Department

Mathematics

Course Code

MAT224H1

Professor

Sean Uppal

MAT224H1a.doc

Page 1 of 17

Complex n-Space Cn, Complex Matrices, Spectral Theorem

COMPLEX NUMBERS

Definition

i=−1 is the imaginary unit.

Definition

R

∈

+

babia,, is a complex number (C). a is the real part, b is the imaginary part.

Properties of a Complex Number

1)

(

)

(

)

ibabia

′

+

′

=

+

iff aa

′

=

and bb

′

=

.

2)

(

)

(

)

(

)

(

)

ibbaaibabia

′

+

+

′

+

=

′

+

′

+

+

.

3)

(

)

(

)

(

)

(

)

ibbaaibabia

′

−

+

′

−

=

′

+

′

−

+

.

4)

(

)

(

)

(

)

(

)

iabbabbaaibabia

′

+

′

+

′

−

′

=

′

+

′

+

.

5)

(

)

( ) ( )

(

)

( ) ( )

i

ba

baba

ba

bbaa

iba

bia

2222 ′

+

′

′

−

′

+

′

+

′

′

+

′

=

′

+

′

+.

6)

(

)

bia

+

is a real number iff 0

=

b (

C

R

⊂

).

Definition

Let us denote biaz

+

=

. The conjugate of z is biaz

−

=

. The absolute value of z is 22 baz+− .

Geometric Interpretation of a Complex Number

(

)

babiaz,

2

R⇔+=

• The absolute value 22 baz+= is just the distance from

(

)

ba, to the origin.

• 21 zz − is the distance from ibaz111

+

=

to ibaz222

+

=

.

Polar Coordinates

•

( ) ( )

θ

θθθθ

θ

θ

i

eririrrbiaz

rb

ra⋅=+=+=+=

=

=sincossincos

sin

cos.

• Note: 22 bazr +== .

• Note: a

b

arctan=

θ

the argument of z.

Definition

(

)

iei

θθ

θ

sincos+= .

(a, b)

real

imaginary

www.notesolution.com

MAT224H1a.doc

Page 2 of 17

Example

Write iz22

+

−

=

in polar form.

• Let

θ

i

rez=.

•

( ) ( )

22822 22 ==+−=r.

•

( )

4

3

1arctan

2

2

arctan

π

θ

=−=

−=

.

• So 4

3

2222

π

i

eiz⋅=+−= .

Theorem: Multiplication In Polar Coordinates

If 1

11

θ

i

erz = and 21

22

θ

i

erz = are complex numbers in polar form, then

(

)

21

2121

ϑθ

+

=i

errzz .

Proof:

•

(

)

(

)

2121 212121

θθθθ

iiii eerrererzz ⋅== .

• Want:

(

)

2121

θθθθ

+

=iii eee .

•

(

)

(

)

( ) ( )

( ) ( )

( )

21

21

2121

21212121

2211

sincos

sincoscossinsinsincoscos

sincossincos

θθ

θθ

θθθθ

θθθθθθθθ

θθθθ

+

=

+++=

++−=

++=

i

ii

e

i

i

iiee

.

Theorem: De Moivre’s Theorem

If

θ

is any angle, then

(

)

( )

θθ

ni

n

iee = holds for all integers n.

Proof (sketch):

• If 0

≥

n, use induction.

• If 0

<

n, then 0

>

−

n. So

( ) ( )

n

i

n

iee

−

−

=1

θθ

. Since

(

)

( ) ( )

( )

θ

θ

θθθθθ

θθ

θθ

θθ

−

−=−+−=−=

−

−

⋅

+

== i

i

ieii

i

i

i

e

esincossincos

sincos

sincos

sincos

11

1, so

(

)

( )

(

)

0,<= −

−nee n

i

n

i

θθ

.

Example

Find

(

)

3

31 i+− .

• Let 3

2

231

π

θ

i

ierei==+− .

•

( )

(

)

231 2

2=+−=r.

•

(

)

3

2

3arctan

π

θ

=−= .

www.notesolution.com

MAT224H1a.doc

Page 3 of 17

• So

( )

88231 2

3

3

2

3==

=+−

π

π

i

i

eei.

Theorem: The nth Root of Unity

If 1

≥

n is an integer, the nth root of unity (the complex numbers z such that

1

=

n

z

) are given by

1,,1,0,

2−== nkezn

k

i

π

.

Proof:

• Let us denote

θ

i

rez=.

• We want r and

θ

such that

( ) ( )( ) ( )

( )

=⇔=

===

=+=

n

k

nr

rrnr

ninrern

nn

ninn

π

θθ

θ

θθ

θ

2

0sin

111cos

1sincos1.

• When 0

=

k, 0

=

θ

, and when nk

=

, 02

=

=

π

θ

. So 1,,0

−

=

nk.

POLYNOMIALS

• Notice that the roots of cbxax++

2 are i

a

bac

a

b

x2

4

2

2

1

−

+

−

= and i

a

bac

a

b

x2

4

2

2

2

−

−

−

=. x2 is the

conjugate of x1, that is 12 xx =.

Complex Polynomials

Let us take

(

)

C∈++= wuwuxxxp,,

2 a quadratic polynomial. Let us assume u1, u2 are the roots of

(

)

xp.

Then:

• 1

u may not be equal to 2

u.

• uuu

−

=

+

21 .

• wuu

=

⋅

21 .

Theorem: Fundamental Theorem of Algebra

Every complex polynomial

(

)

xp of degree 1

≥

n has the form

(

)

(

)

(

)

n

uxuxuxp

−

−

⋅

=

1, where u,

u1,…, un are complex numbers (0

≠

i

u). The numbers u1,…, un are the roots of

(

)

xp. u is the coefficient of

xn.

Corollary

Every polynomial

(

)

xp of positive degree with real coefficients can be factored as a product of linear and

irreducible quadratic factors (in real numbers).

Example

( ) ( )

( )

∏∏ ==

++⋅−=

1

1

2

1

m

j

jj

m

i

icbxaxxp. Using previous theorem,

( ) ( )

( )

( )

∏∏∏ ===

−⋅−⋅−=

11

111

m

j

j

m

j

j

m

i

idxdxaxxp.

www.notesolution.com

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