Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Chapter 9: The Complex Numbers MAT246H1S Lec0101 Burbulla Chapter 9 Lecture Notes The Complex Numbers Winter 2017 Chapter 9 Lecture Notes The Complex Numbers MAT246H1S Lec0101 Burbulla Chapter 9: The Complex Numbers Chapter 9: The Complex Numbers 9.1: What Is A Complex Number? 9.2: The Complex Plane 9.3: The Fundamental Theorem of Algebra Chapter 9 Lecture Notes The Complex Numbers MAT246H1S Lec0101 Burbulla Chapter 9: The Complex Numbers 9.1: What Is A Complex Number? 9.2: The Complex Plane 9.3: The Fundamental Theorem of Algebra Limitations of the Real Numbers Since x 2 ≥ 0 for every real number x, there is no real number that satisfies the simple equation x 2 + 1 = 0. To this end, the number i denotes an imaginary solution to the equation x 2 + 1 = 0, that is i 2 = −1. Surprisingly, it then turns out that any polynomial equation of the form an x n + an−1 x n−1 + · · · + a2 x 2 + a1 x + a0 = 0, with ai ∈ R, will have solutions, either real, or in terms of i. Chapter 9 Lecture Notes The Complex Numbers Chapter 9: The Complex Numbers MAT246H1S Lec0101 Burbulla 9.1: What Is A Complex Number? 9.2: The Complex Plane 9.3: The Fundamental Theorem of Algebra The Quadratic Formula For example, you should be familiar with the formula for solving the quadratic equation: √ b 2 − 4ac −b ± ax 2 + bx + c = 0 ⇒ x = . 2a If b 2 − 4ac < 0, then the roots to the quadratic equation involve the square root of a negative number and i can be used to solve the equation. For example, √ −6 ± 36 − 100 x 2 + 6x + 25 = 0 ⇒ x = √2 −6 ± −64 ⇒ x= 2 ⇒ x = −3 ± 4 i Chapter 9 Lecture Notes The Complex Numbers MAT246H1S Lec0101 Burbulla Chapter 9: The Complex Numbers 9.1: What Is A Complex Number? 9.2: The Complex Plane 9.3: The Fundamental Theorem of Algebra The Set of Complex Numbers For any real numbers a, b the expression a + b i is called a complex number. Let z = a + b i, then I a is called the real part of z, sometimes denoted by Re(z), I b is called the imaginary part of z, denoted by Im(z). The set of all complex numbers is denoted by C. Complex numbers are added and multiplied with the usual operations of addition and multiplication, using the fact that i 2 = −1. Thus if a + b i and c + d i are complex numbers, we define I (a + b i) + (c + d i) = (a + c) + (b + d) i, and I (a + b i)(c + d i) = ac − bd + (ad + bc) i, as you can check. The real number a is equal to the complex number a + 0 i, the imaginary number b i is equal to the complex number 0 + b i. Chapter 9 Lecture Notes The Complex Numbers Chapter 9: The Complex Numbers MAT246H1S Lec0101 Burbulla 9.1: What Is A Complex Number? 9.2: The Complex Plane 9.3: The Fundamental Theorem of Algebra Example 1 Let z = 3 − 2 i and w = −5 + 7 i. Then 1. z + w = (3 − 5) + (−2 + 7) i = −2 + 5 i 2. z − w = (3 + 5) + (−2 − 7) i = 8 − 9 i 3. zw = (3 − 2 i)(−5 + 7 i) = −15 + 10 i + 21 i − 14 i 2 = −1 + 31 i 4. z 2 = (3 − 2 i)(3 − 2 i) = 9 − 12 i + 4 i 2 = 5 − 12 i 5. w 2 = (−5 + 7 i)(−5 + 7 i) = 25 − 49 − 35 i − 35 i = −24 + 70 i Division of complex numbers is a little more complicated. Chapter 9 Lecture Notes The Complex Numbers MAT246H1S Lec0101 Burbulla Chapter 9: The Complex Numbers 9.1: What Is A Complex Number? 9.2: The Complex Plane 9.3: The Fundamental Theorem of Algebra Complex Conjugates and Division of Complex Numbers If z = a + b i is a complex number, define the complex conjugate of z as z̄ = a − b i, and define the modulus of z, or the absolute value of z, as p |z| = a2 + b 2 . Then 1. |z| = |z̄| = √ a2 + b 2 2. |z|2 = z z̄ = a2 + b 2 a bi z̄ = − a2 + b 2 a2 + b 2 |z|2 is called the multiplicative inverse of z. 3. If z 6= 0, then z −1 = If z 6= 0, then z −1 Chapter 9 Lecture Notes The Complex Numbers Chapter 9: The Complex Numbers MAT246H1S Lec0101 Burbulla 9.1: What Is A Complex Number? 9.2: The Complex Plane 9.3: The Fundamental Theorem of Algebra Example 2 To divide by a non-zero complex number w use the fact that |w |2 = w w ⇔ 1 w = . w |w |2 With z and w as in Example 1: z w 3 − 2i −5 + 7 i (3 − 2 i)(−5 − 7 i) = (−5)2 + 72 −15 + 10 i − 21 i + 14 i 2 = 74 29 11 = − − i 74 74 = Chapter 9 Lecture Notes The Complex Numbers MAT246H1S Lec0101 Burbulla Chapter 9: The Complex Numbers 9.1: What Is A Complex Number? 9.2: The Complex Plane 9.3: The Fundamental Theorem of Algebra Properties of Complex Conjugates and Absolute Values Let z = a + b i and w = c + d i be complex numbers. Then √ 1. |z| = |z̄| = a2 + b 2 2. |z|2 = z z̄ = a2 + b 2 3. z ± w = z̄ ± w 4. zw = z̄ w 5. z ∈ R ⇔ z = z̄. 6. z = 0 ⇔ |z| = 0. Proof of 4: z̄ w̄ = (a − bi)(c − di) = ac − bd − (bc + ad)i, and zw = ac − bd + (bc + ad)i = ac − bd − (bc + ad)i. Chapter 9 Lecture Notes The Complex Numbers Chapter 9: The Complex Numbers MAT246H1S Lec0101 Burbulla 9.1: What Is A Complex Number? 9.2: The Complex Plane 9.3: The Fundamental Theorem of Algebra Example 3 Consider again the quadratic equation and its roots: √ −b ± b 2 − 4ac ax 2 + bx + c = 0 ⇒ x = . 2a If b 2 − 4ac < 0, then the roots to the quadratic equation are complex numbers; in fact they will be complex conjugates if a, b, c are all real numbers. Why? az 2 + bz + c = 0 ⇒ az 2 + bz + c = 0̄ = 0 ⇒ ā(z̄)2 + b̄ z̄ + c̄ = 0 ⇒ a(z̄)2 + b z̄ + c = 0 So if z is a solution to the quadratic, so is z̄. Chapter 9 Lecture Notes The Complex Numbers MAT246H1S Lec0101 Burbulla Chapter 9: The Complex Numbers 9.1: What Is A Complex Number? 9.2: The Complex Plane 9.3: The Fundamental Theorem of Algebra Complex Numbers and the Complex Plane √ Let i = −1; let z = a + b i, with a, b ∈ R, be a complex number. The real part of z, denoted by Re(z), is a; the imaginary part of z, denoted by Im(z), is b. Every z ∈ C can be represented by a point in the complex plane. The horizontal axis is called the real axis; 6 b z =r a + b i |z| the vertical axis is called the imaginary axis. z̄ = a − b i real axis a @ - @ |z̄| @ @r z̄ = a − b i Chapter 9 Lecture Notes The Complex Numbers Chapter 9: The Complex Numbers is called the conjugate of z; it is the reflection of z in the real axis. √ 2 |z| = a + b 2 is called the absolute value, or modulus, of z. MAT246H1S Lec0101 Burbulla 9.1: What Is A Complex Number? 9.2: The Complex Plane 9.3: The Fundamental Theorem of Algebra Polar Form of a Complex Number The complex number z = a + b i in the complex plane, can be represented another way, in terms of its ‘polar coordinates.’ Let r = |z| be the distance from the origin to the point (a, b). Let θ be the angle, measured in radians, that the line from the origin to the point (a, b) makes with respect to the positive real axis, √ θ ∈ [0, 2π). θ is called the argument of z. r = |z| = a2 + b 2 , and z = r (cos θ + i sin θ) is called the polar form of the complex number z. Chapter 9 Lecture Notes The Complex Numbers MAT246H1S Lec0101 Burbulla 9.1: What Is A Complex Number? 9.2: The Complex Plane 9.3: The Fundamental Theorem of Algebra Chapter 9: The Complex Numbers Example 1 Find the polar form of z = −3 + 3 i. p √ Solution: r = |z| = (−3)2 + 32 = 3 2. For the argument, use tan θ = 3 = −1; −3 take θ= Thus 3π . 4 √ 3π 3π −3 + 3 i = 3 2 cos + i sin . 4 4 Chapter 9 Lecture Notes The Complex Numbers Chapter 9: The Complex Numbers MAT246H1S Lec0101 Burbulla 9.1: What Is A Complex Number? 9.2: The Complex Plane 9.3: The Fundamental Theorem of Algebra Properties of Complex Numbers in Polar Form If z = |z|(cos α + i sin α), w = |w |(cos β + i, sin β), then z w = |z||w |(cos(α + β) + i sin(α + β)). and z/w = |z|/|w |(cos(α − β) + i sin(α − β)). In particular, you multiply complex numbers by multiplying their moduli and adding their arguments; you divide complex numbers by dividing their moduli and subtracting their arguments. Proof: zw = |z|(cos α + i sin α)|w |(cos β + i sin β) = |z||w |((cos α cos β − sin α sin β) + i(sin α cos β + sin β cos α)) = |z||w |(cos(α + β) + i sin(α + β)) Chapter 9 Lecture Notes The Complex Numbers MAT246H1S Lec0101 Burbulla Chapter 9: The Complex Numbers 9.1: What Is A Complex Number? 9.2: The Complex Plane 9.3: The Fundamental Theorem of Algebra DeMoivre’s Theorem Theorem 9.2.6: (r (cos θ + i sin θ))n = r n (cos(nθ) + i sin(nθ)). Proof: by induction. Its obviously true if n = 1. Now suppose the formula is true for n and calculate (r (cos θ + i sin θ))n+1 , making use of appropriate trig identities, again: (r (cos θ + i sin θ))n+1 = (r (cos θ + i sin θ))n r (cos θ + i sin θ) = r n ((cos(nθ) + i sin(nθ)) r (cos θ + i sin θ) = r n+1 (cos(nθ) cos θ − sin(nθ) sin θ + i (cos(nθ) sin θ + sin(nθ) sin θ)) = r n+1 (cos((n + 1) θ) + i sin((n + 1) θ)) Chapter 9 Lecture Notes The Complex Numbers Chapter 9: The Complex Numbers MAT246H1S Lec0101 Burbulla 9.1: What Is A Complex Number? 9.2: The Complex Plane 9.3: The Fundamental Theorem of Algebra Example 2 Let z = 1 + i. Find z 8 , and a square root of z. √ Solution: put z into polar form. |z| = 2; take θ = π/4. Thus π √ π z = 2 cos + i sin . 4 4 A direct application of DeMoivre’s Theorem gives √ 8π 8π z 8 = ( 2)8 cos + i sin = 16(cos(2π)+sin(2π)) = 16. 4 4 For a square root, we go in ‘reverse:’ π π √ 1/2 1π 1π 1/4 ( 2) cos + i sin =2 cos + i sin . 24 24 8 8 Chapter 9 Lecture Notes The Complex Numbers MAT246H1S Lec0101 Burbulla Chapter 9: The Complex Numbers 9.1: What Is A Complex Number? 9.2: The Complex Plane 9.3: The Fundamental Theorem of Algebra Example 3 Find the cube roots and fourth roots of unity; that is solve (a) z 3 = 1, and (b) z 4 = 1. Solution: note that the argument of the real number 1 is θ = 0. Let z = cos θ + i sin θ. Then z n = cos(nθ) + i sin(nθ). So I (a) z 3 = 1 ⇒ 3θ = 0 + 2kπ ⇒ θ = 2kπ/3. Take θ = 0, 2π/3, 4π/3, and consequently the three distinct solutions are √ √ 1 3 1 3 z = 1, − + i, or − − i. 2 2 2 2 I (b) z 4 = 1 ⇒ 4θ = 0 + 2kπ ⇒ θ = 2kπ/4. Take θ = 0, π/2, π, 3π/4, so the four distinct solutions are z = 1, i, −1, −i. Chapter 9 Lecture Notes The Complex Numbers Chapter 9: The Complex Numbers MAT246H1S Lec0101 Burbulla 9.1: What Is A Complex Number? 9.2: The Complex Plane 9.3: The Fundamental Theorem of Algebra Alternate Calculations for Example 3 We could have just factored: I (a) z 3 = 1 ⇔ z 3 − 1 = 0 ⇔ (z − 1)(z 2 + z + 1) = 0. Thus z = 1 or √ √ 1 −1 + 1 − 4 3 =− ± z= i. 2 2 2 I (b) z 4 = 1 ⇔ z 4 − 1 = 0 ⇔ (z 2 − 1)(z 2 + 1) = 0. Thus z = ±1 or ± i. Chapter 9 Lecture Notes The Complex Numbers MAT246H1S Lec0101 Burbulla 9.1: What Is A Complex Number? 9.2: The Complex Plane 9.3: The Fundamental Theorem of Algebra Chapter 9: The Complex Numbers Example 4 √ Find the all the solutions of z 3 = 1 + 3 i. √ Solution: first put 1 + 3 i into polar form. √ √ |1 + 3 i| = 1 + 4 = 2, and so 1+ √ √ 3 =2 1 3 + i 2 2 ! π π = 2 cos + i sin . 3 3 √ Let z = r (cos θ + i sin θ) and suppose z 3 = 1 + 3 i. Then π π 3 r (cos(3θ) + i sin(3θ)) = 2 cos + i sin . 3 3 Chapter 9 Lecture Notes The Complex Numbers Chapter 9: The Complex Numbers MAT246H1S Lec0101 Burbulla 9.1: What Is A Complex Number? 9.2: The Complex Plane 9.3: The Fundamental Theorem of Algebra Therefore, r 3 = 2 and 3θ = π + 2kπ. 3 Thus r = 21/3 and π 2kπ + , 9 3 for which the three distinct solutions in the interval [0, 2π) are θ= θ= 13π π 7π . , , or 9 9 9 Thus the three values of z are π π 1/3 z1 = 2 cos + i sin , 9 9 7π 7π 13π 13π z2 = 21/3 cos + i sin , or z3 = 21/3 cos + i sin . 9 9 9 9 Chapter 9 Lecture Notes The Complex Numbers MAT246H1S Lec0101 Burbulla Chapter 9: The Complex Numbers 9.1: What Is A Complex Number? 9.2: The Complex Plane 9.3: The Fundamental Theorem of Algebra The Fundamental Theorem of Algebra Having defined the number i to be a solution to the simple equation x 2 + 1 = 0, it may come as a surprise to find that any non-constant polynomial will always have solutions in terms of complex numbers. This is known as the Fundamental Theorem of Algebra. Theorem 9.3.1: every non-constant polynomial with complex coefficients has a complex root. Proof: the proof of this theorem requires results which are beyond the scope of this course. So we will just look at some implications of this result. In particular, we shall investigate how many roots a polynomial can have. Chapter 9 Lecture Notes The Complex Numbers Chapter 9: The Complex Numbers MAT246H1S Lec0101 Burbulla 9.1: What Is A Complex Number? 9.2: The Complex Plane 9.3: The Fundamental Theorem of Algebra Some Definitions 1. An expression of the form an z n + an−1 z n−1 + · · · + a2 z 2 + a1 z + a0 , with coefficients ai ∈ C, and an 6= 0, is called a polynomial of degree n. We will often write it as p(z), or q(z). 2. The complex number r is a root of the polynomial p(z) if p(r ) = 0. 3. The polynomial f (z) is a factor of the polynomial p(z) if there is a polynomial q(z) such that p(z) = f (z)q(z). 4. The root r of the polynomial p(z) has multiplicity k if f (z) = (z − r )k is a factor of p(z), but (z − r )k+1 is not a factor of p(z). Chapter 9 Lecture Notes The Complex Numbers MAT246H1S Lec0101 Burbulla Chapter 9: The Complex Numbers 9.1: What Is A Complex Number? 9.2: The Complex Plane 9.3: The Fundamental Theorem of Algebra Long Division by a Linear Factor The following type of ‘long division’ should be familiar to you: z + 2 ) z4 − z4 + − − z3 z3 2z 3 3z 3 3z 3 − + 3z 2 + 11z 5z 2 − 22 + 10 + − 5z 2 6z 2 11z 2 11z 2 + 22z − 22z − 22z + 10 + 10 + 10 − 44 54 Thus z 4 − z 3 + 5z 2 + 10 = (z + 2)(z 3 − 3z 2 + 11z − 22) + 54. Chapter 9 Lecture Notes The Complex Numbers Chapter 9: The Complex Numbers MAT246H1S Lec0101 Burbulla 9.1: What Is A Complex Number? 9.2: The Complex Plane 9.3: The Fundamental Theorem of Algebra Theorem 9.3.4: if r ∈ C and p(z) is a non-constant polynomial then there exists a polynomial q(z) and a number c such that p(z) = (z − r )q(z) + c. Proof: by complete mathematical induction of the degree of p(z). n = 1 : if p(z) = az + b then check that p(z) = a(z − r ) + ar + b. Now suppose p(z) = an+1 z n+1 + an z n + · · · + a1 z + a0 has degree n + 1. Check that p(z) = (z − r )an+1 z n + (an + ran+1 )z n + an−1 z n−1 + · · · a1 z + a0 . | {z } g (z), deg g (z) ≤n By induction, g (z) = (z − r )h(z) + c, for some polynomial h(z) and constant c. Then p(z) = (z−r )an+1 z n+1 +(z−r )h(z)+c = (z−r )(an+1 z n+1 + h(z))+c. | {z } q(z) Chapter 9 Lecture Notes The Complex Numbers MAT246H1S Lec0101 Burbulla Chapter 9: The Complex Numbers 9.1: What Is A Complex Number? 9.2: The Complex Plane 9.3: The Fundamental Theorem of Algebra The Factor Theorem Theorem 9.3.6: the complex number r is a root of the polynomial p(z) if and only if z − r is a factor of p(z). Proof: if z − r is a factor of p(z), then p(z) = (z − r )q(z), for some polynomial q(z), and p(r ) = (r − r )q(r ) = 0. Now suppose r is a root of p(z). By Theorem 9.3.4, we can write p(z) = (z − r )q(z) + c, for some constant c. Then p(r ) = 0 ⇔ c = 0, implying that z − r is a factor of p(z). Chapter 9 Lecture Notes The Complex Numbers Chapter 9: The Complex Numbers MAT246H1S Lec0101 Burbulla 9.1: What Is A Complex Number? 9.2: The Complex Plane 9.3: The Fundamental Theorem of Algebra Example 1 Consider the polynomial p(z) = i z 3 + z 2 − 4. Since p(2i) = i(2i)3 + (2i)2 + 4 = i(−8i) + (−4) − 4 = 8 − 4 − 4 = 0, (z − 2i) is a factor of p(z). Use long division to check that p(z) = (z − 2i)(i z 2 − z − 2i). Thus p(z) has two more roots: i z 2 − z − 2i = 0 ⇒ z 2 + i z − 2 = 0 √ −i ± i 2 + 8 ⇒ z= √ 2 7 i ⇒ z =± − 2 2 Chapter 9 Lecture Notes The Complex Numbers MAT246H1S Lec0101 Burbulla Chapter 9: The Complex Numbers 9.1: What Is A Complex Number? 9.2: The Complex Plane 9.3: The Fundamental Theorem of Algebra How Many Roots Can a Polynomial Have? Theorem 9.3.8: a polynomial p(z) of degree n has at most n roots; if multiplicities are counted, it has exactly n roots. Proof: p(z) has at least one root, by the Fundamental Theorem of Algebra; call it r1 . By the factor theorem, p(z) = (z − r1 )q(z), where q(z) is a polynomial with degree n − 1. By the Fundamental Theorem of Algebra, q(z) has a root, call it r2 . Thus p(z) = (z − r1 )(z − r2 )r (z), for some polynomial r (z) with degree n − 2. Continuing in this fashion, n times, there will be roots r1 , r2 , . . . , rn , not all necessarily distinct, such that p(z) = k(z − r1 )(z − r2 ) · · · (z − rn ), for some constant k. Chapter 9 Lecture Notes The Complex Numbers Chapter 9: The Complex Numbers MAT246H1S Lec0101 Burbulla 9.1: What Is A Complex Number? 9.2: The Complex Plane 9.3: The Fundamental Theorem of Algebra Example 2 In Example 1, we saw that the cubic polynomial p(z) = i z 3 + z 2 − 4 has three distinct roots. Now consider the fifth degree polynomial q(z) = z 5 + 2z 4 + 2z 3 + 4z 2 + z + 2. Since q(−2) = 0, as you may check, q(z) = (z + 2)(z 4 + 2z 2 + 1) = (z + 2)(z 2 + 1)2 . Thus the distinct roots of q(z) are −2, i and −i. Each of the roots ±i has multiplicity two. So counting multiplicities, q(z) has five roots. Chapter 9 Lecture Notes The Complex Numbers MAT246H1S Lec0101 Burbulla