DHS Complex Numbers (9649) (Revision Solutions)

Complex Numbers H2 Double Math Complex Numbers (Solutions) 1 HCI/2014/I/3 One of the roots of the equation 32 21 3 i 0zz a z    is iz  . Find the complex number a and the other roots. [5] (a)   32Let P 2 1 3i Pi 1 2 1 3 i 0 23 i zz za z a a              Use long division or by comparing coefficient method,       32 2 P2 2 3 i 1 3 i =i 2 i3 i zz z z zz z                  2 2 2i 2i 4 3i2i 3i 0 2 2i 4i 1 or 3 i2 zz z zz             2 MI/2014/I/10 (i) By using de Moivre’s theorem, or otherwise, show that  2 ππ12 c o s s i n . 44 n n nnii    [2] (ii) Using the result in (i) or otherwise, find the least positive integer n for which 1 n i is real and negative. Solution by trial and error will not be accepted. [3] (iii) For the equation 432 0za zb zc z d   where a, b, c and d are real, give a brief explanation and determine the possible number of complex roots the equation can have. [2] (iv) Solve the equation 4 40 ,z  expressing the solutions in the form xi y where x and y are real. [4]

Complex Numbers i  2 2 12 c o s s i n 44 2c o s s i n 44 2c o s s i n44 n n n n ii nn i nn i                        Alternatively,  4 4 2 12 2 2c o s s i n44 n in ni n ie e nn i           Alternatively,   2 12 2 arg 1 arg 1 4 nn n n i in i n       212 c o s s i n 44 n n nnii      ii For 1 n i to be real and negative,  πarg 1 ..., 3 π, π, π,3 π,...4 For least possible , π π4 least 4 n ni n n n          iii Since the coefficients of the equation are real, complex roots will occur in conjugate pairs by Conjugate Roots Theorem. Furthermore, the order of the equation is 4, we would expect 4 roots ie. 0 pair of complex root with 4 real roots or 1 pair of complex conjugate roots with 2 real roots or 2 pairs of complex conjugate roots and no real root.

Complex Numbers iv  4 24 42 40 44 , 0 , 1 , 2 2, 0 , 1 , 2 ik ki z ze k ze k               4 4 3 4 3 4 when 0, 2 1 when 1, 2 1 when 1, 2 1 when 2, 2 1 i i i i kze i kz e i kz e i kz e i                          Alternatively:       4 22 40 22 0 42 12 42 12 z zi zi i zi i zi            Alternatively:   4 43 2 2 3 4 42 2 4 3 3 22 Let . 40 46 4 4 0 64 0 a n d 4 4 0 4 0 wh zx i y xi y xi x y x yi x y y xx y y x y x y xy x y xy              44 en , when , 1 1 1 1 1 1 1, 1, 1, 1 xy x y yy yy xx zi i i i            

Complex Numbers 3 ACJC/2014/II/2 The complex number z satisfies the relations arg 3 3i 4z    and 33 i ,zb  where b is a constant and 13 .b (i) Illustrate each of the above relations on a single Argand diagram. [2] (ii) Find the exact least possible value of 5i .z [1] (iii) Given that the least possible value of z is 18 2 , (a) find the value of b, [1] (b) hence find an exact expression for z, in the form ixy . [2] (c) State the cartesian equation of the locus of the point representing complex variable w such that 1ww z , where 1z is the complex number found in part (b). [1] (i) -5 Locus of Locus of O -3 Locus of M (z1) 3 -3 3 A b C N E

Complex Numbers (ii) (iii) (a) (iii) (b) (iii) (c) Let d be the least possible value of 5i .z Using ONE , 55 2sin 45 2 2 d d     2233 1 8 18 2 2 OA bO A b      Method 1:      1 82c o s s i n 44 1132 2 i 22 32 223i 22 2233 i i 22 2233 i 22 32 32 i zi                                Method 2:    222 2 2 2 18 2 18 2 2 18 2 2 18 2 2 32 xx x x x x          32 32 iz    The locus of 1ww 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. O C x x

Complex Numbers Hence, point M is    32 32 ,22      . Equation of locus:      32 32 122 32 32 22 32 yx yx yx           

Complex Numbers 4 HCI/2014/II/3 (a) (i) Find the fifth roots of –32, expressing the roots in the form ier  , where 0r and ππ  . [2] (ii) The roots representing 1z and 2z are such that    120a r g a r g πzz . State the complex number w in the form ier  where 21zw z . [1] (b) The complex number z satisfies 33 i 1izz   and ππ arg63 z . (i) On an Argand diagram, sketch the re gion in which the point representing z can lie. [3] (ii) Find the area of the region in part (b)(i). [3] (iii) Find the range of values of  arg 5 iz . [2] (a) (i) 5 32z    i5 i255 32e 2e k z z      2i 55Therefore 2e , 0, 1, 2 . k zk     (a) (ii) Since the two complex numbers are in the 1st and 2nd quadrants corresponding to 0k  and 1k  , thus 2i 5ew  . (b) (i) Re 4 4 Im F(2, 2) A B (1, 1) (3, 3) 6  6  O C D

Complex Numbers (b) (ii) Method 1 (Using ½ x base x height)  2222 8 tan tan 12 8t a n 12 1Area of triangle 2 1 8 2 8 tan 8 tan 2.14 (to 3 s.f.)21 2 1 2 OF FAFOA OF FA AB OF                   Method 2 (Using sum of areas of triangles)    22 2 Let Area area area area 11 1sin 2 4 sin 426 2 6 2 83 2 0 44 3 4 4 3 since 0 OA OB x OAB OBC OAD OCD xx xx x xx                Thus, area of shaded region =  221 sin 4 3 1 16 8 326x     (b) (iii) Note that the point (5, –1) lies on the perpendicular bisector. Therefore  131 arg 5 i tan45 z        . If correct to 3 s.f., answer is  2.36 arg 5 i 2.94z   .

Complex Numbers 5 RI/2014/II/4 Do not use a calculator in answering this question. The complex number z satisfies both the relations 23 i 4z  and 5 arg i6 z  . (i) On an Argand diagram, shade the region in which the point representing z can lie. [4] (ii) Find the least possible value of .z [2] (iii) State the cartesian form of the complex number z when iz is greatest. [1] (iv) Find the range of values of arg 4 3 iz . [2] (i) (ii) The least possible z is given by the perpendicular distance from the origin to the half-line  5arg i 6 z  , denoted by h as shown in the diagram. 33 sin or 1.sin 63 2 h  (iii) 43 3 i (iv)     11 20a r g 4 3i 0a r g 4 3i 62 3 zz         4 Locus of z Re Im -1 h Re Im

Complex Numbers 6 RVHS/2014/II/2 (i) Solve the equation 6 64 0z  , giving the roots in the form ier  , where 0r  and ππ   . [3] (ii) Show the roots on an Argand diagram. [2] The roots denoted by 1z and 2z are such that  12 π0a r g a r g 2zz . The complex numbers 1z and 2z are represented by the points 1Z and 2Z in the Argand dia