A complex number is z=a+bi where a,b∈R and i2=−1. We call a=Re(z) the real part and b=Im(z) the imaginary part.
Arithmetic:(a+bi)+(c+di)=(a+c)+(b+d)i and (a+bi)(c+di)=(ac−bd)+(ad+bc)i.
Proposition 1.1 (Properties of Complex Arithmetic). For all z,w∈C:
z+w=w+z and zw=wz (commutativity)
(z+w)+u=z+(w+u) and (zw)u=z(wu) (associativity)
z(w+u)=zw+zu (distributivity)
There exist additive identity 0 and multiplicative identity 1.
Every z=0 has a multiplicative inverse z1=∣z∣2zˉ.
Remark. The complex field C cannot be ordered: there is no total ordering on C Compatible with the field operations. In particular, i2=−1 precludes any such ordering.
1.2 The Complex Conjugate and Modulus
Definition. The complex conjugate of z=a+bi is zˉ=a−bi.
Proposition 1.2. For all z,w∈C:
z+w=zˉ+wˉ and zw=zˉwˉ
zzˉ=∣z∣2
z+zˉ=2Re(z) and z−zˉ=2iIm(z)
zˉˉ=z
Definition. The modulus (or absolute value) of z=a+bi is ∣z∣=a2+b2.
Proposition 1.3 (Modulus Properties). For all z,w∈C:
∣z∣≥0 with equality iff z=0
∣zw∣=∣z∣∣w∣
∣z+w∣≤∣z∣+∣w∣ (triangle inequality)
∣z∣−∣w∣≤∣z−w∣ (reverse triangle inequality)
Proof of (3).∣z+w∣2=(z+w)(zˉ+wˉ)=∣z∣2+zwˉ+zˉw+∣w∣2=∣z∣2+2Re(zwˉ)+∣w∣2≤∣z∣2+2∣z∣∣w∣+∣w∣2=(∣z∣+∣w∣)2. The inequality follows from Re(zwˉ)≤∣zwˉ∣=∣z∣∣w∣. ■
1.3 Polar Form
Every non-zero complex number can be written in polar form:
z=r(cosθ+isinθ)=reiθ
Where r=∣z∣=a2+b2 is the modulus and θ=arg(z) is the argument.
Definition. The principal argumentArg(z) is the unique θ∈(−π,π] Such that z=∣z∣eiθ. The argumentarg(z) is multi-valued: arg(z)=Arg(z)+2πk for k∈Z.
Proposition 1.4. If z1=r1eiθ1 and z2=r2eiθ2Then z1z2=r1r2ei(θ1+θ2) and z1/z2=(r1/r2)ei(θ1−θ2).
Worked Examples: Polar Form Conversions
Solution
Problem. Convert z=−1+3i to polar form and find all arguments.
∣z∣=(−1)2+(3)2=1+3=2.
Re(z)=−1<0 and Im(z)=3>0So z is in the second quadrant.
θ=arctan(−13)=32π (adjusting to second quadrant).
Both real and imaginary parts are negative, so z is in the third quadrant.
θ=arctan(4/3)+π=π+arctan(4/3).
z=5ei(π+arctan(4/3)).
1.4 Euler”s Formula and De Moivre’s Theorem
Euler’s formula:eiθ=cosθ+isinθ.
De Moivre’s theorem:(eiθ)n=einθSo
(cosθ+isinθ)n=cos(nθ)+isin(nθ)
Proposition 1.5. De Moivre’s theorem holds for all integers nIncluding negative values.
Proof. For n≥0It follows by induction from the multiplication law eiαeiβ=ei(α+β). For n<0Write n=−m with m>0: (cosθ+isinθ)n=(cosθ+isinθ)m1=cos(mθ)+isin(mθ)1=cos(−mθ)+isin(−mθ)=cos(nθ)+isin(nθ). ■
Definition. An n-th root of w∈C is a complex number z such that zn=w.
Proposition 1.6. Every non-zero w∈C has exactly n distinct n-th roots. If w=ρeiϕThen
zk=ρ1/nei(ϕ+2πk)/n,k=0,1,…,n−1
Where ρ1/n>0 is the positive real n-th root of ρ.
Proof. If zn=wWrite z=reiθ. Then rneinθ=ρeiϕ So r=ρ1/n and nθ=ϕ+2πk. For k=0,1,…,n−1 these give distinct Values of θ; for k≥n they repeat. ■
Remark. The n-th roots of w lie equally spaced on a circle of radius ρ1/nForming a Regular n-gon.
1.6 Roots of Unity
The n-th roots of unity are the solutions of zn=1:
zk=e2πik/n,k=0,1,…,n−1
They form a regular n-gon on the unit circle in the complex plane.
Proposition 1.7. If ω=e2πi/n is a primitive n-th root of unity, then ∑k=0n−1ωk=0 and ∑k=0n−1ωjk=0 for any j not divisible by n.
Proof. The sum ∑k=0n−1ωk=1−ω1−ωn=1−ω1−1=0 Provided ω=1. For j not divisible by n, ωj is a non-trivial root of unity, So the same argument applies. ■
Solution
Problem. Find all cube roots of −8.
−8=8eiπ. The cube roots are: zk=81/3ei(π+2πk)/3=2ei(π+2πk)/3 for k=0,1,2.