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