Complex Part2

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


2/13

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


3/13

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 ! !


4/13

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)


5/13

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


6/13

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

! !

! ! "

!


7/13

( 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


8/13

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


9/13

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

! !

! !


10/13

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

!


11/13

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 ! ! !

! ! ! !


12/13

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


13/13

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

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Complex Part2