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© John Wiley & Sons Australia, Ltd 2009 1 WorkSHEET 2.2 Number systems: complex numbers Name: ___________________________ 1 Write yi x i i i i + - - - + + form in the 3 2 3 5 2 3 . ( ) ( ) i i i i i i i i i i i i i i i i i i i i i i i i i i i i 65 58 65 4 65 58 4 65 45 95 13 91 65 9 19 5 7 13 13 9 19 5 7 9 4 9 6 15 10 4 2 3 6 3 2 3 2 3 2 3 5 2 2 2 3 3 2 3 5 2 3 2 2 2 2 - - = - - = - - - = + - - = + - - = - - - + - - - + - = + + ´ - - - - - ´ + + = - - - + + 4 2 Express 4 5 cis 2 p in General / Cartesian form. i i i i - - = ÷ ø ö ç è æ - - = ÷ ø ö ç è æ - - = ÷ ø ö ç è æ + = 1 2 1 2 1 2 4 sin 4 cos 2 4 5 sin 4 5 cos 2 4 5 cis 2 p p p p p 3

WorkSHEET 2.2 Number systems: complex numbersthefinneymathslab.weebly.com/.../81042930/11mc_com… ·  · 2017-07-11in the form 2 3 5 3 2 3. i ( ) ( ) i i i i i i i i i i i i i i

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Page 1: WorkSHEET 2.2 Number systems: complex numbersthefinneymathslab.weebly.com/.../81042930/11mc_com… ·  · 2017-07-11in the form 2 3 5 3 2 3. i ( ) ( ) i i i i i i i i i i i i i i

© John Wiley & Sons Australia, Ltd 2009 1

WorkSHEET 2.2 Number systems: complex numbers

Name: ___________________________ 1

Write yixii

ii

+--

-++ form in the

3235

23 .

( ) ( )

i

i

ii

ii

iii

iiii

iiiii

ii

ii

ii

ii

ii

6558

6546558465

4595139165

919571313919

57

94961510

4236

3232

3235

22

23

3235

23

2

2

2

2

--=

--=

---=

+--=

+-

-=

---+

--

-+-=

++

´--

---

´++

=

--

-++

4

2 Express

45cis2 p in General / Cartesian form.

i

i

i

i

--=

÷ø

öçè

æ--=

÷øö

çèæ --=

÷øö

çèæ +=

121

212

4sin

4cos2

45sin

45cos2

45cis2

pp

pp

p

3

Page 2: WorkSHEET 2.2 Number systems: complex numbersthefinneymathslab.weebly.com/.../81042930/11mc_com… ·  · 2017-07-11in the form 2 3 5 3 2 3. i ( ) ( ) i i i i i i i i i i i i i i

Maths Quest Maths C Year 11 for Queensland Chapter 2: Number systems: complex numbers WorkSHEET 2.2

© John Wiley & Sons Australia, Ltd 2009 2

3 Find the modulus and Argument of z = -7 + 7i.

( )

43Argument

27 Modulus4

34

quadrant) (2nd1tan17

7tan

27

249

98

4949

7)7(

1

22

p

p

=

=

=

-=

-=

-=-

=

=

´=

=

+=

+-=

-

π

π

πθ

θ

z

3

4 (a) Sketch z = 5 – 5i. (b) Express z = 5 – 5i in polar form.

(a)

(b)

4

55tan)Arg(

25

50

2525

)5(5

1

22

p-=

÷øö

çèæ -=

=

=

+=

-+=

-z

z

z = 5 2 cis(-4p )

3

Page 3: WorkSHEET 2.2 Number systems: complex numbersthefinneymathslab.weebly.com/.../81042930/11mc_com… ·  · 2017-07-11in the form 2 3 5 3 2 3. i ( ) ( ) i i i i i i i i i i i i i i

Maths Quest Maths C Year 11 for Queensland Chapter 2: Number systems: complex numbers WorkSHEET 2.2

© John Wiley & Sons Australia, Ltd 2009 3

5 Sketch the following complex numbers and express them in polar form: (a) z = 2 + 3i (b) w = –4 – 2i

(a)

0.98 cis 13

98.023tan)Arg(

13

94

32

1

22

=

=

÷øö

çèæ=

=

+=

+=

-

z

z

z

(b)

)68.2( cis 52

68.246.0

42tan)Arg(

52

20

416

)2()4(

1

22

-=

-=-=

÷øö

çèæ--

=

=

=

+=

-+-=

-

w

w

w

p

6

Page 4: WorkSHEET 2.2 Number systems: complex numbersthefinneymathslab.weebly.com/.../81042930/11mc_com… ·  · 2017-07-11in the form 2 3 5 3 2 3. i ( ) ( ) i i i i i i i i i i i i i i

Maths Quest Maths C Year 11 for Queensland Chapter 2: Number systems: complex numbers WorkSHEET 2.2

© John Wiley & Sons Australia, Ltd 2009 4

6 Use Pascal’s Triangle to expand (–1 – i)5.

Fifth row is 1 5 10 10 5 1 (–1 – i)5

= (–1)5 + 5(–1)4(–i) + 10(–1)3(–i)2 + 10(–1)2(–i)3 + 5(–1)(–i)4 + (–i)5

= –1 + 5(1)(–i) + 10(-1)(-1) + 10(1)i + 5(-1)(1) – i = –1 – 5i + 10 + 10i – 5 – i

= 4 + 4i

4

7 Convert z = 4 cis

3p and w = 6 cis

32p into

standard form.

z = 4 cis3p

z = 4(cos3p + isin

3p )

z = 4(21 + i

23 )

z = 2 + 2 3 i

w = 6 cis32p

w = 6(cos32p + isin

32p )

w = 6(-21 + i

23 )

w = -3 + 3 3 i

5

Page 5: WorkSHEET 2.2 Number systems: complex numbersthefinneymathslab.weebly.com/.../81042930/11mc_com… ·  · 2017-07-11in the form 2 3 5 3 2 3. i ( ) ( ) i i i i i i i i i i i i i i

Maths Quest Maths C Year 11 for Queensland Chapter 2: Number systems: complex numbers WorkSHEET 2.2

© John Wiley & Sons Australia, Ltd 2009 5

8 Perform the following operations using

z = 4 cis3p and w = 6 cis

32p .

Convert into standard form: (a) z + w (b) w – z (c) z × w

(d) wz

(e) zw

Using the answers from the previous question:

i

iiwz

351

333322 (a)

+-=

+-+=+

i

iizw

35

)322(333 (b)

+-=

+-+-=-

24)01(24

)sin(cos24 cis 24

32 cis 6

3 cis 4 (c)

-=+-=+=

=

´=´

ppp

pp

i

wz

i

i

wz

33

31

23

21

32

3 cis

32

32

3 cis

32

cis 6 cis 4

(d)3

23

-=

÷÷ø

öççè

æ-=

÷øö

çèæ-=

÷øö

çèæ -=

=

p

pp

p

p

i

i

i

zw

433

43

23

21

23

3sin

3cos

23

3 cis

23

332 cis

23

cis 4 cis 6

(e)3

32

+=

÷÷ø

öççè

æ+=

÷øö

çèæ +=

=

÷øö

çèæ -=

=

pp

p

pp

p

p

7

Page 6: WorkSHEET 2.2 Number systems: complex numbersthefinneymathslab.weebly.com/.../81042930/11mc_com… ·  · 2017-07-11in the form 2 3 5 3 2 3. i ( ) ( ) i i i i i i i i i i i i i i

Maths Quest Maths C Year 11 for Queensland Chapter 2: Number systems: complex numbers WorkSHEET 2.2

© John Wiley & Sons Australia, Ltd 2009 6

9 Sketch the following complex numbers and express them in standard form. Use common trigonometric ratios and special triangles.

(a) z = 2 cis43p

(b) z = 4 cis6p

(c) z = 2 cis ÷øö

çèæ -32p

(a) z = 2 cis43p

z = 2(cos43p + isin

43p )

z = 2(-21 +

21 i)

z = - 2 + 2 i

(b) z = 4 cis6p

z = 4(cos6p + isin

6p )

z = 4(23 +

21 i)

z = 2 3 + 2i

(c) z = 2 cis ÷øö

çèæ -32p

z = 2(cos ÷øö

çèæ -32p + isin ÷

øö

çèæ -32p )

z = 2(-21-23 i)

z = -1 - 3 i

9

Page 7: WorkSHEET 2.2 Number systems: complex numbersthefinneymathslab.weebly.com/.../81042930/11mc_com… ·  · 2017-07-11in the form 2 3 5 3 2 3. i ( ) ( ) i i i i i i i i i i i i i i

Maths Quest Maths C Year 11 for Queensland Chapter 2: Number systems: complex numbers WorkSHEET 2.2

© John Wiley & Sons Australia, Ltd 2009 7

10 Express i52 + in standard and polar form.

Let i52 + = a + bi 2 + 5i = a2 + 2abi – b2) Therefore 2 = a2 – b2 (1) and 5 = 2ab (2)

From (2): a = b25

Substituting a into (1):

2 = 2

25÷øö

çèæb

– b2

2 = 24

25b

– b2

8b2 = 25 – 4b4 0 = 4b4 + 8b2 – 25

b2 = 8

)25)(4(488 2 --±-

= 84648 ±-

» –3.7 or 1.7 (Ignore b2 = –3.7 because b must be real) Therefore, b2 = 1.7 and b » ± 1.3. For b = -1.3, a = -1.9. For b = 1.3, a = 1.9. Therefore i52 + = ±(1.9 + 1.3i) In polar form: (always sketch the number first) | i52 + | = 22 3.19.1 + = 2.3

q = tan–1 9.13.1

= 0.6 Argument = 0.6 or -2.5

i52 + = 2.3 cis 0.6 or 2.3 cis (–2.5)

7