SUCCEEDING IN THE VCE 2017 UNIT 3 SPECIALIST MATHEMATICS STUDENT SOLUTIONS€¦ · · 2017-02-27UNIT 3 SPECIALIST MATHEMATICS STUDENT SOLUTIONS FOR ERRORS AND UPDATES, ... Microsoft

The School For Excellence 2017 Succeeding in the VCE – Unit 3 Specialist Maths Page 1

SUCCEEDING IN THE VCE 2017

UNIT 3 SPECIALIST MATHEMATICS

STUDENT SOLUTIONS

FOR ERRORS AND UPDATES, PLEASE VISIT WWW.TSFX.COM.AU/VCE-UPDATES

QUESTION 2

(a) 01342 xx 4 2 2 13 0 2 4 13 0 2 9 2 3

(b) 0752 2 xx

2 0

2 0

2 0

2 √ √ 0

∴4

315 ix


(c) 233 2 zz

0233 2 zz

03

2

3

32 zz

03

2

6

3

6

322

z

036

21

6

32

z

iz6

21

6

3

iz6

21

6

3

6

)71(3 i

QUESTION 3

(a) 1073 2 xx

01073 2 xx

6

)10)(3(4497 x

6

120497 x

6

717 ix

(b) 0753 2 zz

6

)7)(3(4255 z

6

84255 z

6

595 iz


QUESTION 5 (a) wz wz i)75()23(

i25

(b) wz wz i)75()23(

i121 QUESTION 8 (a) (b)

(c) a 2 i (d) (e)

9 7

5 5i



QUESTION 10


QUESTION 11 (a) z i23 (b) wz i54 (c) wz i2 (d) z3 i69

(e) wz. 3 2 1 3i i (f) 2wi 1 3 1i

i113 i31

(g) zi 23 i (h) 3zi 3 2i i i32

QUESTION 13A (a) iz 22

(b) iz 232

(c) iz 3


QUESTION 13B (a) iz 43

(b) iz 65


QUESTION 14 (a)

2

71

4

722

4

282

)2(2

)3)(2(422

0322

412

2)1(

)1(

2

2

22

22

x

x

xx

xxx

xxz

ixxz

(b)

Arg(z) = 3

x

x 13

3tan

13 xx

13 xx

113 x

13

13

13

1

x

2

13 x


QUESTION 18

(a)

4

32

cisz

4

3sin

4

3cos2

iz

4sin

4cos2

i

2

1

2

12 i

i 1

(b)

6

55

cisz

6

5sin

6

5cos5

iz

6sin

6cos5

i

22

35

i

i

32

5

i

2

5

2

15


QUESTION 21

If

4

32

cisz and

63

cisw find:

(a) wz.

(b) z

w


QUESTION 22

(a) Find 1

2

z

z in exact Cartesian form.

(b) Express 2z in polar form.

2 23

z cis

(c) Given that

4sin

4cos21

iz , find the polar form of

1

2

z

z.

(d) Hence find the exact value of

12

cos

.

From (a):

2

13

2

31

1

2 iz

z

From (c):

12sin2

12cos2

122

1

2 cicis

z

z.

Equate real parts of each expression for 1

2

z

z:

12

cos22

31

4

62

22

31

12cos


QUESTION 24*

QUESTION 25

(b) 2

1

z

(c) 2z


QUESTION 26

i16

1

16

3


Using technology to check your answer:

QUESTION 27 (a) Simplify the following expressions, giving your answer in Cartesian form.

(i) 3(1 3 )z i

= 1

08

i


(ii)

3

4

52

3( 3 )

cisz

i

i4

3

4

1


(iii) z (1 – i)4 ( 3 + i)5

=

4

4cis2

5

6cis2

=

44cis)2( 4

65cis25

= 4 cis (– )

6

5cis32

= 128 cis

6

= 128 3 1

2 2i

= 64 3 – 64i

(b) If

3

2acisz and iz 312

1

find the value of a .


(c) If

6

acisz and

42

bcisw and

122

cis

w

z find the value of a and b .

(d) Given that

3

2acisz and

42

bcisw find the value of a and b if

210823

ciswz .


QUESTION 28* (a) Find 21zz in exact Cartesian form.

1 2

2

(4 3 )(7 )

28 4 21 3

25 25

z z i i

i i i

i

(b) Hence find 21zz in exact Polar form.

2 2 21 2 25 25 2(25) 25 2z z 1 1

1 2

25Arg tan tan 1

25 4z z

1 2 25 2 cis 4

z z

(c) Write both 1z and 2z in exact Polar form.

2 21 4 3 25 5z 1 1

1

3 3Arg tan tan

4 4z

1

1

35 cis tan

4z

2 22 7 1 50 5 2z 1 1

2

1 1Arg tan tan

7 7z

1

2

15 2 cis tan

7z

(d) Hence find the exact value of 1 13 1tan tan

4 7

.

From (b) 1 2 25 2 cis 4

z z

From (c) 1 11 2

3 1 25 2 cis tan tan

4 7z z

Therefore 1 13 1 25 2 cis tan tan 25 2 cis

4 7 4

Therefore 1 13 1tan tan

4 7 4


QUESTION 29*

3 2 cis 3 2 cis 6 6

mm m

i i

3 2 cis 3 2 cis 6 6

mm m

i i

2 cis 2 cis 06 6

m mm m

2 cos sin cos sin 06 6 6 6

m m m m mi i

2 cos sin cos sin 06 6 6 6

m m m m mi i

2 2 sin 06

m mi

2 sin 0 as 2 06

mmi

sin 0 6

m

6

mn n Z

6 m n n Z


QUESTION 30* (a) rw cis

2 2 22 and .n n nw r cis n w a

Equating these equations gives:

2 cis 2nr n a

2 cos 2 sin 2 0nr n i n a i

2 2 cos 2 and sin 2 0n nr n a r n

sin 2 0 as requiredn

(b)

cisw r

2

2 cis 2n

nw r n

2 2 cos 2 sin 2 n nr n r i n

2 2 cos 2 sin 2 n nr n r i n

2 (0) na r i from (a) a (c)

2 2 2 21n n n nw w w a

2 2 2 21 ( ) ( )

n n n nw w w a

Therefore w and w satisfy 2nz a


QUESTION 31*

(b) Hence show that a solution to 052016 35 xxx is

5sin

x .

Substitute sinx into the result given in (a) (i):

xxx 52016)5sin( 35 .

Then the equation 052016 35 xxx given to solve becomes 0)5sin( : 0)5sin( m 5 , where Zm

5

m

Therefore

5sinsin

mx .

Let 1m :

5sin

x is a solution.

(c) Hence show that the exact value of

5sin

is equal to 5210

4

1 .

5 3 4 2 4 216 20 5 0 16 20 5 0 0 or 16 20 5 0x x x x x x x x x

Therefore 2 5 5

8x

from the quadratic formula.

Which gives 5 5 10 2 5

8 4x

.

Now as sin sin sin6 5 4

then 1 2

sin2 5 2

then

10 2 5sin

5 4


(d) Find an exact solution to 152016 35 xxx of the form )( cba , where

ba, and c are integers.

Substitute cosx into the result given in (a) (ii):

xxx 52016)5cos( 35 .

Then the equation 152016 35 xxx given to solve becomes 1)5cos( : 1)5cos( )12(5 m , where Zm

5

)12(

m.

Therefore (2 1)cos cos

5

mx

.

Let 0m :

5cos

x is a solution.

Substitute 52104

1

5sin

from part (c) into 15

sin5

cos 22

and solve

for

5cos

:

4

15

5cos


QUESTION 32

(b) 83 z

(c) iz 13


QUESTION 33 Solution

(b)

Re(z)

Im(z)

Re(z)

Im(z)


QUESTION 34A


(e)

Im(z)

Re(z)1 2 3 4 – 1 – 2 – 3 – 4

1

2

3

4

– 1

– 2

– 3

– 4

QUESTION 34B (a)

(b)

112cis

12

52cis

12

2cis

4


QUESTION 36

Use technology to check your answers:

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SUCCEEDING IN THE VCE 2017 UNIT 3 SPECIALIST MATHEMATICS STUDENT SOLUTIONS€¦ · · 2017-02-27UNIT 3 SPECIALIST MATHEMATICS STUDENT SOLUTIONS FOR ERRORS AND UPDATES, ... Microsoft