt + ys) = (xs − yt) + i(xt + ys). Complex Vector Spaces Integral Transforms - p. 58/78 Conjugate and absolute value of z Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector Before we find the inverse -or reciprocal- of z = x + iy , we need to introduce the complex conjugate: x + iy = x − iy Thus, complex conjugate corresponds to a reflection around the x-axis. spaces • Complex inner product • Complex norm Complex Vector Spaces Integral Transforms - p. 59/78 Conjugate and absolute value of z Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Before we find the inverse -or reciprocal- of z = x + iy , we need to introduce the complex conjugate: x + iy = x − iy Thus, complex conjugate corresponds to a reflection around the x-axis. Now multiply z by z : z·z = (x + iy )(x − iy ) = x2 − (iy )2 = x2 + y 2 . Complex Vector Spaces Integral Transforms - p. 59/78 Conjugate and absolute value of z Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector Before we find the inverse -or reciprocal- of z = x + iy , we need to introduce the complex conjugate: x + iy = x − iy Thus, complex conjugate corresponds to a reflection around the x-axis. Now multiply z by z : spaces • Complex inner product • Complex norm z·z = (x + iy )(x − iy ) = x2 − (iy )2 = x2 + y 2 . √ Thus z z := |z | is the length of the vector (x, y ) or the absolute value of the complex number z . Complex Vector Spaces Integral Transforms - p. 59/78 Reciprocal of z Lemma. Let z Complex Spaces • Introduction • Complex numbers • Operations on = x + iy be a complex number with |z | = 0. Then 1 z x − iy = 2= 2 . z |z | x + y2 C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Complex Vector Spaces Integral Transforms - p. 60/78 Reciprocal of z Lemma. Let z Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector = x + iy be a complex number with |z | = 0. Then 1 z x − iy = 2= 2 . z |z | x + y2 x−iy Proof. We have z x2 +y 2 = (x+iy )(x−iy ) x2 +y 2 = x2 +y 2 x2 +y 2 =1 spaces • Complex inner product • Complex norm Complex Vector Spaces Integral Transforms - p. 60/78 Reciprocal of z Lemma. Let z Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Complex Vector Spaces = x + iy be a complex number with |z | = 0. Then 1 z x − iy = 2= 2 . z |z | x + y2 x−iy Proof. We have z x2 +y 2 = (x+iy )(x−iy ) x2 +y 2 = x2 +y 2 x2 +y 2 =1 Examples • (2 + 3i) + (5 − 2i) = 7 + i. Integral Transforms - p. 60/78 Reciprocal of z Lemma. Let z Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Complex Vector Spaces = x + iy be a complex number with |z | = 0. Then 1 z x − iy = 2= 2 . z |z | x + y2 x−iy Proof. We have z x2 +y 2 = (x+iy )(x−iy ) x2 +y 2 = x2 +y 2 x2 +y 2 =1 Examples • (2 + 3i) + (5 − 2i) = 7 + i. • (2 − 3i)(1 + i) = (2 + 3) + (−3 + 2)i = 5 − i. Integral Transforms - p. 60/78 Reciprocal of z Lemma. Let z Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector = x + iy be a complex number with |z | = 0. Then 1 z x − iy = 2= 2 . z |z | x + y2 x−iy Proof. We have z x2 +y 2 = (x+iy )(x−iy ) x2 +y 2 = x2 +y 2 x2 +y 2 =1 Examples Complex Vector Spaces • (2 + 3i) + (5 − 2i) = 7 + i. • (2 − 3i)(1 + i) = (2 + 3) + (−3 + 2)i = 5 − i. • spaces • Complex inner product • Complex norm 1 2+i = 2−i 5 = 2 5 − 1 i. 5 Integral Transforms - p. 60/78 Reciprocal of z Lemma. Let z Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector = x + iy be a complex number with |z | = 0. Then 1 z x − iy = 2= 2 . z |z | x + y2 x−iy Proof. We have z x2 +y 2 = (x+iy )(x−iy ) x2 +y 2 = x2 +y 2 x2 +y 2 =1 Examples • (2 + 3i) + (5 − 2i) = 7 + i. • (2 − 3i)(1 + i) = (2 + 3) + (−3 + 2)i = 5 − i. • spaces • Complex inner product • Complex norm 1 2+i = • 2+3i 1+5i Complex Vector Spaces 2−i 5 = = 2 5 − 1 i. 5 (2+3i)(1−5i) 26 = (2+15)+(3−10)i 26 = 17−7i 26 . Integral Transforms - p. 60/78 The complex exponential function Complex Spaces • Introduction • Complex numbers • Operations on The idea behind the Fourier transform is to represent a function (or a signal) in the frequency domain using the complex exponential function. C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Complex Vector Spaces Integral Transforms - p. 61/78 The complex exponential function Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Complex Vector Spaces The idea behind the Fourier transform is to represent a function (or a signal) in the frequency domain using the complex exponential function. Definition. Let {zn }n∈N be a sequence of complex numbers. We say that zn converges to the complex number w if for all ǫ N ∈ N such that for all n w. > 0 there exists a |zn − w| < ǫ ≥ N . We write zn → w or limn→∞ zn = w if zn converges to Integral Transforms - p. 61/78 The complex exponential function Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm The idea behind the Fourier transform is to represent a function (or a signal) in the frequency domain using the complex exponential function. Definition. Let {zn }n∈N be a sequence of complex numbers. We say that zn converges to the complex number w if for all ǫ N ∈ N such that for all n w. > 0 there exists a |zn − w| < ǫ ≥ N . We write zn → w or limn→∞ zn = w if zn converges to n zk k=0 k! . Let z ∈ C and define zn = Then it can be shown that the sequence {zn } converges. We denote the limit by ez = ∞ k=0 Complex Vector Spaces zk = lim N →∞ k! N k=0 zk . k! Integral Transforms - p. 61/78 The complex exponential function It can be shown that Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector ez+w = ez ew . In particular 1 = e−z . ez spaces • Complex inner product • Complex norm Complex Vector Spaces Integral Transforms - p. 62/78 The complex exponential function It can be shown that Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Complex Vector Spaces ez+w = ez ew . In particular 1 = e−z . ez Theorem. (The Euler formula) Let z = x + iy ∈ C. Then ez = ex (cos y + i sin y ). Integral Transforms - p. 62/78 The complex exponential function Examples Complex Spaces • Introduction • Complex numbers • Operations on • eπi = cos π + i sin π = −1. C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Complex Vector Spaces Integral Transforms - p. 63/78 The complex exponential function Examples Complex Spaces • Introduction • Complex numbers • Operations on • eπi = cos π + i sin π = −1. • ei 2 = cos π + i sin π = i. 2 2 π C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Complex Vector Spaces Integral Transforms - p. 63/78 The complex exponential function Examples Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector • eπi = cos π + i sin π = −1. • ei 2 = cos π + i sin π = i. 2 2 • e2+i 4 = e2 (cos π + i sin π ) = 4 4 π π e2 √ (1 2 + i). spaces • Complex inner product • Complex norm Complex Vector Spaces Integral Transforms - p. 63/78 The complex exponential function Examples Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Complex Vector Spaces • eπi = cos π + i sin π = −1. • ei 2 = cos π + i sin π = i. 2 2 • e2+i 4 = e2 (cos π + i sin π ) = 4 4 π π Lemma. Let z e2 √ (1 2 + i). = x + iy . Then ez = ez Integral Transforms - p. 63/78 The complex exponential function Examples Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm • eπi = cos π + i sin π = −1. • ei 2 = cos π + i sin π = i. 2 2 • e2+i 4 = e2 (cos π + i sin π ) = 4 4 π π Lemma. Let z e2 √ (1 2 + i). = x + iy . Then ez = ez Proof. ez = ex (cos y + i sin y ) = ex (cos y − i sin y ) = ex−iy = ez Complex Vector Spaces Integral Transforms - p. 63/78 Complex-valued functions Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector Let F : I −→ C, I ⊆ R an interval, be a function. Then we can write F (t) = f (t) + ig (t) where f, g : I −→ R. spaces • Complex inner product • Complex norm Complex Vector Spaces Integral Transforms - p. 64/78 Complex-valued functions Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Complex Vector Spaces Let F : I −→ C, I ⊆ R an interval, be a function. Then we can write F (t) = f (t) + ig (t) where f, g : I −→ R. The function F is continuous if and only if f and g are both continuous. In that case we have lim F (t) = t→t0 lim f (t) + ( lim g (t))i t→t0 t→t0 = f (t0 ) + g (t0 )i = F (t0 ). Integral Transforms - p. 64/78 Integration and differentiation Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector We integrate and differentiate F by integrating f (t) and g (t) (differentiating f (t) and g (t)). Thus b b F (t) dt = a b f (t) dt + a g (t) dt i a df dg dF (t) = (t) + (t) i dt dt dt spaces • Complex inner product • Complex norm Complex Vector Spaces Integral Transforms - p. 65/78 Integration and differentiation Complex Spaces • Introduction • Complex numbers • Operations on We integrate and differentiate F by integrating f (t) and g (t) (differentiating f (t) and g (t)). Thus b F (t) dt = a C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm b b f (t) dt + a g (t) dt i a df dg dF (t) = (t) + (t) i dt dt dt Example 1) 1 2t + 3t2 i dt 0 = t2 ]1 + t3 ]1 i 0 0 = 1 + i. Complex Vector Spaces Integral Transforms - p. 65/78 Integration and differentiation 2) 2π Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Complex Vector Spaces 2π (1+i)t e et cos t + et sin t dt dt = 0 0 We have 2π 2π et cos t dt 0 = cos tet ]2π + 0 = e2π − 1 − et sin t dt 0 2π et cos t dt 0 Integral Transforms - p. 66/78 Integration and differentiation 2) 2π Complex Spaces • Introduction • Complex numbers • Operations on e 0 We have 2π 2π et cos t dt 0 = cos tet ]2π + 0 = e2π − 1 − Thus 2π et cos t dt = 0 Complex Vector Spaces et cos t + et sin t dt dt = 0 C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm 2π (1+i)t et sin t dt 0 2π et cos t dt 0 1 2π (e − 1). 2 Integral Transforms - p. 66/78 Integration and differentiation 2) 2π Complex Spaces • Introduction • Complex numbers • Operations on e et cos t + et sin t dt dt = 0 0 We have C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm 2π (1+i)t 2π 2π et cos t dt 0 = cos tet ]2π + 0 = e2π − 1 − Thus 2π et cos t dt = 0 et sin t dt 0 2π et cos t dt 0 1 2π (e − 1). 2 Similarly we have 2π 0 Complex Vector Spaces 1 et sin t dt = − (e2π − 1). 2 Integral Transforms - p. 66/78 Integration and differentiation Thus, 2π Complex Spaces • Introduction • Complex numbers • Operations on (1+i)t e 0 e2π − 1 (1 − i). dt = 2 C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Complex Vector Spaces Integral Transforms - p. 67/78 Integration and differentiation Thus, 2π Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector (1+i)t e 0 e2π − 1 (1 − i). dt = 2 But we can also have used the rule at e 1 at dt = e + C a where a is any complex number, a = 0. Then spaces • Complex inner product • Complex norm Complex Vector Spaces Integral Transforms - p. 67/78 Integration and differentiation Thus, 2π Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm (1+i)t e 0 e2π − 1 (1 − i). dt = 2 But we can also have used the rule at e 1 at dt = e + C a where a is any complex number, a = 0. Then 2π e(1+i)t dt = 0 = = Complex Vector Spaces 1 (1+i)t 2π e0 1+i 1 [e2π+i2π − e0 ] 1+i 1 [e2π − 1] 1+i Integral Transforms - p. 67/78 Integration and differentiation Furthermore Complex Spaces • Introduction • Complex numbers • Operations on 1 1 = (1 − i) 1+i 2 Thus we have 2π C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector (1+i)t e 0 e2π − 1 (1 − i) dt = 2 spaces • Complex inner product • Complex norm Complex Vector Spaces Integral Transforms - p. 68/78 Integration and differentiation 1 Example. Evaluate the integral 0 (t + 2it2 )(t2 Complex Spaces • Introduction • Complex numbers • Operations on − 3it) dt. C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Complex Vector Spaces Integral Transforms - p. 69/78 Integration and differentiation 1 Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector Example. Evaluate the integral 0 (t + 2it2 )(t2 − 3it) Solution: First we have to carry out the multiplication (t + 2it2 )(t2 − 3it) dt. = t3 + 6t3 − 2it4 − 3it2 = 7t3 − (2t4 + 3t2 )i spaces • Complex inner product • Complex norm Complex Vector Spaces Integral Transforms - p. 69/78 Integration and differentiation 1 Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Complex Vector Spaces Example. Evaluate the integral 0 (t + 2it2 )(t2 − 3it) Solution: First we have to carry out the multiplication (t + 2it2 )(t2 − 3it) dt. = t3 + 6t3 − 2it4 − 3it2 = 7t3 − (2t4 + 3t2 )i Thus 1 1 2 0 2 (t + 2it )(t − 3it) dt = 0 = = 1 3 7t dt − ( 2t4 + 3t2 )i 0 2 7 41 t ]0 − [ t5 + t3 ]1 i 0 4 5 77 − i. 45 Integral Transforms - p. 69/78 Complex vector spaces Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Complex Vector Spaces The axioms for complex vector spaces are the same as those for real vector spaces except the scalars are now complex numbers. Example. Let Cn = {(z1 , . . . , zn ) : z1 , . . . , zn }. The addition is given by u = (z1 , . . . , zn ), v = (w1 , . . . , wn ). u + v = (z1 + w1 , . . . , zn + wn ) The scalar multiplication is given by λu = (λz1 , . . . , λzn ). Integral Transforms - p. 70/78 Complex vector spaces Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector The axioms for complex vector spaces are the same as those for real vector spaces except the scalars are now complex numbers. Example. Let I ⊂ R be an interval. Let V be the space of functions f : I −→ C. The addition and scalar multiplication are given by (f + g )(t) = f (t) + g (t) (λf )(t) = λf (t). spaces • Complex inner product • Complex norm Complex Vector Spaces Integral Transforms - p. 70/78 Inner product on a complex vector space Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Complex Vector Spaces Let V be a complex vector space. A map (·, ·) : V × V −→ C is called an inner product if 1. (u, u) ≥ 0 for all u ∈ V . 2. (u, u) = 0 if and only if u = 0. 3. For fixed v ∈ V , the map u → (u, v ) is linear, i.e., (λu + µw, v ) = λ(u, v ) + µ(w, v ) for all λ, µ ∈ C and all u, w ∈ V . 4. For all u, v ∈ V we have (u, v ) = (v, u). Integral Transforms - p. 71/78 Inner product on a complex vector space Lemma. Let (·, ·) be an inner product on the complex vector space V . Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector Then (v, λu + µw) = λ(u, v ) + µ(v, w) for all λ, µ ∈ C and all u, v, w ∈ V . spaces • Complex inner product • Complex norm Complex Vector Spaces Integral Transforms - p. 72/78 Inner product on a complex vector space Lemma. Let (·, ·) be an inner product on the complex vector space V . Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Complex Vector Spaces Then (v, λu + µw) = λ(u, v ) + µ(v, w) for all λ, µ ∈ C and all u, v, w ∈ V . Proof. We have (v, λu + µw) = (λu + µw, v ) = λ(u, v ) + µ(w, v ) = λ (u, v ) + µ (w, v ) = λ(v, u) + µ(v, w) Integral Transforms - p. 72/78 Examples Example (1). Let V Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Complex Vector Spaces = Cn . Define for u = (z1 , . . . , zn ) and v = (w1 , . . . , wn ) (u, v ) = z1 w1 + · · · + zn wn . Then (u, u) = z 1 z1 + · · · + z n zn = |z1 |2 + · · · + |zn |2 ≥ 0. Integral Transforms - p. 73/78 Examples Example (1). Let V Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm = Cn . Define for u = (z1 , . . . , zn ) and v = (w1 , . . . , wn ) (u, v ) = z1 w1 + · · · + zn wn . Then (u, u) = z 1 z1 + · · · + z n zn = |z1 |2 + · · · + |zn |2 ≥ 0. If (u, u) = 0, then we must have |z1 | = |z2 | = · · · |zn | = 0, so u = 0. Let y = (t1 , . . . , tn ) ∈ V and λ ∈ C. Then (u + y, v ) = (z1 + t1 )w1 + · · · + (zn + tn )wn = (z1 w1 + · · · + zn wn ) + (t1 + · · · + tn )wn = (u, v ) + (y, v ). Complex Vector Spaces Integral Transforms - p. 73/78 Examples Similarly, Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector (λu, v ) = (λz1 )w1 + · · · + (λzn )wn = λ(z1 w1 ) + · · · + λ(zn wn ) = λ(u, v ). spaces • Complex inner product • Complex norm Complex Vector Spaces Integral Transforms - p. 74/78 Examples Similarly, Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Complex Vector Spaces (λu, v ) = (λz1 )w1 + · · · + (λzn )wn = λ(z1 w1 ) + · · · + λ(zn wn ) = λ(u, v ). Finally, (u, v ) = z1 w1 + · · · + zn wn = z 1 w1 + · · · + z n wn = (v, u). Integral Transforms - p. 74/78 Examples Similarly, Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm (λu, v ) = (λz1 )w1 + · · · + (λzn )wn = λ(z1 w1 ) + · · · + λ(zn wn ) = λ(u, v ). Finally, (u, v ) = z1 w1 + · · · + zn wn = z 1 w1 + · · · + z n wn = (v, u). Example (2). Let V be the space of piecewise continuous functions f : [0, 1] −→ C. Define 1 (f, g ) = f (t)g (t) dt. 0 Complex Vector Spaces Integral Transforms - p. 74/78 Norm in a complex vector space Definition. Let V be a complex vector space with inner product (·, ·). Complex Spaces • Introduction • Complex numbers • Operations on Then the norm (or length) of a vector u u := ∈ V is defined by (u, u). C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Complex Vector Spaces Integral Transforms - p. 75/78 Norm in a complex vector space Definition. Let V be a complex vector space with inner product (·, ·). Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector Then the norm (or length) of a vector u u := Remark. Notice that ∈ V is defined by (u, u). u = 0 if and only if u = 0 and that λu = |λ| u . spaces • Complex inner product • Complex norm Complex Vector Spaces Integral Transforms - p. 75/78 Norm in a complex vector space Definition. Let V be a complex vector space with inner product (·, ·). Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm Complex Vector Spaces Then the norm (or length) of a vector u u := Remark. Notice that ∈ V is defined by (u, u). u = 0 if and only if u = 0 and that λu = |λ| u . Example 1. V = C2 and u = (1, i). Then u 2 = 1 + ii = 1 + i(−i) = 1 + 1 = 2. Integral Transforms - p. 75/78 Norm in a complex vector space Definition. Let V be a complex vector space with inner product (·, ·). Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector Then the norm (or length) of a vector u u := Remark. Notice that ∈ V is defined by (u, u). u = 0 if and only if u = 0 and that λu = |λ| u . Example 1. V = C2 and u = (1, i). Then spaces • Complex inner product • Complex norm u 2 = 1 + ii = 1 + i(−i) = 1 + 1 = 2. 2. V = C2 and u = (1 + i, 2 + 3i): u 2 = (1 + i)(1 − i) + (2 + 3i)(2 − 3i) = 1 + 1 + 4 + 9 = 15 or u = Complex Vector Spaces √ 15. Integral Transforms - p. 75/78 Norm in a complex vector space Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm 3. Let V be the space of piecewise continuous functions on [0, 1]. Let a ∈ R and f (t) = eait = cos (at) + i sin (at). Then f (t)f (t) = |f (t)|2 = (cos (at) + i sin (at))(cos (at) − i sin (at)) = (cos (at))2 + (sin (at))2 = 1. Hence, 1 f= 0 Complex Vector Spaces 1 |f (t)|2 dt = 1 dt = 1. 0 Integral Transforms - p. 76/78 Norm in a complex vector space Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm 4. Let V be the space of piecewise continuous functions on [0, 1]. Let f (t) = t + it2 . Then |f (t)|2 = (t + it2 )(t − it2 ) = t2 + t4 . Hence, 1 t2 + t4 dt = f= 0 Complex Vector Spaces 11 += 35 8 . 15 Integral Transforms - p. 77/78 Norm in a complex vector space Complex Spaces • Introduction • Complex numbers • Operations on C • Conjugate 1 •z • ez • Complex-valued • Integration • Complex vector spaces • Complex inner product • Complex norm 5. Let V be the space of piecewise continuous functions on [0, 1]. Let f (t) = t2 + 1 + 2i(t − 3). Then |f (t)|2 = (1 + t2 )2 + 4(t − 3)2 = 1 + 2t2 + t4 + 4t2 − 24t + 36 = t4 + 6t2 − 24t + 37. Hence, 1 f 2 = 0 = = Complex Vector Spaces t4 + 6t2 − 24t + 37 dt 1 + 2 − 12 + 37 5 136 . 5 Integral Transforms - p. 78/78 ...
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2025lecture_11 - Chapter 11 Complex Numbers And Complex...

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