Complex Numbers
imaginary unit
it
I
by
i
denoted as
which
is
defined
its property
to complex numbers
numbers
be used to extend reel
can
by
using addition and multiplication
powers of i
in
ison
i
ciyiow.cn
her Number
i3
Y
iz i
i
i
ice
Thi
the
PI
a
real
of
b
i
i
i
i
of the form
number
where
fycoco
are
i
atibT
Z
reel numbers
all
pat and the imaginary
2
respectively
The imaginary port Does Not
include the imaginary unit
i
Below
3
I
some
are
examples of complex numbers
C Red port
4
imaginary part
3
4
21 5 i
7
8
4
part
0
imaginary port
red part
8
imaginary part
red
Gi
3
6
u
SET of Numbers
complex Numbers
Exempts
1
It
Solve
o
Xx I
Solution
compare
b
to
9
a
l
tea c
there
is No solution here
if
belongs to Reel numbers
7 bx
b
1C
L
CT
0
C
1
4G
3
The equation hes
i
co
No
reel number solution
2
It
Solve
if
so
xx
belongs to complex
numbers
Solution
a
x
l
b
I
C
I
ITE
b
12
2cg
53ft
II
2
fs z
i
I 1
I
Gi
Z
Z
Ei
EqueCityofcomplexNumt
Two
if
complex
only if
and
Z
numbers
and
2
refuel
Reczz
Recze
to each other
In
aired
Example
let 2
Find
p
2
Zz
8
if
of P and q
value
the
Qi
t
peg
2
Zz
Solution
Re
Tr
Retta
Zi
th
a
p
E
z
p
g
peg
z
6
sp
t
p
From
3
8
Z
p
2
3
I
Inte
ptg
e
ImAy
Addition
of complex numbers
Subtraction
and
Example
let
2 13
Z
2
Find
3
e
tar
a
as
lb
272
32
Solution
9
Z
12
2
2
13
5
b
32
22
3
t
ti
i
3
2
or
2t3i
6
I
9
O
Hi
2
G
t
171
na
3
18
ei
Multiplicationofcompleanumi
EM
Gt3
3
Si
6
ei
9i
I
iz
18 t i
Conjugateofacomplexnumbery
The conjugate of
E
is
given
2
at
denoted
it
axis
by E
at
as
fists
Example
374
In
Fto
I
z
3
Gi
3
5
5
Fi
5
5
Toti
X
wrong
7C i
7 its
7
15
its
Theorem
e
I
Tf
Z
real
a
E
Z
or
then
2
is
number
Example
e
Determine
ei
2
whether
7
e
i
is
Solution
I
I
2C i
e
zi
me
21
ei
e
i
Z
21
So
2
ei
I
e
i
is
a
real
number
a
real number
Exercise
at
z
b
at
Compute
and
I
2
simplify
Solution
z
I
Cat i
of
A
ib
iab
iab
ib
cancelled
Itb
n
General
Non negative reel number
ealpaA5t
z.z
2mport
Divisionofcompleanumtersy
at
youneed
ib
I d
to eliminate
i from the denominator
C do
something
A
t
i
B
C standard form
atib
at
ib
I
C e id
ib
a
Ct
c
id
c
id
e
id
A
conjugate of the
denominator
rn
Carib G id
d
Et
2
I
iadxibc
ac
i
b Et _agIyd.ti
CEIL
A
t
i B
Example
3
2
i
2
3t
z
5i
2
5
i
155
i
It
C
55
Igi
Exercise
e
I
x
i
23 i
f
X
3
i
2
4
Let 2 a
2
if Lz
in
5
3i
tf
2
I 12
n
Izz
Solve for
Find
2
Leave your answer
form of
the
i
b
at
2
32
2
1
2
Leave your answer
I
in
tri
the
form of
at
b
Solution
IIIT
i
i
e
C
3i
z
3
Z
5 135
73
3
3 145
3
3 4
Ii
zti
Ii
6
31 t
C
reel
at
IT
z
sit
4C y
15
imaginary part
Ii
21
2
3i
2
Iz
I
Z
z
1
n
21
I
2
1
2
22
21
Zz
21.72
2
172
k
3ij.CH
3
i
3i
Zt4i
6C c
i
3
3ti
Sti
Fi
3ti
mt8it3i
3
C
Tft
I
if
i
5
32
1
I 121
2 g 2
32
1 I
Ita
7
2
32
22
2
2
z
z
42
2
2
ziz
Z
14in
I
iz
2
4
I
14
Itai
zi
Itai
Z
Zizi
2
Zi
2ft
i
21
ArgandDiagramy
An argand diagram
of
c
complex numbers
points
on
a
is
a
as
vectors
plot
complex plane
Imaginary axis
complex plane
q
Example
Construct
2
Argand diagram for
an
3
4 i
Im
In
s
For this example
i
i
i
4
7
6
3 4i
2
3
Recz
3
r
3 14in
z
i
Re
r
s
r
O
Iz l
arg.cz
o
modulus of Z
argument of 2
iso
Iai Est
126 870
Exercise
Zi
let
let 2
Construct
Argand diagram for
an
Solution
2
iz
and
iz
2
I
Im
4
A
y
3
j
et Zi
µ
I
g
z
i
r
l
i
y
i
i
Re
i
a
z
ai
t 2C
a
z
n
ETT
I
ie 12
i
it
e
Y
y
zi
q
Notes
nectar
the
is obtained
of iz
by rotating
the vector
7 900 in counterclockwise
the
vector
about
of
the
E
Real
is
the
axis
reflection of
vector
af I
f
z
PolarformofacomplexNumt
2
x
Im
a
standard form
ing
tf
coso
where
if
121
o is in
188 tan Bal
O
1800
368
0
Er
x
5
D
andi
if
tan Bal
tan 1
1
QI
o is in
if
if
Sino
G is
r
QE
in
2
it
2
i
into
QTL
polar form
I
µ
o
E
e5
re
is
I 6 570
I
Nfp
T
QI
0 is in
rc
Example
Convert
y
I
poterorm
tan
i
i rsino
reoso
r
2
Ft
Now
2
C
i
2
r
cosotisino
G
form
standard
I t
cos cab 570
i sin
t
u6
SF
Exercia
Convert
2
i
3i
into
polar form
Solution
r
Em
C
To
157 C 35
a
fit
O
2
µ
l hi
re
tariff
1
251.570
3
in
2
l 3 i
2
r
Caso
fo
cos
standard form
ti Sino
1.570 c is incest 570
polar form
MulliplicationandDivisioninpolarformy
Gf
Z
Zz
r
Corso
rz
r
2 i Zz
ascent
Caso
r h
r
i Sinor
AT
cozoitisino
rare
two complex
Numbers
cos
cosontisinoy
Ee
L
isino cosy
Sinoisin
Sirico Or
Oy
t
tor
i sin
Og
1
I
measles
multiply
thelengths
Similarly
cos
when divide
complex numbers
0
tiGinOicoso tcoso.sinoD
coso.coson sinaiSin0
cosCOc
Emling
Coosontisino
r
ti Sina
cosacosozticoso since
rz
r.ru
2 i Zz
Fol
isin Ol
t
o
Y Yeah
A
tisinco
Or
Subtract the angles
Eixample
I t z
2
let
l
3
2c
22
2
a
Compute
final answers
the
convert
Zi
calculate
Re
and
using polar form formula Then
2
and
the
verify
and
Zn
answers
standard form
to
2
Yz
using
obtained in
standard form
port
a
Solution
e
a
i
Z
12
G
2
n
isinccib.SI
cosccib.s77
Si
l
l 57YtisinCnT
foCco3
57o Zi.Z
Js Io
7
Iz
t
I i
an
Eg
0.5
i sin 368.1403
68.140
cos
C
of
O 5
i
a
Isin
C
135
s
b
Zi
C
Zz
I
fl
e
6C
Zi
i
L
C
121
I
3i
i
et
t3i
7
Z
7C
zi
it3i
3i
i 13 i
t
1ft C 35
l
fo
10
o 5
0.5
Exercise
7
let
cos 1230
24
2
2
2
compute
answer
cos 120 t
3
2
cos 210
2
3
t
z
to
i sin in30
isin ni
t
isin 210
using polar form
standard form
then
convert
the
fine
Powers of
let
complex number
a
2
r
Casa
eisino
then
I
Z
Z
r r
f
I
3
2
2
2
r
73
Oto
cos
Oton
tisin
coke tisintroD
z
cos
r
r3
Goto
costa
a
isin Cueto
Tisin Ceo
a
rnccoscnoj
Y
lznn
De
isir.cn
Moivre's
theorem
5
Example
let
2
compute
G
29
i
by
using De Moine's theorem
Solution
Izu
Jt
r
4
2
360
0
Re
G i
C
15
z
I
tan
3300
2
2
cos 330
so
79
29
t
isin 335
cos 3300
0
512
9
t
3300
isin
97
t i ca
512 i
solution
a
Exercise
if
2
l
cos
aft
zone I'm
osfesox.ro cisinfefx.oov
i
i
I
isin450
oi
nth roots
of
let say
complex number
a
4
X
then
Tx
Now
became
z
let say
2
Gi
742
because
4
1
then
zte
free
X
Wrong
Crete 3
8 Tfi 116L l
Catti
4
12 1lb
4i
Im
2
4
accossootisingoo
ne
Suppose
then
Y
sac
SGT
524
494
4900
490C
5
2h
1
i
5
e
54
a
go
sac
E
54
G
Gi
l
ee
SG
suppose
then
G
368
y
i
t
450
it
e
2
is in
T
Cosas
2
9
yay
s
J
5K
E
4
the
5
900 3600
22
45
2
i
Czk
a
54
5
4
11800
SG
k
See
2
z
cosmic
z
Ig
f
t
is in 2257
i
f ta
Gi
Exercise
Find
all
the
3
answers
of
cube roots
of
8
Solution
e
8 1 Oi
2
f
Seol
2
243
z
843
reccesfoot is.info
Zm
G
I t
FTpe
8
St
2
ft't
i
8
0
2
6
2Castfootisinis
2
In
Sto
z
re
1
8
i
e
si
Goo 100
2
Zon
23
2
Cassweisin3od
l
Re
Gi
Is
E
T.EE Iire
3
3
236
nth roots
let
of
a
complex number
rccosotisino
2
GO
then
for
Z'T
ve
any
O t
rt
h
integer
368k
K
0,1 2
r s
h
l
r
n
1
OR
Zk
2TK
O 1
r'T
K
o I
2
Example
Given
Find
all
86 i
8
2
the
4 values
244
of
Solution
r
O
t
JC sj
boot
o
f.SE
JTT
tariff
l
neo
2
Sfi
re
z
0
if
g
16
Bared
on
2 th
Formia
the
OT 3600k
rt
0
I 2
r
n
n
i
2
3
Now
zke
240013600K
16
k
o
a
i
a
neo 136043
2
26
Im
zoos
a
Helu
hey
4
9
z
Re
AT
605
z
Tsin 657
Tsin 4507
Casa
isinero
zoos
2
2M
2
Cars
I
Tsin 6357
r
I t
Gi
l
Gi
It
Gi
l
Gi
Form
Exponential
z
xx
Euler Identity
standard form
iy
r
2
cosoeisino
polar form
Itis
where O is in
ra
eio
2
re
O
C exponential Form
Example
let
3tei
2
Express
in
2
Form
exponential
Solution
r
p
o
1 3547
Iso
tan
in
870
126.87
5
1 11
degrees
93
thou
z
3T 41
5
e.io
exponential Form
2
72
radian
Exercise
f
Compute
2
of
value
the
compute
of
value
the
t l
e
ans
i
e
n
z
Leave your
Form
Using exponential
in
terms
let
r
2
2
Zi
derive the formulae
and powers
of complex numbers
can
modulus and
its
of
ion
he
r
Zz
t
arguments
il
e
u
e
O
reei0nam.an_amt
Cri.ryeiO.eiO
i.ryeioitid
Zc.z
eiC0
Cir
when multiply
two complex
numbers
E
easily
we
division
for multiplication
Ars
standard form
in
answers
o
op
multiply
the length
add the angles
7
AM
O
he
Fei
ioi
e
r
a
m
ein
o
2
e
z
I
subtract the angles
when divide
divide
complex
numbers
Lef
re
z
z
I
2
O
then
Creign
ioyn
r
rheino
n
when rise z
to the
the lengths
nth
power
t
µ
rise
r
to the
nth power
multiply 0
by
n
n
ExpansionofsincnojandaBCnosinternsefcosoania
Sino
De Moi ure's
By
cos Cno
theorem
isino
of
T isin nd
Binomial Theorem
corot
nc
ncr
cos Cno
as Cna
Re
sirena
In
t
oI rCisinoJt.r
since
i Im
Re
tisincno
of Lisino
Example
Express
sin
o
and cosGo
in terms
of
Cera and or Sino
Solution
cosC3O3tisinL3oj
Ccosotisinuj
CcosoPt3G WCis.no
t
t
34600 Cising
nvm
Cisnop
coda
3 as
3 usoc since5
of isino
Goto 3cososino
ti
3030Sino
is Po
Sino
cos
3coso Sino
coda
o
3cosoci code
as3o
a
Sin za
cop a
3C DO
Sino
3,50
Sino
3
Sino since since
i
cesin30
3 Sino
Exercise
Express Sinceto
Ars
of
Sino and
or
coso
8 Coto 80050 11
CBCleo
atb
in terms
y cofosino ecososin30
Sin yo
Binomial
and eosCeo
Mr
Theorem
nco ambo
nc
ncr an rb
a
t
ng
a
bn
an b
t
t
t
b't
ncaa
NG
a
b
Expreesigapowerofcosoorsinoarasumofter
containingthecosinesandsinescfmultipleangi
Suppose
cosotisino
2
cero
1
isin cosotisino
Cosa isino
Iz
isi.no
ceso
co5he
i since
ceso
2
t
t'z
BO
ri since
Cosotisino
zn
2
nose
a
tz
z
sinU
n
cos
t
a
De Maure's theorem
i sincno
n
z
coho
1
Cisco
I
Zn
isin og
cos Cno
isin
Cesaro
no
isin o
isin Go
2 ht
t
Zh
In
2
Iz
zcoso
Z
tz
zisino
2
t
Za
zn
Iz
cosCno
a
Zi Sir no
zascno
2isincno
Example
Find
an
expression for coda
of cosinesof
in terms
multiple
angles
Solution
Caso
I
coda
2Gt'z
tg 3Gz3
tg
3
2
t 3C
32
1 3
3
E
t
3
t3GGIGzjt3g jf
3
f Est
3GHz
t
3
Leo
Ig
t.no
3
2 cos 30
tacos
o
zcoso
Zg COO
t
Exercise
Find
an
exprees
on
for Sinko
terms
in
of
cosines
of multipleangles
Solution
e
since
Gino
z.CZ
K.EC I 5
ztj
eGEC
4674ftj
64 EC EY
IT
2
4
ez't
If
cosCeo
f cosGeo
E
zn
Eez
e
IT Zetzte
e
6
e
Gooseo
CH
t
Ig
Ie
t
t
6
6
4
z'ftp
Exercise
Find
an
exprees
on
for
5 n5o in
terms
sines of multipleangles
of
Solution
Sino
Coin of
Ii La ED
2
5
3
5
75 tfs
2isin Go
Ty Sin
not
52
to
5
o
Es Es
s
Iz
t
5
t.dz
zisin Go
Eg Sinko
Is
Is
t
to
Ig since
Gisino
Loci Problems
Loci
plural of locus
The locus is
a
For example
a'is
set of
the locus
complex numbers
this would be
satisfies
refer
such that
to
a
9
where
set of
points
the modulus
circle of radiusF
a
certain condition
a
12 1
the equation
of
red number
eve
a
points that
a
equals to
centered at o.o
In
a
Ca
a
otai
Eat
Gtz
Aqi
the
q Ei
o
V
ai
_Xtiy
l ZI E
a
jay
lo
12 139
Im
a
Re
a
xtigs
545
th
a
Re
a
a
IZ
Zo
a
point 2 and point
fixed point let say Zo
distance between
Zo
is
2
is
a
any
that
complex number
2o
ht i k
Satifies
Zn
i
IZ
locus of
the
aaron
the
win
equations
2 ol
main
µ
Z
let 2
I
a
Zo
xtiy
Xt i
g
Ch ti k l
a
i
X HJ
H
Cy
KJ
hJ Cy k5
a
a
a
is
Example
in
Write down the equation
b
I
2
X 3
t
i
J
I
2
3ft
Z
I
distance btw 2
and
3
ti
3
tyJ
X
Zi
the locus defined
3
F
J
3
ly
J of
3
I
3 ti
iy
it
l
3 ti
X and
y
3
15
Z
32
Za
and z
distance between 2
where
2
Zi
2
is
are
u
any
2
fixed points
the
I
Im
Z
Ct
id
the egnetion
satisfies
point that
u
X
2
a
locus of
Zit H
if
perpendicular
bisector
and
i
Z
of point
point
2
z
A
a
The
Example
Find
of
the
Iz
quota
3
tail
in
and y
X
I
2 t
l
of
3
the
I
Coas
Z
Solution
Iz
3 zi
I
I
C
2
it 3
1
ZM
n
1
Ex Fo
y
3
n
i
12 11
C fl T
re
3 27
12 31 251
3
xtiy
37
12
ril
1
3
lxtiyti
3i cx
i.ly 1231
2T
JRet
jcx
1
5 G.tn
35 4 25
Xxl
icy
D1
JReiI
It
6 5
Cy
1
35
Cy
35
31
I
I
txt frug
SX
te
It
ti
t
51 by
3
toy
Fo
Ix
y
2 27
12 271 12
In
a
a
iii
17
Z
I
f
distance between
2
and z
I
t
E
Zn
l
Hmu between
2 and 2
en
if
Zet
Z
the locus
left
2
a
circle
is
21
where
k
1
K
Example
e
Determine the locus
Z
of
if
212
1
i
1
Solution
2
il
xtiy
lxtiyti cx
iyf
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