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MAT224H1 (51)
Sean Uppal (45)
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Department
Mathematics
Course Code
MAT224H1
Professor
Sean Uppal

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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 Moivres 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.
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Description
MAT224H1a.doc n Complex n-Space C , Complex Matrices, Spectral Theorem C OMPLEX N UMBERS Definition 1 = i is the imaginary unit. Definition a +bi,a,bR is a complex number (C). a is the real part, b is the imaginary part. Properties of a Complex Number 1) (a +bi = a +b i ) iff a = a and b = b. 2) (a +bi + a +b i = a + a + b+b i ) . 3) (a +bi a +b i = a a + b b i ) . 4) (a +bi a +b i = aa bb + ab +ba i ) . a bi aa + bb) (a b ab ) 5) a +b i = 2 2 + 2 2 i. a )+ b ( )( a )+ b ( ) 6) (a+bi ) is a real number iff b = 0 (R C ). Definition Let us denote z = a+bi . The conjugate of z is z = abi. The absolute value of z is z a +b . 2 Geometric Interpretation of a Complex Number z = a +bi R 2(a,b 2 2 The absolute value z = a +b is just the distance from imaginary (a,b to the origin. (a, b) z1 z2 is the distance from z1= a1 + b1i to z 2= a 2+ b2i . real Polar Coordinates a = rcos i b = rsin =z+ a bi = r cos + r sin i=)r cos(+ isin = r e . Note: r = z = a2 +b 2 . b Note: = arctan the argument of z. a Definition ei = cos + (sin)i . Page 1 of 17 www.notesolution.com MAT224H1a.doc Example Write z = 2 2i in polar form. z =re i Let . r = ( 2 2)+ 2 2( ) 8 = 2 2 . = arctan 2 =arctan = 3 2 4 . i3 z = 2+ 2i= 2 2 e 4 So . Theorem: Multiplication In Polar Coordinates i i i + If z1 = r1e 1 and z 2= r2e 21 are complex numbers in polar form, then z z1 2 = 1 2 e 1 2 . Proof: z z = r e i1 (r ei2 )= r r ei1e i2 . 1 2 1 2 1 2 Want: e i1ei2 = ei 1 +2 . ei1ei2= (cos 1+ isin 1)cos 2 + isin 2) = (cos 1cos 2 sin1 sin 2 + sin 1c) ( 2+ cos 1sin 2 i. ) = cos ( + +i)in + ( ) 1 2 1 2 = e (1+2 ) Theorem: De Moivres Theorem i n i n) If is any angle, then (e) = e holds for all integers n. Proof (sketch): If n 0 , use induction. n i n i If n < 0 , then n > 0 . So (e =) e ( . Since i 1 1 1 cos i sin i ) (e) = i = = cos isin = cos + )i sin (= e ) , so e cos +isin cos isin i n i )n (e) ( = e ) ,n < 0 . Example 3 Find ( + 3i . 2 i i3 Let 1 3i = re = 2 . 2 2 r = (1 )+ ( 3 = 2 . 2 = arctan() 3 = 3 . Page 2 of 17 www.notesolution.com
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