# Complex Number and Bi

Original Notes adopted from November 13, 2001 (Week 10)

© P. Rosenthal , MAT246Y1, University of Toronto, Department of Mathematics typed by A. Ku Ong

Polynomial Equations with Integer Coefficients

Eg. 3x + 2 = 0.

No Solution in Z, solution in Q

Every linear equation with coeffients in Z, has a root in Q. Eg. x2 – 2 = 0. No root in Q. Has root in R.

Eg. x2 + 1 = 0, no root in R.

To define set of complex numbers, C

C = { a + bi, a,b ∈ R }

i2 = -1

Define ( a + bi ) + (c + di) = (a + c) + ( b + d) i

Define ( a + bi ) ( c + di ) = ( ac – bd ) + ( ad + bc) i

Given a + bi, we say " a is the real part" of a + bi, & write R ( a + bi ) , & we say "b is the imaginary part" of a + bi, & write Im( a + bi) = b.

a + 0i, write as a,

R is embedded in C:

0 + bi, write as bi

to a ∈ R , corresponds a + 0i.

Numbers of form bi for b ≠ 0 are "pure imaginary" numbers

0 = 0 + 0i

a + bi + 0 = a + bi ∀ a + bi.

Given a + bi, there is an additive inverse.

(a + bi ) + (-a + (-b)i) = a + (-a) + ( b + (-b)) i = 0.

1 = 1 + 0i

1( a + bi) = ( a + bi) 1 = a + bi.

Eg. ( 2 + 3 i ) ( a + bi ) = 1.

Want 1/2+3i to be a + bi, some a, b ∈ R.

Note: a + bi = c + di is equivalent to a = c & b = d.

Given a + bi, define its modulus | a + bi | to bi √a2 + b2 |a + bi| = distance from a + bi to 0.

Argand diagram

a + bi ⇒ (a, b) in R2

The complex conjugate of a + bi is a = bi.

Conjugate of 3 - √2i is 3 + √2i

Notation: a + bi

= a – bi

(a+bi)(a+bi) = (a + bi) ( a – bi) = a2 + b2 = |a + bi|2

Eg. Want

1

= 1

2 – 3 i = 2 – 3i

2 + 3i 2+ 3i

2 -3i

13

Generally,

if a + bi ≠ 0

1

= 1

a + bi a+ bi

= 2/13 – 3/13 i

a-bi = a

a-bi

a2 + b2

=

∴ (a + bi) (

1

a2 + b2

-b

a2 + b2

i

(a – bi)

1 ( a – bi) )

a2 + b2

1 (a2 + b2 ) = 1.

2

a + b2

∴ Every complex number ≠ 0 has a multiplicitive inverse. The complex numbers with the operations are a "field" (as Q,R). The argument of a + bi is the angle that line from (0, 0) to (a,b) makes with positive x –axis.

If a + bi has modulus r & argument θ,

r = √ a2 + b2

θ = arctan b/a

a = r cos θ, d = r sinθ

a + bi = r cosθ + irsinθ = r ( cos θ + isin θ)

|cos θ + i sinθ| = √cos 2 θ+ sin2 θ =1

[ r (cosθ + isinθ) ] [ s (cos φ + isinφ)]

= r s [( cosθcosφ - sinθsinφ ) + i (sinθcosφ + cosθsinφ)] = r s ( cosθ + φ ) + i sin( θ + φ)

To multiply complex numbers, multiply modulus & add arguments. In Particular,

(r (cosθ + isinθ))2 = r2 ( cos2θ + isin2θ)

De Moivre's Theorem (For n ∈ N)

[r (cosθ + isinθ))n = rn ( cosnθ + isin nθ)

Proof: By Induction, n = 1 – Obvious.

Assume n = k.

To show (r (cosθ + isinθ))k+1 = rk+1(cos(k+1) θsin (k+1) θ) [r (cosθ + isinθ)]k+1 = [r(cos θ + isin θ)] k[r(cos θ + sin θ)] = r k (cosk θ + isin kθ)[r(cos θ + isin θ)]

= r k r(cos(k θ +θ) + isin(kθ + θ) )

= rk+1(cos(k+1) θsin (k+1) θ)

8

Eg. (1 + i)

(1+ i)

π/4

1 + i = √2( cos π/4 + isinπ/4)

(1 + i) 8= [√2 cos π/4 + isinπ/4)] 8

= 24 (cos 2π + isin2π)

= 16( 1 + i0)

= 16

(1 + i) 100= [√2 cos π/4 + isinπ/4)] 100

= 250 (cos 25π + isin25π)

= 250 (cos π + isinπ)

= 250(-1 + 0) = -250

2

Eg. z = -1

=i

=-i

Eg. z 2= 1

= 1, -1.

-1

1

Eg. z 3= 1

z =1

|z 3| = |z |3= 1

∴ |z| = 1.

if |z| = 1, z = cosθ + sinθ , some θ.

z 3= cos3θ + isin3θ = 1 = cos ( 0 + k2π) + isin( 0 + k2π) Can have 3θ = 2πk for any k ∈ Z.

In particular, 3θ = 0 z = cosθ + isinθ = 1

θ=0

k = 1: 3θ = 2π

θ = 2π/3

z = cos 2π/3 + isin 2π/3

= -1/2 + i √3/2

k = 2: 3θ = 4π

θ = 4π/3

z = cos 4π/3 + isin 4π/3

= -1/2 - i √3/2

k = 3: 3θ = 6π

θ = 2π

z = cos 2π + isin 2π

=1

Get only 3 solutions to z3 =1.

Eg. The fourth roots of unity are the solutions of z4 =1.

z = r (cosθ + isinθ). Know r =1.

z4 = cos4θ + isin4θ = 1 = cos(0 + 2πk) + isin(0 + 2πk)

4θ = 0 ⇒ θ = 0.

4θ = 2π ⇒ θ = π/2

4θ = 4π ⇒ θ = π

4θ = 6π ⇒ θ = 3/2 π

Eg. The fifth roots of unity are the solutions of z5 =1.

z = r (cosθ + isinθ). Know r =1.

z5 = cos5θ + isin5θ = 1 = cos(0 + 2πk) + isin(0 + 2πk)

5θ = 0 ⇒ θ...

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