Complex Part2
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8/6/2019 Complex Part2
1/13
===== 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
8/6/2019 Complex Part2
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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 ! !
8/6/2019 Complex Part2
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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)
8/6/2019 Complex Part2
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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
8/6/2019 Complex Part2
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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