3 The Field of Complex Numbers Recall the standard description of the complex numbers: C = a + ib : a,b R,i = √ 1 . { ∈ − } Any complex number z = a + ib can be thought of as having two components, a real part, Re(z) = a, and an imaginary part, Im(z)= b. A real number is then just a complex number with imaginary part equal to zero. A complex number is called pure imaginary if its real part is zero. We add and multiply numbers in C as follows (compare this with the previous description of C). (a + ib) + (c + id) = (a + c)+ i(c + d) and (a + ib)(c + id) = (ac ad)+ i(ac + bd). − Then C is an ID, with zero 0 = 0+0i, unity 1 = 1 + 0i and the negative of z = a + ib is z = ( a)+ i( b)= a ib. (We leave it as an exercise to show that the associative (G1,R1) and distributive− laws− hold− (R2), and− − that both operations are well-defined and commutative.) Suppose we’d like to solve the equation (2+3i)z =4 i. If (2+3i)−1 exists in C, then we can solve for unique z C by − ∈ z =(2+3i)−1(4 i). − Note that (a + ib)(a ib)= a2 + b2 is a positive real number for any non-zero complex number z = a + ib. Consider the following:− 1 1 2 3i 2 3i 2 3 2 3 = − = − = + i − = + i− . 2+3i 2+3i 2 3i 4+32 4+32 4+32 13 13 − Then (2+3i)−1 C, since 2 , −3 R. So we can solve for z, and in fact ∈ 13 13 ∈ 2 3 5 14 z =(2+3i)−1(4 i) = ( i )(4 i)= i . − 13 − 13 − 13 − 13 We can compute the inverse of an arbitrary complex number z = a+ib using the same technique. In general, if z = 0 then a2 + b2 is a positive real number and 6 a b z−1 = (a + ib)−1 = i , a2 + b2 − a2 + b2 is well-defined in C. It follows that C is a field. There is a particular map on C called complex conjugation: θ : C C : z = a + ib z¯ = a ib. → 7→ − This is an example of a special kind of map called a field automorphism. Note that zz¯ = a2 + b2 R+ ∈ and so z(a2 + b2)−1z¯ =1= zz−1. In particular, the inverse of z in C is a multiple of its conjugate,z ¯, i.e. z¯ z−1 = . a2 + b2 The quantity a2 + b2 is the square of the modulus of a + ib. Definition 3.1. Let z = a + ib be a complex number. The modulus of z is denoted by z and is given by z = √a2 + b2. | | | | 8 In this notation, we may write z−1 =z/ ¯ z 2 for any non-zero z C We note the following facts relating | | ∈ to z andz ¯ Lemma 3.1. Let z,z ,z C. Then 1 2 ∈ 1. zz¯ = z 2, | | 2. z1 + z2 =z ¯1 +z ¯2, 3. z1z2 =z ¯1z¯2, 4. cz = cz for any c R. ∈ 3.1 The Geometry of Complex Numbers We’ve remarked before that z = a + ib can be identified with the vector (a,b) R R. Note that the Euclidean distance between (a,b) and the origin is given by √a2 + b2 = √zz¯, so the∈ modulus× of z, denoted by z , can be interpreted as its distance to the origin in the complex plane. |Two| complex numbers have the same modulus d iff they both lie on the same circle of radius d about the origin. The argument of a complex number, denoted arg(z), is the angle that the vector representing z makes with the positive real axis, measured in the anti-clockwise direction. Definition 3.2. Let z = a + ib C, let r = z , and let θ = arg(z). The polar form of z is given by ∈ | | z = rcos θ + irsin θ. If z = a + ib with a,b > 0 and arg(z)= θ then 0 θ π/2 and z is positioned in the 1st quadrant of the argand diagram. Its relationship with the numbers≤ a≤ ib is shown in the table below. ± ± z arg(z) interval quadrant polar form a + ib θ [0,π/2] 1st r(cos θ + isin θ) a + ib π θ [π/2, π] 2nd r( cos θ + isin θ) −a ib π +− θ [π, 3π/2] 3rd r(−cos θ isin θ) −a − ib 2π θ [3π/2, 2π] 4th r(cos− θ −isin θ) − − − Lemma 3.2. Let z C and let c be a positive real number. Then ∈ 1. cz = c z , | | | | 2. arg(cz)=arg(z) Example 3.1. Let z =2i 2√3. Then z =4 and arg(z)=2π/3. The polar form of z is given by − | | z = 4(cos(2π/3)+ isin(2π/3)). It conjugate is z¯ = 4(cos(2π/3) isin(2π/3)) = 4(cos(4π/3)+ isin(4π/3)) − We recall the following trigonometric identities. cos(A + B) = cosAcosB sinAsinB − sin(A + B) = sinAcosB + cosAsinB These can be applied for fast multiplication in C. 9 Let z = r(cos A + sin A) and let w = (cos B + sin B). Then zw = rs(cos(A + B) + sin(A + B)). In words, the modulus and argument of the product of a pair of complex numbers is the product of their moduli and the sum of their arguments, respectively. This leads to the following theorem. Theorem 3.1. (De Moivre’s Theorem) Let z = r(cosθ + sinθ) and let n be a non-negative integer. Then zn = rn(cosnθ + sinnθ). Proof. We apply an inductive proof. It’s clear the statement is true for the case n = 1. Suppose it is true for n = k. Then zk+1 = zzk = r(cosθ + sinθ)rk(coskθ + sinkθ)= rk+1(cos(k + 1)θ + sin(k + 1)θ). Example 3.2. Let z =2i 2√3 and let w = 1+i. Then z =4, w = √2, arg(z)=2π/3, and arg(w)= π/4. The product zw can be computed− as | | | | zw =4√2(cos(2π/3+ π/4)+ isin(2π/3+ π/4)) =4√2(cos(11π/12) + isin(11π/12)) = (2+2√3) + i(2√3 2). − − De Moivre’s theorem is very useful for finding nth powers of a complex number. (zw)6 =4623(cos(11π/2)+ isin(11π/2)) =215((cos(π/2) + isin(π/2))) = i215. Remark: Note that cos(2π+θ) =cosθ and sin(2π+θ) = sinθ. Also cos(π+θ)= cosθ and sin(π+θ)= sinθ. − − 3.2 Complex Roots of Unity If z has modulus 1 then so does any power of z. Note that if z = cosθ + isinθ then zn = cosnθ + isinnθ from De Moivre’s theorem, so if θ =2πk/n for some non-negative integer k, we get zn = 1. Example 3.3. Let z = 1/2+ i√3/2. Then z = 1 and arg(z) = π/6. The polar form of z is z = cos(π/6) + isin(π/6). Then z12 = cos2π + isin2π| =1| . We say that z is a 12th root of unity. We list the distinct powers of z as follows. z = cos(π/6)+ isin(π/6) =1/2+ i√3/2 z2 = cos(π/3)+ isin(π/3) = √3/2+ i1/2 z3 = cos(π/2)+ isin(π/2) = i z4 = cos(2π/3)+ isin(2π/3) = √3/2+ i1/2 − z5 = cos(5π/6)+ isin(5π/6) = 1/2+ i√3/2 − z6 = cosπ + isinπ = 1 = z6 = z12 − − z7 = cos(7π/6)+ isin(7π/6) = 1/2 i√3/2 = z5 = z − − − z8 = cos(4π/3)+ isin(4π/3) = √3/2 i1/2 = z4 = z2 − − − z9 = cos(3π/2)+ isin(3π/2) = i = z3 = z3 − − z10 = cos(5π/3)+ isin(5π/3) = √3/2 i1/2 = z2 = z4 − − z11 = cos(11π/6)+ isin(11π/6) = 1/2 i√3/2 =¯z = z5 − − z12 = cos(2π)+ isin(2π) =1 = z12 = z6 − 10 Note that z7z9 = z16 = z12z4 = z4 since z12 = 1. In fact every product zszt = zs+t gives another complex 12th root of unity for any s,t 1, ..., 12 , so complex multiplication is an operation on the the set of all such 12th roots of unity. It is not∈ { hard to} see that the set C = z,z2,z3, ..., z11,z12 =1 12 { } k forms a group wrt complex multiplication: the identity element 1 is contained in C12, and c has inverse 12−k 3 9 12 c in C12 for each k (for example z z = z =1). Definition 3.3. Let n be a positive integer. A complex number z is called an nth root of unity if zn =1. We construct these nth roots as follows. Let z = r(cos(θ)+ isinθ) be a complex number. Suppose that z is an nth root of unity. Then zn = rn(cos(nθ)+ isin(nθ)) = 1 = cos(2kπ)+ isin(2kπ), for any non-negative integer k. Set r = 1. Then zk = cos(2kπ/n)+ isin(2kπ/n) is an nth root of unity for each non-negative integer k. Observe that z = z iff k t modulo n (i.e. iff k t is a multiple of n). To see this, note that k t ≡ − zt = zk ztzn−k =1 znzt−k = zt−k =1 t k = mn ⇔ ⇔ ⇔ − for some integer m. This means that z = z = z = = z = , 0 n 2n · · · tn · · · z = z = z = = z = , 1 n+1 2n+1 · · · tn+1 · · · . z − = z − = = z − = , n 1 2n 1 · · · tn t+1 · · · so in particular there are exactly n distinct complex nth roots of unity.