3 The of Complex Numbers

Recall the standard description of the complex numbers:

C = a + ib : a,b R,i = √ 1 . { ∈ − } Any 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 . 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 . 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 . 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.

The set of these nth roots is given by

2 3 n−1 n C = z = z,z = z ,z = z , ..., z − = z ,z = z =1 . n { 1 2 3 n 1 n } Lemma 3.3. The complex nth roots of unity form a group under complex multiplication. Proof. If z, w are both complex nth roots of unity then

(zw)n = znwn =1 1=1, · so the product of any pair of complex nth roots of unity is another nth root, and hence complex multiplication is an operation on the set of all such nth roots of unity. Multiplication is assoc in C, and hence on any subset of C, so G1 is immediately satisfied. The identity 1 is trivially an nth root of unity for any positive integer n, so G2 holds. Finally, z has inverse zn−1 for any complex nth root of unity z so G3 is satisfied.

Remark: Cn is a non-empty subset of C 0 , and is also a group wrt complex multiplication. We thus say that C is a subgroup of C 0 . We’ll examine\{ } substructures of groups, rings and fields in the next section. n \{ }

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