COMPLEX NUMBERS
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Complex.nb
1
COMPLEX NUMBERS
Consider the quadratic equation;
x2
1
0
It has no solutions in the real number system since
x2
j
Similarly x 2
1
or
x
1
ie.
j2
16
1
j
1
gives x
0
16
or x
16
Powers of j
j
j2
j3
j4
1
1
j 2. j
j.j 3
j
j
j
j5
j4j
j6
j 4. j 2
j2
j7
j 4. j 3
j3
!
j2
1
j
Clearly all the powers of j can be described by j, j 2 , j 3 , j 4.
e.g. j 2307
#
j 2304. j 3
"
$
j4
576.
%
'
)
j
&
'
j
(
1
4j
Complex.nb
2
General form of complex numbers
Consider now the quadratic equation
x2
2x
2
x
10
4
0
b2
4 ac
40
2
2
36
2
2
6
j
2
1
2
3j
This leads us t oa general form for complex numbers
z a bj
a and b are real
a real part of z
imaginary part of z Im z
b
Re z
In the previous example;
roots
z
hence a
"
#
1, b
"
$
a
a
&
4, 2
7 j, 16 j are all special cases
&
bj is purely real if b 0 ,
a .
bj is purely imaginary if a 0 ,
bj .
'
+
%
3 j,
!
3
For complex numbers :
3 2 j,
%
1
(
,
)
*
-
.
/
Complex.nb
3
Basic Rules of algebra :
Consider :
z1 a1 b1 j,
z2 a2 b2 j
z1
Sum :
Difference : z1
z2
z2
a1
a1
a2
a2
b1
b1
b2 j
b2 j
Example :
for : z1
z1
z1
z2
z2
3
Product :
3
3
4 j and : z2
z1. z2
4
4
4
4
7j
7 j
7 j
4
1 3j
7 11 j
a1 b1 j a2 b2 j
a1 a2
a1 b2 j
b1 a2 j
b1 b2 j2
a1 a2
b1 b2
a1 b2 a2 b1 j
Example :
if z1
2
3 j and z2
1
j find
z1 2 z2
z1 2 z2 1
2 z1 z2
z1
2 2 3
2 3 j
2
2 1 5j
2 3j
4 7j
3j
1
Complex.nb
4
Complex Conjugate :
For any complex number a
a
bj there corresponds a complex number
bj
obtained by changing the sign of the "imarigary part".
a
bj is the Complex Conjugate of a
Notation :
If
a
then
bj
z
_
Clearly
and
hence
z
z z
z z
2a
2 bj,
1
a
_
z
z
Re z
2
1
b
_
z
z
Im z
2
Also
zz
a
bj
hence
zz
a2
a bj
a2
j2 b2 ,
!
!
b2 ,
a
bj
bj.
Complex.nb
5
Quotient :
z1
To find
where z1
z2
z2
z1
z1
z2
z1 z2
z2
a2
a1
z2
b1 j,
a1 b1 j a2 b2 j a2 b2 j a2 b2 j b2 j
z2
z2 z2
a1 a2 b1 b2 b1 a2 a1 b2 j
a2 2
a1 a2 b1 b2
a2 2
a2 b1 a1 b2 j
b2 2
b2 2 j2
a2 2
b2 2
Complex number in general form
z1
NB : to find the quotient in general a bj form we multiply above
z2
and below by the complex conjugate of the denominator
ie. by z2
Equality
Let
and
z1
z2
then
z1
ie
and
a1
a2
b1 j
b2 j
z2 when
Re z1 Im z1 a1
a2
and
b1
b2
Re z2 Im z2 Complex Number zero :
0 0j
a
0 and b
0!
ie. both real part and imaginary
part are zero.
Example :
given that z1
z1
,
z2
-
3
find the real numbers " and #
( j and z2 ) * + 5 j.
&
3'
.
/
0
1 2
3
54 j 5 0
and
if 3 6
7
8
:
9
such that z1
$
0 or 7 8
5 ; 0 or <
5
3
9
=
z2
%
0,
Complex.nb
6
For complex numbers,
z1 , z2 and z3 :
Commutative property :
z1
z2
z2
z1
z1 z2
and
z2 z1
Associative property :
z1
z2
z3 z1
z2
z1 z2 z3 z1 z2 z3
z3
and
Distributive property :
z1 z2
z3
z1 z2
z2 z3
Geometrical Representation:
A complex number a b j can be represented either as a point a, b
or as a position vector of a, b in the x y plane complex plane
or z plane.
x axis is called real axis
y axis is called imaginary axis.
This geographical representation is called
Argand Diagram
Example :
Modulus of a complex number
z
Let
a
bj, the modulus of z is denoted by z and
z
Example :
z
!
'
*
z
,
-
Also
!
!
(
(
!
(
!
(
!
(
(
)
!
!
!
!
!
!
!
"
(
b2
#
!
3
!
!
$
!
!
!
4
!
!
!
4j
2
!
!
%
16
(
(
(
(
(
25
(
(
(
(
(
(
5
*
,
(
(
For z
9
(
+
32
&
a2
length of the vector for z and is always
zz
/
0
1
a
bj a
a2 b2
z 2
2
3
8
7
7
9
9
4
5
bj
6
.
0
Complex.nb
7
Polar form of z
Point x, y represents z
thus
z
r cos
r cos
Here
r
x2
r cos , y
y2
z
#
$
Clearly r2
)
%
&
x2
'
2k
tan
+
,
x
1
./0
1
1
1
1
1
0
-
y
x
Also cos
7
5
7
7
7
x
7
7
6
7
7
7
7
7
7
7
7
7
7
7
8
9
9
9
9
9
9
9
9
7
?
?
?
9
7
7
9
7
9
x2
:
y
?
=
7
7
7
7
7
7
7
7
7
7
7
7
7
7
6
r
sin
9
9
9
9
9
9
9
9
9
y2
;
9
<
y
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
>
A
r
!
arg z
k an integer produces
gives the principle value of arg z .
(
y2 and
*
arg z is not unique since
The
another value of arg z.
However
r sin
This is called the polar form of z
is called the "argument of z" denoted by
yj, where x
j r sin
j sin
x
B
B
B
@
C
B
B
B
B
B
x2
B
B
B
D
B
B
B
B
B
B
B
y2
B
B
E
B
?
?
23
4
4
"
Complex.nb
8
Determination of pricipal
Argument of z :
1.
z
x
jy in first quadrant x
arg z
tan
x
1
z
r
2
determine the modulus and argument of
32j
z
3
!
2j
"
(
arg z
,
0
#
$
+
0
y
Example :
0, Y
-
/
*
2
1
tan
.
)
4
4
4
56
&
&
&
&
&
&
&
&
&
&
'
22
&
&
&
&
&
13
*
*
*
*
*
*
*
7
7
8
0.588 radians
123
4
&
32
%
4
3
0
3
2. z
9
x
zy in second quadrant x
:
;
?
0, y
<
arg z
@
0
=
A
B
>
tan
C
D
x
1
FGH
I
I
I
I
I
H
E
y
Example : Determine
r
Y
Z
z
Z
z and arg z for z
S
S
U
V
W
1
X
j
[
Z
1
\
j
]
b
_
e
T
f
g
a
a
a
a
^
12
b
a
a
a
a
a
a
a
a
a
a
a
c
a
a
a
a
d
12
a
a
a
a
a
2
g
g
g
1
jkl
l
Since tan
`
1
op
q
q
r
tan
r
m
m
m
m
m
m
1
r
i
m
r
m
i
h
r
1
t
w
u
w
w
w
w
v
i
i
n
i
i
1
4
r
r
rr
s
i
ii
thus
3
x
y
arg z
z
{
|
}
~
4
4
rad.
JK
L
L
,
P
N
N
N
N
N
M
2
O
Q
R
Complex.nb
9
3. z
x
zy in third quadrant x
arg z
x
1
tan
0, y
0 :
,
y
Example
Determine the polar form of complex number
z
r
!
z
"
$
%
&
Also tan
6
*
7
,
*
*
'
'
'
%
<
<
<
7
7
'
'
'
$
2
-
*
*
*
*
*
62
*
*
*
*
*
*
*
*
.
*
*
*
*
*
/
*
*
0
*
*
*
*
1
*
*
*
2
*
2
*
3
*
3
*
3
*
*
*
*
*
3
4
7
tan
<
;
8
y
'
8
9:;
<
2 j
&
x
1
2 j
6
'
+
)
5
"
#
(
6
2
1
BCD
H
H
H
H
H
H
H
H
H
H
H
H
H
H
D
A
D
D
D
=>
?
E
F
G
I
J
K
6
?
@
G
G
G
K
K
K
LM
N
N
N
N
N
tan
O
1
1
QRS
T
T
T
T
T
T
T
T
T
T
T
S
P
S
U
V
V
V
V
SS
3
WX
Y
Y
Y
YY
Z
[
\
\
\
\
\
6
rad.
2
Complex.nb
10
Example :
z1
z2
Verify the inequailities
z1
z
Triangle inequality
2
5
Also
z1
z2
2
5j
2
29
5
2
$
%
z
1
!
z2
"
#
29
&
&
&
&
&
&
&
Polar form of z :
5
z
'
(
)
8 cos
)
)
)
0
*
0
0
0
12
5
3
3
4
/
0
0
0
0
j sin
0
+,-
-
:
:
:
7
5
@
A
8 cos
A
A
B
G
G
G
G
HI
5
J
J
G
G
G
G
K
j sin
G
CDE
P
P
P
x
>
:
P
QR
S
S
T
O
P
P
P
P
P
6
jy in fourth quadrant
V
:
N
6
U
:
LMN
E
4. z
=
=
:
6
F
A
;<
9
:
8
6
?
:
567
.
W
x
X
, y
Y
0 :
Z
[
x
def
f
\
arg
]
^
z
_
`
tan
a
1
hi
j
m
j
k
k
k
g
g
g
g
g
n
y
c
k
k
kk
l
c
cc
find the priciple argument of z
x
{|}
}
r
arg
u
z
v
w
1
tan
x
1
r
s
z
~
~
~
~
~
z
y
z
y
z
z
z
z
zz
1
tan
1
1
4
rad.
n
n
n
k
c
c
t
n
c
c
Example :
0
k
c
b
tan
1
1
j.
2
o
p
q
Complex.nb
11
Exponential form of z :
Recall
refer to "Tables of mathmatical formulas"
x2
ex
1
x
2
Let x
j
e
j
j
1
1
j
1
Thus z
z
5
0
r
r ej
6
1
cos
2
3
...
4
3
j
4
j
3
4
...
2
4
4
3
%
...
j
4
"
"
has a new form
the exponential form
&
&
&
&
&
&
&
5
%
&
&
&
&
&
&
&
&
...
&
!"
"
/
jsin
2
jsin
2
.
3
-
cos
2
2
,
x4
j
x3
#
$
3
'
(
5
)*
+
+
+
+
'
(
Complex.nb
12
Important Points :
1. For the correct value in exponential form
in radians not in degrees.
2. Negative
ej is measured in the clockwise direction.
Replacing by in
cos we get
e
ej or
jsin j
3. ej and e z
must be
cos
cos j
sin
,
jsin are complex cojugates hence
rej
and
Consider ej" #
e + j,
z
cos $ %
cos .
-
&
/
re !
sin ( )
sin 2
'
0
j"
1
and
*
3
4
Adding and substracting
ej 5
j5
e6
1
7
2 j sin 3
gives two important results :
1
cos 8
9
: : : ::
;
ej <
=
e>
j?
@
D
ej E
2
1
sin A
B
C C C CC
F
jH
eG
I
2
Log of complex numbers :
rej 2 k
Let z
J
ln z
S
[
b
K L
M
N
ln T rej U V W 2 kX Y Z
ln r \ j ] ^ _ 2 k `
ln r c j d e c 2 k f a
The principal value of ln z is
O
a
P
k
0, 1, 2, ...
Q
ln e
ln z
ln r
g
h
ji
j
k
g
Example :
Principal value of
is
ln 2 ej
l
ln 2
u
j
opn o o o
m
w v w w ww
4
4
q
2 kr
0.693
x
y
s
t
0.785 j
0k
R
Complex.nb
13
Multiplication in polar form
Let z1
r1 cos 1 j sin 1 z2 r2 cos 2 j sin 2
then
r1 r2 cos z1 z2
1
r1 r2 cos j sin cos 1
r1 r2 cos 1
2
1
cos sin 1 sin 2
j sin 1 cos 2 jsin r1 r2 e j z1
#### ###
Similarly
r1 &
%%%% %%%
$
z2
'(
cos
r2
2
e 4 j5
3 3 3 r3 3 13 3
r2
1
!
16
1
cos 2
1
sin 2 2"
) * 2 + , jsin - .
1
j sin 2 2
1
/ . 20 1
28
7
Clearly
9
z1 z2
9 :
A B
D D D D D D D D D D zD D D CD D D 1D D
EEE
EEE
EE
z2
and
arg z1 z2
arg
N N N zN N 1N N N
z2
:
; <
z=
>?> z2 @@@
@@@
@@
G H
H
D D D rD D 1D D
D D G D D D D H D D D zD D D ID D D1D D D HD D
F r2 F
zI 2
r1 r2
1
F J
1
K L
2
M arg z1 K arg z2
O P
1
Q R
2
S arg z1 Q arg z2
Example :
for
z1
T 2 U cos W V W W W
and z2 \ 3 ] cos _ ^ _ _ _
3
z1 z2 \
2a 3
b cos c e d e e e
4
X
4
`
j sin
Z YZ Z Z [
4
j sin _ ^ _ _ _
3
e de e e
k kk k
j sin i k j k k k
f 3g h
m n o
4 l 3
wx y z
7 ~
p 3 q cos rst v 7v v v uv v v
j sin {|}
12
z
and
1
z2
2
j sin
3
cos
12
12
[
2
Complex.nb
14
De Moivre ' s Theorem :
If
z1
z2
zn
r1 ej
r2 ej
rn ej
1
2
n
then
z1 z2 zn
r1 r2 rn ej 1
r1 r2 rn cos
..
2
n
1
2
.
n
j sin
Set
z1
then,
zn
z2 . ..
zn
z
r ej
rn cos
j sin n
n jn
r e
rn cos n
j sin n
!
"
!
r cos
j sin
#
%
$
$
!
Set r
"
!
#
1 to give De Moivre' s Theorem
$
n
&
cos
'
j sin
(
)
+
cos n
,
*
-
j sin n
.
Application :
If
z
1
1 then z
1
2
1
cos
3
1
4
j sin
5
cos
4
j sin
9
;
ej
3
<
8
7
7
7
7
7
7
z
7
7
7
7
7
7
7
7
7
7
cos
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
j sin
9
:
cos
7
<
7
=
=
=
=
=
=
=
=
=
cos
9
j sin
;
7
=
<
=
=
=
=
=
=
;
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
<
=
=
=
=
=
=
2
cos2
cos
<
sin
?
<
<
j sin
;
<
>
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
1
e
>
zn
Hence
and
z
j
n
A
@
ejn
B
e
B
@
cos n
B
A
jn
@
j sin n
D
cos n
B
A
C
C
E
Adding and subtracting gives
2 cos n
and
zn
B
C
2 j sin n
B
C
1
F
F
F
F
F
F
F
D
zn
zn
1
F
F
F
F
F
F
F
E
zn
=
=
=
=
=
=
j sin
>
=
=
C
j sin n
C
=
=
=
<
=
=
6
/
0
1
2
.
n
Complex.nb
15
Example :
Evaluate
3
i
cos
j sin
6
ii
1
j
6
25
,
1
iii
.
1
j
Solution :
8
By De Moivre ' s Theorem
3
i
cos
j sin
6
cos 3
6
j sin 3
6
cos
!
j sin
2
6
2
0
!
j
j
"
!
ii
Let z
#
1
$
j, then
%
and
*
r
arg z
+
$
tan
+
z
&
1
&
'
-
(
2
)
)
)
)
.
/
/
/
/
/
,
4
25
Thus z25
0
1
2
3
2
3
3
cos
3
4
6
6
6
6
j sin
6
5
4
25
<
=
>
2
>
>
25
cos
?
D
D
D
D
D
2
L
25
2
L
L
L
L
L
9
9
:
:
25
C
D
D
D
D
D
D
j sin
D
4
K
9
4
@AB
>
9
8
7
B
;
9
G
G
G
G
G
HI
J
J
F
G
G
G
G
G
G
4
E
U
L
M
cos
M
6
Q
N
Q
O
Q
Q
j sin
Q
P
4
W
G
212
X
2
Y
Y
Y
Y
Z
cos
\
\
R
\
\
6
T
V
S
S
j sin
\
[
4
_
_
_
_
_
^
4
]
`
qr
12
2
a
1
b
c
2
c
c
g
g
g
g
g
g
i
i
y
Set z
{
1
j thus from ii
|
}
and
r
arg z
g
j
g
i
n
n
y
tanh
1
2
4
Thus
8
1
z
z8
8
2
cos
j sin
4
4
p
s
s
s
s
s
s
ss
ss
tu
n
n
n
p
p
2
mm
n
p
j
z
n
o
2
j
n
klm
m
x
n
i
m
212 1
~
g
f
w
iii
g
f
ff
z
g
def
h
v
1
c
n
n
V
V
V
V
U
4
R
R
Complex.nb
16
8
8
2
cos
8
j sin
4
4
1
cos 2
24
"
j sin 2
#
$
%
!
1
&
'
'
'
'
'
'
'
16
Powers of cos
Example :
and sin
(
1
cos6
)
*
+
1
012
2
z
,-.
/
/
/
/
/
.
4
4
:
(
4
4
56
7
7
8
6
4
3
2
z
Binomial expansion gives
1
cos6
9
z6
:
;
;
;
;
;
;
;
26
1
E
D
D
D
6 z4
15 z2
=
H
D
D
1
FGH
C
D
=
=
20
15 z
=
2
6z
?
>
4
D
z6
KL
J
J
J
J
J
M
J
1
M
N
6 z4
I
J
TU
S
S
S
S
V
S
S
z6
1
V
W
15 z2
R
S
OPQ
64
b
b
b
b
b
2 cos 6
b
64
1
z4
Q
i
i
i
i
i
d
i
i
32
cos6
i
j
i
i
i
i
12 cos 4
e
f
30 cos 2
e
15
i
i
k
16
cos 4
i
j
i
i
i
i
i
i
k
32
5
cos 2
i
j
i
i
i
i
i
i
k
16
f
_
_
[
\
\
\
\
\
\
\
z2
Z
c
3
h
B
@
]^
[
XYZ
1
a
b
6
z
A
@
<
e
20
g
20
`
Complex.nb
17
Example :
Express cos 6
Example :
and sin 6
1
Sketch z
2
and evaluate z
3
z
1
4
arg z
j on an Argand Diagram
24
1
4
and sin
3
in terms of cos
1
6
Polar form of z is
z
1
cos
j sin
6
cos
$
"
"
"
"
j sin
"
!
6
e
$
Hence z
24
(
)
j
'
'
'
"
"
"
"
"
!
6
"
"
"
"
#
#
"
!
6
'
6
&
%
*
e
j
+
24
-
-
-
-
6
,
.
/
0
ej 4
1
2
cos 4
3
4
j sin 4
5
6
1
Complex.nb
18
Roots of complex numbers :
Recall de Moivre' s theorem;
For
z
r cos
j sin
zn
rn cos n
j sin n
fFor any value of n integer or fraction, positive or negative
This is a very important result.
When n is a fraction we are finding roots of complex numbers
Consider wn
z
integer
The n different solutions w0 , w1, w 2, , ..., wn
are the ' nth roots of z' denoted by
n
Let
z
and
Equation wn
1
or
z
zn
r cos
j sin
w
cos
j sin
of this equation
1
!
z
"
n
gives
cos n
$
#
%
j sin n
&
'
r cos
(
)
*
+
j sin
,
-
.
Equality of two complex numbers in polar form means that;
/
i
their modulus are equal
0
n
i.e.
5
ii
1
or
r
2
1
rn
4
2
3
4
4
4
their arguments, arg wn and arg z may differ by a
multiple of 2 , say 2 k ,
k 0, 1, 2. ..
2k
6
7
7
8
Thus n
9
:
;
<
=
2k
or
B
>
?
B
B
B
B
@
B
B
B
B
B
B
B
B
B
B
B
=
B
B
B
,
B
A
k
0, 1, 2, ..., n
C
1
D
n
E
all other choicesof k duplicate the values of
Thus nth roots of z,
w0 , w1, w 2, , ..., wn
2k
1
wk
rn
J
I
J
J
J
J
J
K
cos
R
R
R
c
c
c
R
O
R
P
1
c
R
R
R
R
R
LMN
N
r n exp j
c
b
R
c
h
d
h
e
h
h
h
f
h
h
h
h
2k
n
h
h
h
h
h
h
h
h
h
h
R
R
R
R
R
R
ST
H
are given by
1
U
2k
U
V
Q
R
R
R
j sin
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
^_
`
`
\
]
]
]
WXY
Y
n
]
Z
[
n
g
h
h
h
h
h
i
k
j
0, 1, 2. .., n
k
1
a
F
G
Complex.nb
19
Example :
Find all the cube roots of
Set
then
arg
Hence
and
8
r
8
8
8
8,
8,
1
3
z
8 cos
8
j sin
1
8 3 expj
3
2k
3
2k
8
cos
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
%
&'
2k
(
(
)
$
%
%
%
j sin
%
!
!
"
#
0
0
0
0
0
0
0
0
,
-
There are 3 cube roots of w3
Thus the 3 cube roots of z
w0
2 cos
;
?
?
=
?
?
z
9
;
9
8
:
j sin
?
>
3
since n
6
8 are
<
B
B
B
B
B
A
3
@
C
M
1
2
D
H
H
H
H
3
j
H
EFG
2
G
G
G
L
L
L
L
L
L
J
L
K
K
L
K
L
L
L
K
2
I
N
O
O
O
G
O
O
1
P
w1
Q
R
3 j
S
S
S
S
2 cos
U
T
2
Y
Z
1
V
[
j sin
W
0
\
[
5
w2
2 cos
^
V
X
]
2
^
c
c
c
c
5
b
c
c
c
c
j sin
c
_`a
f
f
f
f
gh
f
3
d
s
1
2
j
n
n
n
n
3
j
n
klm
2
m
m
m
o
r
r
p
r
r
r
r
q
r
q
2
r
q
r
r
r
q
t
u
u
u
m
u
v
u
1
w
x
y
3 j
y
y
y
i
i
e
f
a
3
0
0
3
0, 1, 2
5
0
.
3
k
0
*+,
f
f
f
7
3
8
0
0
0
0
0
12
3
3
/
0
0
0
4
Complex.nb
20
2
w0 , w1, w2 are equally spaced
centered at 0, 0
For wn
radians
120 ° apart on circle of radius 2
3
2
z roots are spaced by
rad and lie on a circle of radius
n
Example :
2
Find the values of j 3
2
There are two ways depending upon how you express j 3
2
i
j3
j2
1
3
1
2
ii
j3
#
!
"
#
#
1
3
1
2
1
j3
i.e find 3 cube roots of
i.e find the 3 cube roots of j
#
$
and square them .
Here it appears sensible to use method i
%
'
i
&
polar form of
r
)
*
wk
1
+
5
6
1
7
3
:
;
1 and
,
-
9
1
<
<
<
.
arg
/
1
0
1
2
3
4
1
3
9
9
9
8
2k
=>?
<
&
1:
(
?
?
cos
?
B
E
FG
2k
H
H
I
D
j sin
C
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
@A
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
QR
M
S
P
P
3
k
b
c
1
cos
X
X
V
X
X
j sin
X
W
w1
3
cos
e
f
Y
]
[
]
]
]
]
`
3
j sin
f
e
`
`
`
c
c
h
^
c
3
`
2
_
d
d
d
d
d
d
d
d
d
d
j
d
2
a
1
~
5
w2
cos
i
3
n
n
n
op
q
5
q
r
m
n
n
jkl
n
n
n
j sin
l
j
w
w
w
w
xy
z
w
w
w
w
1
z
{
v
w
|
|
|
|
stu
Solve the equation
3
2
3
|
u
3
Exercise :
w2
n
0, 1, 2
U
\
Z
g
T
P
N
3
U
S
O
P
JKL
L
w0
n
}
2
j
r
3
8 ,
