F08 - Further Complex Numbers - Assignment 2 (Solutions)

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FMaths / Further Complex Numbers (Assignment 2 Solutions) Page 1 of 6 National Junior College 2016 – 2017 H2 Further Mathematics Topic F7: Further Complex Numbers (Assignment 1 Solutions) Qn No. Solution Marking Scheme 1 B1: circle with centre at (1,0) and radius 2; B1: perpendicular bisector that passes through the centre of circle B1: shading the correct region with the correct boundaries 1 Consider triangle ABC or ACD where AB and AD are tangents to the circle. Then 3AB AD  and 2 2 2 2 2 3 13AC AB BC     12 2sin 2sin 13 13 BAC BAD              M1: construct appropriate right angled triangle B1: correct length of AC A1: Working to get angle BAD www.KiasuExamPaper.com 804

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FMaths / Further Complex Numbers (Assignment 2 Solutions) Page 2 of 6 Qn No. Solution Marking Scheme 2(a) (i) (ii)     i 3 1 i 3 1 i 3 2 z w z z z z                 arg( ) arg 1 i 3 arg 1 i 3 arg 3 z w z z           M1 : Use of 1 2 1 2z z z z A1 : 2 M1 : Use of      1 2 1 2arg arg argz z z z  A1 : 3   2 (b) (i) 1 i 3 3z   is a circle with centre  1, 3 and radius 3 units.         arg 2 i4 3 3 arg 2 arg i2 3 3 arg i2 3 3 z z z              arg i2 3 3z   is a half -line with starting point at  0, 2 3 that makes angle of 3  with the positive real axis. Note that 1. the half line passes through the centre of circle. 2. The real axis is a tangent to the circle. B1: correct sketch of circle (or the line segment must appear to be confined in the circle) B1: correct sketch of half line (hollow circle is not required at end point) B1: Required locus is a diameter of the circle (must be clearly and correctly labeled) 2b www.KiasuExamPaper.com 805

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FMaths / Further Complex Numbers (Assignment 2 Solutions) Page 3 of 6 Qn No. Solution Marking Scheme (ii) B is the point that represents the complex number z = x + iy that gives the greatest possible value of arg( ) .z From the diagram,     31 3 cos 1 , 3 2 33 3 sin 3 3 2 3 31 i 32 2 x y z                           B1: choose the correct point M1: use of trigonometry A1: 3 31 i 32 2z                www.KiasuExamPaper.com 806

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FMaths / Further Complex Numbers (Assignment 2 Solutions) Page 4 of 6 Qn No. Solution Marking Scheme 3 (i), (ii) (i) B1: Circle centered at A with radius 5 (must pass through origin) B1: Correct shaded region (ii) B1: Circle centered at B with radius 5 (must pass through origin) B1: Correct shaded region and relative position of A 3 (iii) A 90 anti-clockwise rotation about the origin B1 3 (iv) Since ( i ) ( i ) i 5,a a a a a a      thus z a ia satisfies the equations z a  5 and z ia  5. Thus C lies on the loci of z a  5 and z ia  5. B1: Or any other appropriate explanations 3(v) Since triangle OBC is an isosceles right angled triangle, 2 21 1Required area = 2 (5) (5)4 2 25 ( 2)2           M1 A1 3 (vi) Let D represents ia a on the Argand diagram. By circle properties, 4ODC   . Since i 7 ia a   , thus maxarg(z ia a)  4  tan1 1 7     . OR Since DC is parallel to OA, 1 4max arg( i ) arg( ) tan 3z a a a          M1: Identify point D correctly A1 A B C D O Re Im Locus for (i) Locus for (ii) www.KiasuExamPaper.com 807

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FMaths / Further Complex Numbers (Assignment 2 Solutions) Page 5 of 6 Qn No. Solution Marking Scheme 4 (i) B1: correct sketch of half-line. B1: correct sketch of circle with its interior shaded. 4(ii) Let d be the least possible value of 5i .z Using ONE , sin 4 5 5 5 2 22 d d     B1 4 (iii) (a)   2 23 3 18 18 2 2 OA b OA b        B1 www.KiasuExamPaper.com 808

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FMaths / Further Complex Numbers (Assignment 2 Solutions) Page 6 of 6 4 (iii) (b) Method 1:           18 2 cos sin 4 4 1 13 2 2 i 2 2 3 2 223 i 2 2 2 23 3i i 2 2 2 23 3 i 2 2 3 2 3 2 i z i                                                     Method 2:       22 2 2 2 2 18 2 18 2 2 18 2 2 18 2 2 3 2 x x x x x x              3 2 3 2 iz      M1 A1 4 (iii) (c) The locus of 1w w z  is a perpendicular bisector of the line segment joining the points O and C. Since gradient of line segment OC is -1, Hence, gradient of perpendicular bisector = 1. Let point M be the midpoint of OC. Hence, point M is    3 2 3 2 ,2 2          . Equation of locus:           3 2 3 2 12 2 3 2 3 2 2 2 3 2 y x y x y x                     B1 O C x x www.KiasuExamPaper.com 809