X2 t01 06 geometrical representation (2013)

Geometrical Representation of Complex Numbers


Complex numbers can be represented on the Argand Diagram as vectors.


Complex numbers can be represented on the Argand Diagram as vectors.y

x



x

iyxz



x

iyxz

The advantage of using vectors is that they can be moved around the Argand Diagram



x

iyxz

The advantage of using vectors is that they can be moved around the Argand DiagramNo matter where the vector is placed its length (modulus) and the angle made with the x axis (argument) is constant



x

iyxz

The advantage of using vectors is that they can be moved around the Argand DiagramNo matter where the vector is placed its length (modulus) and the angle made with the x axis (argument) is constant

A vector always

represents

HEAD minus TAIL

Addition / Subtraction


y

x


y

x

1z


y

x

1z

2z


y

x

1z

2z

To add two complex numbers, place the vectors “head to tail”


y

x

1z

2z


21 zz


y

x

1z

2z


21 zz

To subtract two complex numbers, place the vectors “head to head” (or add the negative vector)


y

x

1z

2z


21 zz


21 zz


y

x

1z

2z


21 zz


21 zz 2121 and diagonals; twohas vectors

addingby formed ramparallelog the:

zzzz

NOTE


y

x

1z

2z


21 zz




zzzz

NOTE

Trianglar InequalityIn any triangle a side will be shorter than the sum of the other two sides


y

x

1z

2z


21 zz




zzzz

NOTE


BCABACABC ;In


y

x

1z

2z


21 zz




zzzz

NOTE


BCABACABC ;In (equality occurs when AC is a straight line)


y

x

1z

2z


21 zz




zzzz

NOTE


BCABACABC ;In (equality occurs when AC is a straight line)

2121 zzzz

AdditionIf a point A represents and point Brepresents then point C representing

is such that the points OACB form a parallelogram.

1z2z

21 zz

AdditionIf a point A represents and point Brepresents then point C representing

is such that the points OACB form a parallelogram.

1z2z

21 zz

SubtractionIf a point D represents and point E represents then the points ODEB form a parallelogram.

Note:

1z12 zz

12 zzAB

12arg zz

1995..ge y

x

P

O

Q

The diagram shows a complex plane with origin O.The points P and Q represent the complex numbers z and w respectively.Thus the length of PQ is wz wzwzi that Show

1995..ge y

x

P

O

Q


zOP is oflength The

1995..ge y

x

P

O

Q


zOP is oflength ThewOQ is oflength The

1995..ge y

x

P

O

Q



wzPQ is oflength The

1995..ge y

x

P

O

Q



wzPQ is oflength Thewzwz

OPQ

on inequalityr triangulatheUsing

1995..ge y

x

P

O

Q




OPQ


(ii) Construct the point R representing z + w, What can be said about thequadrilateral OPRQ?

1995..ge y

x

P

O

Q




OPQ



R

1995..ge y

x

P

O

Q




OPQ



R

OPRQ is a parallelogram

?about said becan what , If zwwzwziii


wzwz


wzwz i.e. diagonals in OPRQ are =


wzwz i.e. diagonals in OPRQ are =rectanglea is OPRQ



2argarg

zw



2argarg

zw

2arg

zw



2argarg

zw

2arg

zw imaginarypurely is

zw



2argarg

zw

2arg


zw

Multiplication



2argarg

zw

2arg


zw

Multiplication

2121 zzzz



2argarg

zw

2arg


zw

Multiplication

2121 zzzz 2121 argargarg zzzz



2argarg

zw

2arg


zw

Multiplication


21212211 cisrrcisrcisr



2argarg

zw

2arg


zw

Multiplication


21212211 cisrrcisrcisr

2

2121

by multiplied islength its and by iseanticlockwrotated isvector the,by multiply weif ..

rzzzei

If we multiply by the vector OA will rotate by an angle of in an anti-clockwise direction. If we multiply by it will also multiply the length of OA by a factor of r

1z cis

rcis


1z cis

rcis

ii 2

sin2

cos Note: 1iz will rotate OA anticlockwise 90 degrees.


1z cis

rcis

ii 2

sin2


2by iseanticlockwrotation a is by tion Multiplica i


1z cis

rcis

ii 2

sin2


REMEMBER: A vector is HEAD minus TAIL

2by iseanticlockwrotation a is by tion Multiplica i

y

x

A

O

C

B

)1(D

y

x

A

O

C

B

)1(D

DC DA i

y

x

A

O

C

B

)1(D

DC DA i

1 1C i

y

x

A

O

C

B

)1(D

DC DA i

1 1C i

1 1

1

C i

i i

y

x

A

O

C

B

)1(D

DC DA i

B A DC

1 1C i

1 1

1

C i

i i

y

x

A

O

C

B

)1(D

DC DA i

B A DC

1B C

1 1C i

1 1

1

C i

i i

y

x

A

O

C

B

)1(D

DC DA i

B A DC

1B C

1 1C i

1 1

1

C i

i i

1

1

B i

i i

y

x

A

O

C

B

)1(D

DC DA i

B A DC

1B C

OR

1 1C i

1 1

1

C i

i i

1

1

B i

i i

y

x

A

O

C

B

)1(D

DC DA i

B A DC

1B C

OR

1 1C i

1 1

1

C i

i i

1

1

B i

i i

B C DA

y

x

A

O

C

B

)1(D

DC DA i

B A DC

1B C

OR

1 ( 1)B i i

1 1C i

1 1

1

C i

i i

1

1

B i

i i

B C DA

y

x

A

O

C

B

)1(D

DC DA i

B A DC

1B C

OR

1 ( 1)B i i

1 1C i

1 1

1

C i

i i

1

1

B i

i i

B C DA

1i i

y

x

A

O

C

B

)1(D

DC DA i

B A DC

1B C

OR

1 ( 1)B i i

OR

1 1C i

1 1

1

C i

i i

1

1

B i

i i

B C DA

1i i

y

x

A

O

C

B

)1(D

DC DA i

B A DC

1B C

OR

1 ( 1)B i i

OR

24

DB cis DA

1 1C i

1 1

1

C i

i i

1

1

B i

i i

B C DA

1i i

y

x

A

O

C

B

)1(D

DC DA i

B A DC

1B C

OR

1 ( 1)B i i

OR

24

DB cis DA

1 2 cos sin ( 1)4 4

B i

1 1C i

1 1

1

C i

i i

1

1

B i

i i

B C DA

1i i

y

x

A

O

C

B

)1(D

DC DA i

B A DC

1B C

OR

1 ( 1)B i i

OR

24

DB cis DA

1 2 cos sin ( 1)4 4

B i

1 1C i

1 1

1

C i

i i

1

1

B i

i i

B C DA

1i i

1 ( 1) 1B i

y

x

A

O

C

B

)1(D

DC DA i

B A DC

1B C

OR

1 ( 1)B i i

OR

24

DB cis DA

1 2 cos sin ( 1)4 4

B i

1 1C i

1 1

1

C i

i i

1

1

B i

i i

B C DA

1i i

1 ( 1) 1B i 1 1i i

y

x

A

O

C

B

)1(D

DC DA i

B A DC

1B C

OR

1 ( 1)B i i

OR

24

DB cis DA

1 2 cos sin ( 1)4 4

B i

1 1C i

1 1

1

C i

i i

1

1

B i

i i

B C DA

1i i

1 ( 1) 1B i 1 1i i

1i i

2000..ge y

x

A

O

C

In the Argand Diagram, OABC is a rectangle, where OC = 2OA.The vertex A corresponds to the complex number

B

2000..ge y

x

A

O

C


B

? toscorrespondnumber complex What Ci

2000..ge y

x

A

O

C


B


2OC OA i

2000..ge y

x

A

O

C


B


2OC OA i

2C i

2000..ge y

x

A

O

C


B


2OC OA i

(ii) What complex number corresponds to the point of intersection D of the diagonals OB and AC?

2C i

2000..ge y

x

A

O

C


B


2OC OA i


diagonals bisect in a rectangle

2C i

2000..ge y

x

A

O

C


B


2OC OA i



2C i

midpoint of D AC

2000..ge y

x

A

O

C


B


2OC OA i



2C i

midpoint of D AC

2A CD

2000..ge y

x

A

O

C


B


2OC OA i



2C i

midpoint of D AC

2A CD

22

iD

2000..ge y

x

A

O

C


B


2OC OA i



12

D i

2C i

midpoint of D AC

2A CD

22

iD

Examples

Examples

3OB OA cis

Examples

3OB OA cis

13

B cis

Examples

3OB OA cis

iB23

21

13

B cis

Examples

3OB OA cis

iB23

21

13

B cis

OD OB i

Examples

3OB OA cis

iB23

21

iBD

13

B cis

OD OB i

Examples

3OB OA cis

iB23

21

iBD

iD21

23

13

B cis

OD OB i

Examples

3OB OA cis

iB23

21

iBD

iD21

23

13

B cis

OD OB i C B OD

Examples

3OB OA cis

iB23

21

iBD

iD21

23

1 3 3 12 2 2 2

C i i

13

B cis

OD OB i C B OD

Examples

3OB OA cis

iB23

21

iBD

iD21

23

1 3 3 12 2 2 2

C i i

iC

231

231

13

B cis

OD OB i C B OD

( ) i AP AO i

P A i O A

( ) i AP AO i

P A i O A

11 0 zizP

( ) i AP AO i

P A i O A

11 0 zizP

11 izzP

( ) i AP AO i

P A i O A

11 0 zizP

11 izzP 11 ziP

( ) i AP AO i

P A i O A

11 0 zizP

11 izzP 11 ziP

( ) i AP AO i

( ) ii QB BO i

P A i O A

11 0 zizP

11 izzP 11 ziP

Q B i O B

( ) i AP AO i

( ) ii QB BO i

P A i O A

11 0 zizP

11 izzP 11 ziP

Q B i O B iBBQ

( ) i AP AO i

( ) ii QB BO i

P A i O A

11 0 zizP

11 izzP 11 ziP

Q B i O B iBBQ

21 ziQ

( ) i AP AO i

( ) ii QB BO i

P A i O A

11 0 zizP

11 izzP 11 ziP

Q B i O B iBBQ

21 ziQ 2

QPM

( ) i AP AO i

( ) ii QB BO i

P A i O A

11 0 zizP

11 izzP 11 ziP

Q B i O B iBBQ

21 ziQ 2

QPM

2

11 21 ziziM

( ) i AP AO i

( ) ii QB BO i

P A i O A

11 0 zizP

11 izzP 11 ziP

Q B i O B iBBQ

21 ziQ 2

QPM

2

11 21 ziziM

2

2121 izzzzM

( ) i AP AO i

( ) ii QB BO i

Cambridge Ex 1E; 2 to 8, 10 to 15, 18, 19, 21, 22

Terry Lee: Exercise 2.6

X2 t01 06 geometrical representation (2013)

Education

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