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===== The square root of a complex number =========================
To find the two square roots of the
complex number a ib do t
e followi g:
(i) Let t
e sq are root be x iy ie
a ib x iy ! (ii) Sq are bot
sides, t
e apply t
e eq ality property to get two
sim
lta
eo
s eq
atio
s i
x a
d y, solve t
em.EXAMPLE: Fi d t
e two sq are roots of t
e complex mber 5-12i
SOLUION:
2 2
2 2
2 2
2 2
( i ) let 5-12i x iy
( ii ) Sq
are bot
e sides: 5-12i=x y 2 xyi
x y 5...........( 1)
2 xy 12......( 2 )
( 1 ) ( 2 )
x y 13........( 3 )
Solvi
g ( 1 )&( 3 ) : x 3 & y 25 12i ( 3 2 y ) A
s
!
@ !!
!
! s ! s@ ! s
Ot
er sol tio :
2 2 2 2
2 2
From eq
atio
s (1),(2)
e q
a
tities x y ;2 xy; x y form t
e t
ree sides of a
rig
t a
gled tria
gle i
w
ic
x y is t
e
ypoti
eo
s. So we ca
form a rt. a
gled tria
glew
ose legs are 5;12 a
d get t
2 2
2 2
e
ypoti
eo
s
eq
als 13, a
d write
x y 5..................( 1 )
2xy 12.............( 2 )x y 13...............( 3 )
Solvi
g( 1 ) &( 3 ) x 3 ; y 2 5 12i ( 3 2i ).
!
! !
! s ! ! s m
Ot
er sol tio :
2 2 2 2 2
1T
i
k abo t t
o
mbers
ose prod ct 6 ( t
e imagi
ary part )2
a
d t
e differe
ce of t
eir sq ares is 5 (t
e real part); t
e
mbers
ill
be 3 a d 2
5-12i
3 12i 2 i 9 12i 4i ( 3 2i )
5 12i ( 3 2i )
s.
@ ! !
@ !
Now, try t
ese problems:
3 4i ; 7 24i ; 8 15i
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EXAMPLE: Solve t
e followi
g eq atio
s i
C(i) !z z2 1 0 (ii) x ( i )x !22 5 6 0
SOLU ION
1 1 4 1 1 1 3i 1 3( i )z i
2 2 2 2
s v v s s
2
2 2 2 2
1 2
( 5 i ) ( 5 i ) 4 2 6 ( 5 i ) 24 10i ( ii ) x
4 424 10i l mi l m 24 ;l m 26
l 1 ; m 5
24 10i ( 1 5i )
( 5 i ) ( 1 5i ) 3 3i x x ; x 1 i
4 2
s v v s
s s
s s
EXAMPLE: Solve t!
e followi" g eq# atio" i" C2
( 1 i )z ( 1 3i )z 2( 2 3i ) 0 SOLU ION:
2
2
1 2
Divi de t$
e eq%
atio&
by (1+i) a&
d simplify
z ( 2 i )z ( 1 5i ) 0
( 2 i ) ( 2 i ) 4( 1 5i ) ( 2 i ) 7 24i z
2 2
7 24i ( 4 3i )
( 2 i ) ( 4 3i ) ( 2 i ) ( 4 3i )z 3 2i ; z 1 i
2 2
s s
s
@
EXAMPLE:
Form t!
e q#
adratic eq#
atio" wit!
real coefficie" ts, if o" e of its roots is 3+i
SOLU ION:
2
2
3 i is a root 3-i is t $
e ot$
er roots
%
m of roots = 3+i + 3-i=6
prod%
ct of roots=(3+i)(3-i)=10
t$
e eq%
atio&
is
x ( s%
m )x prod %
ct 0
i .e . x 6 x 10 0
@
@
Q
EXAMPLE:
1 2i 1 2i If x ; y ,then find 5x ' 3y .
1 i 1 i
! !
SOLU(
ION:1 2i 1 2i 1 i 1 3i x
1 i 1 i 1 i 21 2i 1 2i 1 i 1 3i
y1 i 1 i 1 i 2
2 3 25 x 3 y 4 3i ( i )
2 2
! ! v !
! ! v !
! ! s
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Exercise 2
2 2
4 2 2
2
2
1 If x ) then find the solution set of
a. x 25 0 b.x 4 x 13 0
c.x 7 x 12 0 d .x 6 x 9 2i 0
e.x ( 5 i )x ( 8 i ) 0
f .( 2 i )x ( 9 7 i )x 5( 3 2i ) 0
20
ind the square roots of the follo1
ing complex numbersa.z 5
! !
! !
!
!
!
3-2
1 1
2 2
2 2
12i b.z 3 4i
c.z 7 24i d .z 1 i
13 65i e.z i f .z
5 i
3 If x 3 4i , then find x
4 If x 2 21 3 20i, then find the value of x 29x .
5 If x and y are real values, find these values if
y iy 6
! ! !
! !
!
2 2 2
2
2 2
*
i 2 x ix
( x i ) ( 2 y i ) 4( 3 1 )i 2 y x
6 If l and m are the roots of the equation x ( 4 6i )x ( 10 20i ) 0 ,then find
then find the equation1
hose roots are l and m .
74
olve the simultaneously the follo1
ing equ
! !
!
ations, 1 here x and y are real
2 5i and x y i
x y ! !
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Geometric Represe5 tatio5 Of Complex N6
mbers
Arga5
d Fig6 res
7
8
e Fre5 c8
mat8
ematicia5 Arga5 d establis8
ed t8
e geometric represe5 tatio5
of t8
e complex5 6 mber x + y i as a5 ordered pair (x , y) R2,8
e called t8
e x-axis as
t8
e real axis a5
d t8
e y-axis as t8
e imagi5
ary axis. So, t8
e Cartesia5
pla5
e is5
amed
as Arga
5
d pla
5
e a
5
d t
8
e fig6
res t
8
at represe
5
t t
8
e complex
5
6
mbers or a
5
yoperatio5 performed o5 t8
em as Arga5 d fig6 res.
For t8
is, t8
e complex5 6 mbers will be represe5 ted i5 Arga5 d pla5 e by a poi5 t
(x , y), ( sometimes t8
ey call it vector), as yo6
see i5 t8
e fig6
re:
9
@
e complexAB
mberz1 = 3 + 4i is represeA ted by poiA t A(3,4), z2 = -1 + 2i is
represeA ted by t@
e poiA t B(-1,2) aA dz3 = -2 3i is represeA ted by t
@
e poiA t C(-2,-3),
aA d so oA .
RepreseA tatioA of SB
m
Ifz1=(x1,y1) aA dz2=(x2,y2) t@
eA
z1 + z2 = (x1 + x2 , y1 + y2). From t@
e
figB re we fiA d t@
at t@
e poiA ts (x1,y1) ;
(x2,y2); aA d (x1+x2 ,y1+y2) are t@
ree vertices
of t@
e parallelogram OACB, t@
at is t@
e s B m of
two complexAB
mbers is t@
e foB
rt@
vertex C. A(x1,y1)
B(x2,y2)
C(x1+x2,y1+y2)
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TheC
D mbers izaC
d iz
RepreseC t theC
Dmberz= 1+2i, the
C
represeC t the two C D mbers izaC d iz. What do
yoD
C otice.
1
2
1
1
2
z1 2i is t
he poi
E
t A (1,2)iz=-2+i is t he poi
E
t A (-2,1)
-iz=2-i is t he poiE
t A (2,-1)
UsiE
g the slope, we fiE
d that OA OA
aE
d OA OA . This meaE
s that we rotate OA
aE
ticlock 90 . OE
the otherhaE
d OA OA
OA
!
B
!
r B
2OA ,This mea
F
s that we rotate OA
clockwise90
!
r
ModG lG s- ArgG meH t-TrigoH ometric Form
of a ComplexHG
mber
We kH ew that the complexH
G mberz = x + iy caH be represeH ted iH the arga
H d
plaH
e by the poiH
t A(x , y). This poiH
t caH
also fG lly determiH
ed as sooH
as we kH
ow
the distaH ce OA aH d the polar a
H gle betweeH OA aH d the (+)ve x-axis (Ur), meas G red
aH ticlockwise.
The distaH ce OA is called the modG lG s ofz, a
H d is deH oted by r
2 2r x y@ ! The aH gle U is called the ArgG meH t ofz, aH d is deH oted by Arg(z) where
ytaH
x!U
SiH
ce if we kH
ow taH
U, theH
U willhave maH
y valG es ( 2FU Ts ), it is agreed
that [0 ,2 [ aH
d is called the priH
cipal ArgG meH
t ofzU T .
the priH
cipal ArgG meH
t [0,2 [ U T@
From the figG re we fiH
d that
xcos x r cos
ry
siH y r si H
r
U U
U U
! !
! !
z r(cos i si H )U U!
A(1,2)
A1(-2,1)
(2,-1)A2
U
A
x
y
O
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The previo I s form is called the TrigoP
ometric Form of the complexP
I mber.
YoI
mI
stP otice that the previoI
s form is the priP cipal form, which mI
st be, wheP
weP
eed to fiP
d its modI lI s aP
d ArgI meP
t, aP
d yo I o I ght to modify aP
y other form
before yoI
determiP e the modI
lI
s aP d the Arg.
SpecialTrigoP
ometric forms
z 1 z cos 0 i si Q
0
z i z cos i si Q
2 2z 1 z cos i si
Q
3 3z i z cos i si
Q
2 2
T T
T T
T T
! R ! r r
! R !
! !
! !
EXAMPLE:
FiS
d the modT lT s aS
d the priS
cipal Arg. of the followiS
g complexS
T mbers
theS write the trigoS ometric form of eachST
mber.
( i )2 2i ; 3 3i ; 1 3i ; 3 i
( ii )5 ; 2i ; 4 ; 4i
SOLUTION:
2 2
2 2
2 2
( i ) z 2 2i r 2 2 2 2
2ta
U
1 ( x 0 , y 0 )2 4
z 2 2(cos i si U
)4 4
z 3 3i r 3 3 3 2
3 3ta
U
1 ( x 0, y 0 )
3 43 3z 3 2(cos i si
U
)4 4
z 1 3i r ( 1 ) ( 3 ) 2
3 4ta
U
3 ( x 0, y 0 )1 3
4 4z 2(cos i si
U
)3 3
z 3
TU U
T T
TU U
T T
TU U
T T
! ! !
! ! ! " "
!
! ! !
! ! ! "
!
! ! !
! ! !
!
!
X
Y
Z
[2 2
i r ( 3 ) ( 1 ) 2
1 11ta
V
( x 0 , y 0 )63
11 11z 2(cos i si
V
)6 6
TU U
T T
! !
! ! "
!
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( i i ) z 5 5 1 5(cos 0 i si W
0 )
r 5 ; Arg( z) 0
z 2i 2 i 2(cos i si W
)2 2
r 2 ; Arg( z)2
z 4 4 1 4(cos i si W
)
r 4 ; Arg( z)
3 3z 4i 4 i 4(cos i si
W
)2 2
3r 4 ; Arg( z) .
2
T T
T
T T
T
T T
T
! ! v ! r r! !
! ! v !
! !
! ! v ! ! !
! ! v !
! !
X
Y
Z
[
ModificatioX
s iX
TrigoX
ometric Form
A. ChaY ge of sigY of real or imagiY ary parts or both
WheY
the sigY
of either real or imagiY
ary parts or both chaY
ge the Arg(z) ismodified accordi
Y
g to the followiY
g table, aY
d all sigY
s became positive.
EXAMPLE:
Fi d the moda la s a d pri cipal Arga me t for each of the followi g complex
a mbers:
1 z 2(cos i si ) 2 z 2( cos i si )3 3 4 4
3 z 2(cos i si
) 4 z 4(cos i si
)6 6 3 3
T T T T
T T T T
! !
! !
SOLUTION
cos 0,sin 0 r 2 ;b
rg(z)32
cos 0 ; sin 0 r 2 ;b
rg(z)3
2 2z 2(cos i sin )
3 3
T
T
T T
" " ! !
" ! !
!
X
Y
Arg(z) =UArg(U) = T- U
Arg(U) = T+ U Arg(z) =2T- U
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4cos 0 ;si
c
0 r 2 ; Arg( z)3
4 4z 2(cos i si c )
3 35
cos 0 ,si c
0 r 4 ; Arg( z)3
5 5
z 4(cos i si
c
)3 3
T
T T
T
T T
! !
!
" ! !
!
Z
[
B. Co-chad ge with chad ge id sigd of real ad d imagid ary parts or both
Whee co-chae ge occ f rs( ie cos chae ges to sie or vise versa) witho f t or with sige
chae ge the Arg(z) is cha
e ged relative to3
or2 2
T Taccordig g to the followi
g g table:
EXAMPLEWrite the followi
g g complexgh
mbers ig stag dard form, theg fig d the
modh lh s ag d prig cipal Argh meg t for each:
7 71 z 4(si
g
i cos ) 2 z 4( si g
i cos )6 6 6 6
5 5 3 33 z 3 2( si g i cos ) 4 z 4(sig i cos )
3 3 4 4
T T T T
T T T T
! !
! !
SOLUTION
z 4(cos( ) i si i
( )) 4(cos i si i
)2 3 2 3 6 6
r 4 ; Arg( z)6
7 7 5 5z 4(cos( ) i si
i
( )) 4(cos i si i
)2 6 2 6 3 3
5r 4 ; Arg( z)
3
T T T T T T
T
T T T T T T
T
! !
@ ! !
! !
@ ! !
X
Y
2
TU
2
Tp
3
2
TU
3
2
Tq
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3 5 3 5 2 2z 3 2(cos( ) i si
r ( )) 3 2(cos( ) i si r ( ))2 3 2 3 3 3
4 4 43 2(cos i si r ) r 3 2 ; Arg( z)
3 3 33 3 3 3 9 9
z 4(cos( ) i si r ( )) 4(cos i si r )
2 4 2 4 4 4
4(cos i si
r
) r 4 ; Arg( z)4 4 4
T T T T T T
T T T
T T T T T T
T T T
! !
! ! !
! !
! ! !
Z
[
Ms ltiplicatiot , Divisiot at d Powers of complext s mbers
it trigot ometric form
1 1 1 1 2 2 2 2
1 2 1 2 1 2 1 2
1 1
1 2 1 22 2
u u
Givev
thatz r (cos i si v
) ;z r (cos i si v
),the followiv
g is
trw
e:
z z r r [cos( ) i si v
( )]
z r
[cos( ) i si v
( )]z r
1 1[cos( ) i si
v
( )]z r
z r [cos( v
) i si v
(v
)] ;v
Z
U U U U
U U U U
U U U U
U U
U U
! !
!
!
!
!
X
Y
Z
[
EXAMPLE
1 2
51
1 2 2
2 1
3 3Give
x
thatz 2(cos i si x
) ax
d z 2(cos i si x
) fi x
d6 6 4 4
z 1( i )z z ( ii ) ( iii ) ( iv )z
z z
T T T T ! !
SOLUTION
1 2
1
2
1
3 3 11 11( i )z z 2 2 [cos( ) i si
x
( )] 2 2(cos i si x
)6 4 6 4 12 12
z 2 3 3 7 7 ( ii ) [cos( ) i si
x
( )] 2 [cos( ) i si x
( )]z 6 4 6 4 12 122
17 17 2(cos i si
x
)12 12
1 1 1 11 11( iii ) [cos( ) i si
x
( )] (cos i si x
)z 2 6 6 2 6 6
( i v )
T T T T T T
T T T T T T
T T
T T T T
! !
! !
!
! !
5 5
2
3 3 7 7 z ( 2 ) [cos( 5 ) i si
x
( 5 )] 4 2(cos i si x
)4 4 4 4
T T T T ! v v !
Remember that:
cos( ) cos( 2 ) si x
( ) si x
( 2 )
cos 2 1 1 cos 2cos si
x
2 2
U T U U T U
U UU U
! !
! !
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EXAMPLE
1 2
2 3
1 2 1
1 2
y ivenz 1 3i and z 2 2 2 2i; find the modulus and argument of
the folo
ing numbers
1(i) z z ( ii ) ( iii )1 z
z z
! !
SOLUTION
1 1 1
2 2 2
2 3
1 2
1 2
5 5 5z 1 3i r 2 ; Arg( z ) z 2(cos i sin )3 3 3
5 5 5z 2 2 2 2i r 4 ; Arg( z ) z 4(cos i sin )
4 4 410 10 15 15
( i )z z 4(cos i sin ) 64(cos i sin )3 3 4 4 13 13
256(cos i sin )12 12
35 35( i i )z z 8(cos i sin ) 8(co
12 12
T T T
T T T
T T T T
T T
T T
! ! ! !
! ! ! !
! v
!
! !
1 2
1
11 11s i sin )
12 12
1 1 11 11 1 13 13(cos i sin ) (cos i sin )z z 8 12 12 8 12 12
( iii )1 z 1 ( 1 3i ) 3i 3(cos i sin )2 2
T T
T T T T
T T
! !
! ! !
EXAMPLE
1 2
1
2
If z 18(cos 3 i sin 3 ) ,and z 6(sin 2 i cos 2 ),
here [270 ,360 [
z-3and tan .
ind the trigonometric and algebraic forms of .4 z
U U U U U
U
! ! r r
SOLUTION
2
2
1
2
1
2
z 6(sin2 i cos 2 )is not in standard form, so
e modify it to3 3
z 6 cos 2 i sin 22 2
z 18 3 3cos i sin 3 cos i sin
z 6 2 2 2 2
zr 3 ; Arg
z
U UT T
U U
T T T T U U U U
!
!
@ ! !
@ !
Q
1
2
2
z 3 4 9 12z 3 si
i cos 3 i i ( A
s )
z 5 5 5 5
TU
U U
!
! ! ! v v !
Notice that:
2
U
TU
2
T UT
U
3
2
T U
TU
2
3
2
T UT
U
!
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D'Moivre's Theorem
If Q (ratio al mber), the
cos i si cos i si U U U U !
(proof is ot req
ired)
From r le 4 (power r le), we k
ow that, for kZ+:
k
1
k
1
k
cos i si
cos i si
k k
cos i si
is o
e of the val
es for cos +i si
k k
cos i si
cos 2
i si
2
2
2
cos i si
cos i si
k kwhere
{ 0,1,2,3,...........,k 1}
U UU U
U UU U
U U U T U T
U T U TU U
!
@
!
!
@
Q
The last relatio is very importa t, beca se we se it to determi e the differe t roots
of a certai complex
mber.
EXAMPLE
Use D'Moivre's theorem to fi d the val
e of si 2U a d cos2U i
terms of si U
a
d cosU.
SOLUTION
2
2 2
2 2
cos i si cos 2 i si 2
cos si
i (2si
cos ) cos 2 i si
2
ompairi g real a d imagi ary parts:
cos2
cos si a d si 2 2 si cos (
s )
U U U U
U U U U U U
U U U U U U
!
@ !
@ !
Q
EXAMPLE
Use D'Moivre's theorem to solve the eq atio x4=1 i C.
SOLUTION
4
1
4
1 2
3 4
x 1 cos 0 i sin0 cos 2n i sin 2n
x cos 2n i sin 2n
2n 2nx cos i sin ,n 0 ,1,2,3
4 4
x cos 0 i sin0 1 x cos i sin i 2 23 3
x cos i sin 1 x cos i sin i 2 2
T T
T T
T T
T T
T TT T
! ! r r !
!
! !
! r r ! ! !
! ! ! !
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EXAMPLE
1 2
3 3
Fi
d the differe
t val
es of the followi
g
(i) z=(1+ 3i ) ( ii )z ( 1 i )
the
represe
t the mber 1+ 3i a
d the differe
t val
es o
the same Arga
d
fig
re.
!
SOLUTION
1
1
1 133 3
1
3
3 3
11 12
3
13
( i ) let z 1 3i r 2 ; Arg( z) 60
z 2 cos 60 i si
60
z z 2 cos 60 i si
60
60
360 60
3602 cos i si
,
0,1,23 3
z 2 cos 20 i si
20 z 2 cos 140 i si
140
z 2 cos 260 i si
260
! ! ! r
@ ! r r
! ! r r
r v r r v r ! !
! r r ! r r
! r r
Q
112 1 32 33 3
3
3 3
1 2
3
3
3 3( i i )z ( 1 i ) 1 i 2i 2 cos i sin
2 2
270 n 360 270 n 3602 cos i sin
3 3
z 2 cos 90 i sin90 z 2 cos 210 i sin 210
2 cos 330 i sin 330
T T ! ! ! ! r v r r v r !
! r r ! r r
! r r
z1
z11z12
z13
z
z1
z2 z3
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EXAMPLE
1
2 2
se
'Moivre's theorem to find the solution set of the equation
x i ( 1 i )
here x !
SOL TION
112
2
2 2
4 4
1 3
4
2 4
1 i 2 cos i sin4 4
5 5i cos i sin cos i sin or cos i sin
2 2 4 4 4 4
3 3x 2 cos i sin x 2 cos i sin
2 2 2 2
3 3x 2 cos i sin x 2 cos i sin
4 4 4 4
5 5x 2 cos i sin x
4 4
T T
T T T T T T
T T T T
T T T T
T T
!
! !
@ ! @ ! ! ! ! !
4
4 4 4 4
7 72 cos i sin
4 4
S .S { 8 1 i ; 8( 1 i ); 8( 1 i ); 8( 1 i )}
T T
!
EXAMPLE
1
12 15 21 i 1 i
If x ; y ,then find the value of 3x 4 y1 i 1 i
! !
SOL
TION
12 15 12 15 3
1 112 15 2 2
1 2
1 i 1 i 2i x i
1 i 1 i 21 1
y ix i
3x 4 y 3i 4( i ) 3 4i 3 4i
2n 2nl 3x 4 y 3 4i 5 cos i sin n 0 ,1
2 2
4l 5 cos i sin ; l 5 cos i sin ,tan
2 2 2 2
U T U T
U U U UT T U
! v ! !
! ! !
@ ! ! !
@ ! ! ! !
@ ! ! !
1 2
3
311 cos 4 2 15cos sin2 2 2 5 25 5
l 2 i ; l 2 i
U U U! ! ! ! !
@ ! !
Q