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Complex Numbers and the Complex Exponential Complex Numbers and the Complex Exponential Frank R. Kschischang The Edward S. Rogers Sr. Department of Electrical and Computer Engineering University of Toronto September 15, 2005; updated January 10, 2017 1 Numbers and Equations Numbers have often been invented to solve equations. For example, by introducing negative numbers the natural numbers N = f0; 1; 2;:::g can be extended to the integers Z = f:::; −2; −1; 0; 1; 2;:::g so that we may solve simple equations such as x+2 = 0. Likewise, by introducing integer quotients, the integers can to be extended to the rational numbers Q = fa=b : a; b 2 Z; b 6= 0g so that we may solve simple equations such as 2x = 1. The rationals seem like a very nice set in which to do arithmetic. There is a well-defined ad- dition operation and a well-defined multiplication operation, and Q is closed with respect to these operations. Addition and multiplication are associative (i.e., for all x; y; z 2 Q; x(yz) = (xy)z and likewise for addition) and commutative (i.e., for all x; y 2 Q; x + y = y + x and likewise for multiplication). Every element x 2 Q has an additive inverse −x (from which we may define a subtraction operation) and every nonzero element x 2 Q; x 6= 0, has a mul- tiplicative inverse 1=x (from which we may define a division operation). Furthermore, multi- plication and addition satisfy the distributive law, i.e., for all x; y; z 2 Q; x(y +z) = xy +xz. Mathematically speaking, the rational numbers form a field. Who could ask for anything more? The trouble is that certain simple equations such as x2 − 2 = 0 (1) 1 have no solutions in Q, i.e., no rational number x satisfies (1). However, following the progression from N to Z to Q, we might try to get around this problem by extending Q, i.e., by adjoining an element|let's call it θ for now|that satisfies θ2 − 2 = 0, or, equivalently, θ2 = 2. We will demand that θ be combinable with ordinary rational numbers (and with itself) via addition and multiplication, while satisfying all of the formal arithmetic properties (such as closure with respect to addition and multiplication, associativity, commutativity, the distributive law, etc.) that we have grown to expect. If we denote this extended set by Q[θ], then certainly Q[θ] must contain all numbers of the form a + bθ, where a; b 2 Q. Numbers involving higher powers of θ do not arise, since any such higher power can be reduced to a multiple of a lower power, i.e., θ2 = 2, θ3 = 2θ, θ4 = 4, etc. Indeed, the sum, difference, product and quotient of any two elements of this form is another element of this form (provided that we don't attempt to divide by zero). To see this, observe that (a + bθ) ± (c + dθ) = (a ± c) + (b ± d)θ; and since a±c and b±d are rational when a; b; c and d are rational, we have another number of the same form. Likewise (a + bθ)(c + dθ) = ac + adθ + bcθ + bdθ2 = (ac + 2bd) + (ad + bc)θ: Again, since ac+2bd and ad+bc are rational when a; b; c and d are rational, we have another number of the same form. Finally, to show that we can form quotients, it is enough to show that we can form reciprocals. Note that if a and b are not both zero then 1 a − bθ = a + bθ (a + bθ)(a − bθ) a − bθ = a2 − b2θ2 a b = − θ: a2 − 2b2 a2 − 2b2 Now, a2 − 2b2 6= 0 (why?), so a=(a2 − 2b2) and −b=(a2 − 2b2) are rational when a; b; c and d are rational; thus once again we have another number of the same form. Indeed, we have shown that Q[θ] is a field, and unlike Q, in this field (1) does indeed have a solution. Since every element of Q[θ] can be written as a + bθ, with a; b 2 Q, we could associate such an element with the ordered pair (a; b). We might refer to the first component, a, as the \rational part" of a + bθ and the second component, b, as the \irrational part" of a + bθ. (Note that the \irrational part" is in fact itself a rational number!) The arithmetic in Q[θ] could be defined via operations that occur on such ordered pairs, e.g., (a; b) + (c; d) = (a + c; b + d) (a; b) × (c; d) = (ac + 2bd; ad + bc) 2 From this viewpoint there is no mention of the \irrational" number θ; indeed, from this viewpoint one can simply regard θ as a \tag", pointing out which of the two rational numbers in question is the designated \irrational part." p By the way, a more usual notation for θ is 2. 2 Complex Numbers and the Complex Plane p So far we have \extended" Q by adjoining 2, thus taking one small step toward the con- struction of the real numbers R. Let usp now proceed to the point where we have constructed 3 all of R. Certain real numbers (e.g., 3) appear as the zeros of a polynomial withp rational coefficients and can be adjoined to Q in the same manner that we used to adjoin 2; others (e.g., π) appear as the limit points of certain sequences. Unfortunately, even in R, many polynomials have no zeros, e.g., there is no real number x that satisfies x2 + 1 = 0: (2) Emboldened by our experience in extending Q, let us extend R by introducing a new element, a purely \imaginary" number, j, that satisfies j2 +1 = 0. Once again, we will demand that j satisfy all of the formal arithmetic properties (such as closure with respect to addition and multiplication, associativity, commutativity, the distributive law, etc.) that we have grown to expect in R. Let us denote the extended set by C. Certainly, as we compute all possible sums and products involving real numbers and j, we find that C must contain all numbers of the form a + b j, where a; b 2 R. Numbers involving higher powers of j do not arise, since any such higher power can be reduced to a multiple of a lower power, i.e., j2 = −1, j3 = − j, j4 = 1, etc. We have, for all a; b; c and d in R, (a + b j) ± (c + d j) = (a ± c) + (b ± d) j (a + b j) × (c + d j) = (ac − bd) + (ac + bd) j and, provided a and b are not both zero, 1 a − b j a − b j = = : a + b j (a + b j)(a − b j) a2 + b2 This shows that the sum, difference, product and quotient of any two elements of this form is another element of this form (provided that we don't attempt to divide by zero), and so C is a field. The elements of C are called complex numbers and C is referred to as the complex field. 3 We usually denote individual complex numbers with a single symbol, such as z. Now, since every such element can be written in the form z = a + b j, with a; b 2 R, we can associate such an element with the ordered pair (a; b) of real numbers. We refer to the first component, a, as the \real part" of z (written Re(z)) and the second component, b, as the \imaginary part" of z (written Im(z)). Thus for all z 2 C, we have z = Re(z) + Im(z) j : Note that the \imaginary part" is in fact itself a real number! If Im(z) = 0, then z is referred to as a (purely) real number; likewise if Re(z) = 0, then z is referred to as a purely imaginary number. Thus, in particular, j itself is purely imaginary. Two complex numbers z1 and z2 are equal if and only if their real and imaginary parts agree, i.e., z1 = z2 if and only if [Re(z1) = Re(z2)] and [Im(z1) = Im(z2)]: Thus every complex equality can always be regarded as a pair of real equalities. The complex conjugate z∗ of a complex number z = Re(z) + Im(z) j is the complex number z∗ = Re(z) − Im(z) j : Note that z∗ = z if and only if Im(z) = 0, i.e., if and only if z is real. The complex conjugate obeys the following properties for all w; z 2 C: (w ± z)∗ = w∗ ± z∗; (wz)∗ = (w∗)(z∗); (w=z)∗ = (w∗)=(z∗)(z 6= 0): Furthermore, for all z 2 C we have z + z∗ = 2 Re(z); z − z∗ = 2 j Im(z); (z∗)∗ = z: 2 d It is possible to show (see the problems) that if f(x) = a0 + a1x + a2x + ··· + adx is a polynomial with real coefficients (i.e., a0; a1; : : : ; ad 2 R) and f(z) = 0 for some complex value z, then we must also have f(z∗) = 0. In other words, if z is a zero of a polynomial with real coefficients then so is its complex conjugate z∗. Note that, unlike the real numbers, complex numbers are not in general ordered, i.e., it makes no sense to ask which is larger: 2 + 3 j or 3 + 2 j. However, we can always compare the magnitudes of two complex numbers. The magnitude (or absolute value or modulus) jzj of a complex number z is defined as p q jzj = zz∗ = Re2(z) + Im2(z): 4 y y b a + b j = r(cos θ + j sin θ) 2 + 3 j r j θ x x −4 0 0 a unit circle 2 − 3 j −4 − 4 j (a) (b) Figure 1: (a) Several points in the complex plane.
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