Complex Number

COMPLEX NUMBER INTRODUCTION Indian mathematician Mahavira (850 A.D.) was first to mention in his work 'Ganitasara Sangraha'; 'As in nature of things a negative (quantity) is not a square (quantity), it has, therefore, no square root'. Hence there is no real number x which satisfies the polynomial equation x2 + 1 = 0. A symbol 1 , denoted by letter i was intrdouced by Swiss Mathematician, Leonhard Euler (1707-1783) in 1748 to provide solutions of equation x2 + 1 = 0. i was regarded as a fictitious or imaginary number which could be manipulated algebrically like an ordinary real number, except that its square was – 1. The letter i was used to denote 1 , possibly because i is the first letter of the Latin word 'imaginarius'. DEFINITION 1 . It is denoted by z i.e. Complex numbers are definited as expressions of the form a + ib where a, b R & i = z = a + ib. ‘a’ is called as real part of z (Re z) and ‘b’ is called as imaginary part of z (Im z). EVERY COMPLEX NUMBER CAN BE REGARDED AS Purely real if b = 0 Purely imaginary if a = 0 Imaginary if b  0 (A) The set R of real numbers is a proper subset of the Complex Numbers. Hence the complete number system is N  W  I  Q  R  C. (B) Zero is purely real as well as purely imaginary but not imaginary. (C) i =  1 is called the imaginary unit. Also i² =  1; i3 =  i ; i4 = 1 etc. (D) a b = a b only if atleast one of a or b is non - negative. (E) If z = a + ib, then a – ib is called complex conjugate of z and written as z = a – ib (F) Real numbers satisfy order relations where as imaginary numbers do not satisfy order relations i.e. i > 0, 3 + i < 2 are meaningless. Ex. Write the following as complex number (i) Sol. 16 (ii) x , (x > 0) (iii) –b + (i) 0 + 4i (ii) x + 0i (iii) –b + i 4ac , (a, c> 0) 4ac Ex. The value of i57 + 1/i125 is :- Sol. i57 + 1/i125 = i56. i +   4 = i =i  14 i 1 i1 24 .i 1 i  4 31 i 1 i  i 2  ii  0 i i ALGEBRAIC OPERATIONS Fundamental operations with complex numbers In performing operations with complex numbers we can proceed as in the algebra of real numbers, replacing i2 by – 1 when it occurs. 1. Addition (a + bi) + (c + di) = a + bi + c + di = (a + c) + (b + d) i 2. Subtraction (a + bi) – (c + di) = a + bi – c – di = (a – c) + (b – d) i 3. Multiplication (a + bi) (c + di) = ac + adi + bci + bdi2 = (ac – bd) + (ad+ bc)i 4. Division a  bi c  di = = a  bi c  di ac  adi  bci  bdi 2 . = c  di c  di c 2  d 2i2 ac  bd  (bc  ad)i 2 c d 2 = ac  bd 2 c d 2 + bc  ad c2  d 2 i Inequalities in imaginary numbers are not defined. There is no validity if we say that imaginary number is positive or negative. e.g. z > 0, 4 + 2i < 2 + 4 i are meaningless. In real numbers if a2 + b 2 = 0 then a = 0 = b however in complex numbers, z12 + z 22 = 0 does not imply z 1 = z 2 = 0. (i) The algebraic operations on complex numbers are similar to those on real numbers treating i as a polynomial. (ii) Inequalities in complex numbers (non-real) are not defined. There is no validity if we say that complex number (non-real) is positive or negative. e.g. z > 0, 4 + 2i < 2 + 4i are meaningless. (iii) In real numbers, if a2 + b2 = 0, then a = 0 = b but in complex numbers, z12 + z22 = 0 does not imply z1 = z2 = 0. COMPLEX NUMBER Ex. Sol. Find multiplicative inverse of 3 + 2i. Let z be the multiplicative inverse of 3 + 2i. then  z . (3 + 2i) = 1  z= 3  2i 1 =  3  2i   3  2i  3  2i  z= 3 2 – i 13 13 3 2    i  13 13  Ex. 3  2isin  will be purely imaginary, if  = 1  2isin  Sol. 3  2isin  will be purely imaginary, if the real part vanishes, i.e., 1  2isin    3  4 sin 2   i  8 sin   (3  2 isin  ) (1  2i sin  )  = (1  2isin ) (1  2i sin ) 1  4 sin 2  3  4 sin 2  0 1  4 sin 2     3 – 4 sin2  = 0 (only if  be real) 2   3   sin  =  =  sin    3  2    = n ± 2 2  ,nI 3 EQUALITY IN COMPLEX NUMBER Two complex numbers z1 = a1 + ib1 & z2 = a2 + ib2 are equal if and only if their real and imaginary parts are equal respectively i.e. z 1 = z2  Re(z1) = Re(z2) and m (z1) = m (z2). Ex. Sol. Find the value of x and y for which (2 + 3i) x2 – (3 – 2i) y = 2x – 3y + 5i where x, y  R. (2 + 3i)x2 – (3 – 2i)y = 2x – 3y + 5i  2x2 – 3y = 2x – 3y  x2 – x = 0  x = 0, 1 and 3x2 + 2y = 5 5  if x = 0, y = and if x = 1, y = 1 2 5  x = 0, y = and x = 1, y = 1 2 5 are two solutions of the given equation which can also be represented as  0,  & (1, 1)  2  5  0,  , (1, 1)  2 Ex. If x  5  2 4 , find the value of x4 + 9x3 + 35x2 – x + 4. Sol. We have , x = –5 + 2 4   x + 5 = 4i x2 + 10x + 25 = –16   (x + 5)2 = 16i2 x2 + 10x + 41 = 0 Now,    x4 + 9x3 + 35x2 – x + 4 x2(x2 + 10x + 41) – x(x2 + 10x + 41) + 4(x2 + 10x + 41) – 160 x2(0) – x(0) + 4(0) – 160 –160 Ex. Find square root of 9 + 40i Sol. Let x + iy = 9 (x + iy)2 = 9 + 40i  x 2 – y2 = 9 and xy = 20 .....(i) ......(ii) squaring (i) and adding with 4 times the square of (ii) we get x4 + y4 – 2x2 y2 + 4x2 y2 = 81 + 1600  (x2 + y2)2 = 1681  x2 + y2 = 41 ......(iii) from (i) + (iii) we get x2 = 25 and y2 = 16 from equation (ii) we can see that x & y are of same sign  x + iy = (5 + 4i) or – (5 + 4i)  sq. roots of 9 + 40i = ± (5 + 4i)   x=±5 y=±4 and ± (5 + 4i) CONJUGATE OF A COMPLEX NUMBER If z = a + ib then its conjugate complex is obtained by changing the sign of its imaginary part & is denoted by z . i.e. z = a  ib. IMPORTANT PROPERTIES OF CONJUGATE (A) z + z = 2 Re (z) (B) z  z = 2 i Im (z) (E) z1  z2 = z1  z2 (D) z1  z2 = z1 + z 2 (F) z1 z2 = z1 . z2 . In general z1 z2 .........zn  z1 .z2 .........zn (G)  z1  z1  z  = z ; z2  0 2 2 (H) If f( + i) = x + iy (C) (z) = z  f( – i) = x – iy COMPLEX NUMBER Note that (iii) z z = a² + b², which is purely real (iv) If z is purely real, then z – z = 0 (v) If z is purely imaginary, then z + z = 0 –z (vi) Im z If z lies in the 1st quadrant, then z lies in the 4th quadrant and Re –z z  z lies in the 2nd quadrant. Modulus If P denotes complex number z = x + iy, then the length OP is called modulus of complex number z. It is denoted by |z|. OP = |z| = x 2  y2 Geometrically zrepresents the distance of point P from origin. ( z0) IMPORTANT PROPERTIES OF MODULUS (A) z  0 (B) z  Re (z) (C) z  Im (z) (D) z =  z  = z– z  (E) z z = z (F) z1 z2 = z1.z2 . In general | z1 z2 .........zn | | z1 | .| z2 | .........| z n | (G) z1 z1 = , z2  0 z2 z2 (H) zn = z n, n  I (I) |z1 + z2|2 = |z1|2 + |z 2|2 + 2 Re  z1 z2  (J) |z1 + z2|2 = |z1|2 + |z2|2 + 2|z1||z2| cos( – ), where ,  are arg(z1), arg(z2) respectively. (K) z1  z2 2 + z1  z2 2 = 2  z1 2 2 2  z2   (L) z1 z2  z1 + z2  z1+ z2 [Triangle Inequality] (M) z1 z2  z1 – z2  z1+ z2 [Triangle Inequality]  z if z  0 Unlike real numbers, z =  is not correct.   z if z  0 (C) Argument or Amplitude : If P denotes complex number z = x + iy and if OP makes an angle  with real axis, then  is called one of the arguments of z.  = tan1 y (angle made by OP with positive real axis) x IMPO7RTANT PROPERTIES OF AMPLITUDE (A) amp (z1. z2) = amp z1 + amp z2 + 2 k k  I (B) z  amp  1  = amp z1  amp z2 + 2 k k  I  z2  (C) amp(zn) = n amp(z) + 2k; n,k  I where proper value of k must be chosen so that RHS lies in (, ]. (i) (ii) (iii) Argument of a complex number is a many valued function. If  is the argument of a complex number, then 2n +  ; n  I will also be the argument of that complex number. Any two arguments of a complex number differ by 2n The unique value of  such that   <   is called Amplitude (principal value of the argument). Principal argument of a complex number z = x + iy can be found out using method given below : Im y   such that    0,  . x  2 (A) Find  = tan 1 (B) (iv) Use given figure to find out the principal argument according as the point lies in respective quadrant. Unless otherwise stated, amp z implies principal value of the argument. (v) The unique value of  = tan1 (vi) If z = 0, arg(z) is not defined (viii) If z is real & positive, arg(z) = 0     Re y such that 0    2  is called least positive argument. x (vii) If z is real & negative, arg(z) = . (ix) If    , z lies on the positive side of imaginary axis. 2  , z lies on the negative side of imaginary axis. 2 By specifying the modulus & argument a complex number is defined completely. Argument impart direction & modulus impart distance from origin. For the complex number 0 + 0i the argument is not defined and this is the only complex number which is given by its modulus only. (x) If    COMPLEX NUMBER Ex. Find the modulus, argument, principal value of argument, least positive argument of complex numbers (A) 1 + i 3 Sol. (A) (B) –1 + i 3 For z = 1 + i 3 (D) –1 – i 3 (C) 1 – i 3 y (1, 3 ) | z|  1 2  ( 3 )2  2 3  arg (z) = 2n + , n  I 3 60° x 1  Least positive argument is 3 If the point is lying in first or second quadrant then amp(z) is taken in anticlockwise direction. In this case amp(z) = (B)  3 For z = –1 + i 3 (–1, 3 ) y |z| = 2 2 , n I 3 arg (z) = 2n + Least positive argument = amp(z) = (C) 3 2 3 60° 1 120° x 2 3 For z = 1 – i 3 |z| = 2 arg (z) = 2n –  ,nI 3 5 3 If the point lies in third or fourth quadrant then consider amp(z) in clockwise direction. Least positive argument = In this case amp(z) = – (D)  3 For z = –1 – i 3 |z| = 2 arg (z) = 2n – 2 n I , 3 Least positive argument = amp(z) = – 2 3 4 3 2 Ex. Sol.  (3  4 i)(1  i)(1  3i)  Find amp z and |z| if z    .  (1  i)(4  3i)(2i)  amp z = 2 amp (3  4 i)  amp(1  i)  amp(1  3 i)  amp (1  i)  amp(4  3i)  amp(2i)  2k where k  I and   k chosen so that amp z lies in (–,].   4      3   amp z  2  tan 1        tan 1       2 k 3 4 3 4    4  2    amp z = 2  tan   amp z    3 1 4 4     cot 1   + 2k  amp z  2     2 k 3 3 3 2 3  [at k = –1] Also, | z|   3  4i 1  i  1  3i 1  i  4  3i2i  2   3  4 i 1  i 1  3i   | z|   | 1  i| | 4  3i| | 2i|       2    1 |z| =   2 5 2  2 2 z 1 is purely imaginary, then prove that | z | = 1 z 1 Ex. If Sol. Re   z 1   =0  z 1   z  1  z 1  +  =0 z  1  z 1   z 1 z 1 + =0 z 1 z 1  zz – z +z–1+zz –z+ z –1 =0  zz =1   | z |2 = 1 |z|=1 Hence proved COMPLEX NUMBER Ex. z1 and z2 are two complex numbers such that z1  2 z2 2  z1 z2 is unimodular (whose modulus is one), while z2 is not unimodular. Find |z1|. Sol. Here z1  2 z 2 z1  2 z 2 1 =1  2  z1 z 2 2  z1 z2 2 z1  2 z2  2  z1 z2   z1  2 z2   z1  2 z2   2  z1 z2  2  z1 z2    z1  2 z2   z1  2 z2   2  z1 z2 2  z1 z2   z1 z1  2 z1 z2  2 z2 z1  4 z2 z2  4  2 z1 z2  2 z1 z2  z1 z1 z2 z2  z1  z 2  4 z2 2 1 2  4  z1   4 1  z2 2  z1  2 z2 2  2 z2 2  2  z1 z2  z1 2  z1 2 z2 2  4 z2 2 4 0 0 But |z2|  1 (given)  |z1|2 = 4 Hence, |z1| = 2. DISTANCE, TRIANGULAR INEQUALITY If z 1 = x1 + iy1, z2 = x2 + iy2 , then distance between points z1, z2 in argand plane is |z1 – z 2| = ( x1  x 2 )2  ( y1  y 2 )2 In triangle OAC OC  OA + AC OA  AC + OC AC  OA + OC using these in equalities we have ||z 1| – |z2||  |z 1 + z2|  |z 1| + |z 2| Similarly from triangle OAB we have ||z1| – |z2||  |z1 – z2|  |z1| + |z2| (A) ||z1| – |z2|| = |z1 + z2|, |z1 – z2| = |z1| + |z2| if origin, z1 and z2 are collinear and origin lies between z1 and z2 . (B) |z 1 + z2| = |z1| + |z2|, ||z1| – |z2|| = |z 1 – z2| if origin, z1 and z2 are collinear and z1 and z2 lies on the same side of origin. Ex. z 2 = 1 then find the maximum and minimum value of |z| z Sol. z 2 =1 z |z| 2 z  z 2 2 |z|+  2 z Let | z | = r r  r+ 2 2 1  r+ r r 2 1 r  r  R+  –1  r – and r  2 1 r  r  [1, 2]  from (i) and (ii) r  [1, 2] ......(i) 2 1 r ......(ii) r  [1, 2] |z|max = 2, |z|min =1 Ex. 4 If z  = 2, then the greatest value of z is z Sol. 4 4 4 4 4 We have z  z    z   =2+ z z z z z  z 2 z 4   z 1  5  2  z  1 2 5 z  5 1 Therefore, the greatest value of z is 5 +1. REPRESENTATION OF A COMPLEX NUMBER Cartesian Form (Geometric Representation) Every complex number z = x + i y can be represented by a point on the cartesian plane known as complex plane (Argand diagram) by the ordered pair (x, y). length OP is called modulus of the complex number denoted by z &  is called the argument or amplitude . eg. z =  = tan1 x 2  y2 & y (angle made by OP with positive xaxis) x  COMPLEX NUMBER  z if z  0 is not correct  z if z  0 (i) z is always non negative . Unlike real numbers z =  (ii) Argument of a complex number is a many valued function . If  is the argument of a complex number then 2 n+  ; n  I will also be the argument of that complex number. Any two arguments of a complex number differ by 2n. (iii) The unique value of  such that –  <   is called the principal value of the argument. (iv) Unless stated, amp z implies principal value of the argument. (v) By specifying the modulus & argument a complex number is defined completely. For the complex number 0 + 0 i the argument is not defined and this is the only complex number which is given by its modulus. (vi) There exists a one-one correspondence between the points of the plane and the members of the set of complex numbers. Trigonometric / Polar Representation z = r (cos  + i sin ) where z = r ; arg z =  ; z = r (cos  i sin ) Note : cos  + i sin  is also written as CiS  Euler's formula : The formula eix = cosx + i sin x is called Euler's formula. It was introduced by Euler in 1748, and is used as a method of expressing complex numbers. Also cos x = e ix  e  ix e ix  e  ix & sin x = are known as Euler's identities. 2 2i Exponential Representation Let z be a complex number such that  z  = r & arg z =  , then z = r.ei VECTORIAL REPRESENTATION OF A COMPLEX NUMBER (A) y In complex number every point can be represented in terms of position vector.  P(z)  If the point P represents the complex number z then, OP = z &  OP  = z O (B) If P(z1) & Q(z2) be two complex numbers on argand plane then x y Q(z ) P(z )  PQ represents complex number z2 – z1. O x (i)     If OP = z = r ei  then OQ = z1 = r ei ( + ) = z. e i. If OP and OQ are ^ ^ of unequal magnitude then OQ  OP e i  i.e. z1 z  e i z1 z y (ii) C(z ) In general, if z1, z2, z3 be the three vertices of ABC then z 3  z 1 | z3  z 1 |  z  z1  i  arg  3 . z2  z1 | z2  z1 | e . Here  z2  z1   B(z ) A(z ) x (iii)  z  z1  Note that the locus of z satisfying arg     is:  z  z2  Case (A) 0 <  < /2 Locus is major arc of circle as shown excluding z1 & z2 Case (B)   2 Locus is minor arc of circle as shown excluding z1 & z2 (iv) If A, B, C & D are four points representing the complex numbers z4  z3 z1 , z2 , z3 & z4 then AB  CD if z  z 2 1 AB  CD if (v) is purely real ; z4  z3 is purely imaginary.. z2  z1 If z1, z2, z3 are the vertices of an equilateral triangle where z0 is its circumcentre then (i) z 12 + z 22 + z 23  z1 z2  z2 z3  z3 z1 = 0 (ii) z 12 + z 22 + z 23 = 3 z 20 COMPLEX NUMBER GEOMETRICAL REPRESENTATION OF FUNDAMENTAL OPERATIONS (i) Geometrical representation of addition. z2 If two points P and Q represent complex numbers z 1 and z2 respectively in the Argand plane, then the sum z 1 + z 2 is represented by the extremity R of the diagonal OR of parallelogram OPRQ having OP and OQ as two adjacent sides. (ii) Geometric representation of substraction. (iii) Modulus and argument of multiplication of two complex numbers. Theorem For any two complex numbers z 1, z 2 we have |z 1 z2| = |z 1| |z2| and arg (z 1z2) = arg (z 1) + arg (z 2). Proof z1 = r1 e i1 , z2 = r2 ei2 z1z2 = r2r2 e i( 1   2 )  |z1z2| = |z1| |z2| arg (z1z2) = arg (z1) + arg (z2) i.e. (i) to multiply two complex numbers, we multiply their absolute values and add their arguments. P.V. arg (z1z2)  P.V. arg (z1) + P.V. arg (z2) (ii) |z1 z2 .... zn| = |z1| |z2| ..... |zn| (iii) arg (z 1z2 .... zn) = arg z1 + arg z2 + ..... + arg z n (iv) Geometrical representation of multiplication of complex numbers. Let P, Q be represented by z 1 = r 1 e i1 , z 2 = r2 e i2 repectively. To find point R representing complex number z1z2 we take a point L on real axis such that OL = 1 and draw triangle OQR similar to triangle OLP. Therefore OR OP = OQ OL  OR = OP.OQ i.e. OR = r1r2 and ˆ = QOR 1 ˆ = LOP ˆ + POQ ˆ ˆ = + – + = + LOR + QOR 1 2 1 1 1 2 Hence, R is represented by z1z2 = r1r2 e i( 1   2 ) (v) Modulus and argument of division of two complex numbers. Theorem : Note : (vi) If z1 and z2 (0) are two complex numbers, then z1 z2 = | z1 | z  and arg  1  = arg (z1) – arg (z2) | z2 |  z2  z  P.V. arg  1   P.V. arg (z1) – P.V. arg (z2)  z2  Geometrical representation of the division of complex numbers. Let P, Q be represented by z1 = r1e i1 , z 2 = r2 e i 2 respectively. To find point R representing complex number z1 z 2 , we take a point L on real axis such that OL = 1 and draw a triangle OPR similar to OQL. Therefore and OP OR = OQ OL  OR = r1 r2 ˆ = LOP ˆ – ROP ˆ = – LOR 1 2 Hence, R is represented by z1 r1 i( 1   2 ) = . e z2 r2 CONJUGATE OF A COMPLEX NUMBER Conjugate of a complex number z = a + ib is denoted and defined by z = a – ib. In a complex number if we replace i by – i, we get conjugate of the complex number. z is the mirror image of z about real axis on Argand's Plane. COMPLEX NUMBER Geometrical representation of conjugate of complex number. |z| = | z | arg (z) = – arg (z) General value of arg (z) = 2n – P.V. arg (z) Properties zz zz ,y= 2 2i (i) If z = x + iy, then x = (ii) z= z  z is purely real (iii) z+ z =0  z is purely imaginary (iv) Relation between modulus and conjugate. |z|2 = z z (v) zz (vi) (z1  z 2 ) = z1 ± z 2 (vii) (z1 z 2 ) = z1 z 2 (viii) (z1 )  z1  (z2  0)   =  z 2  (z2 ) Theorem Imaginary roots of polynomial equations with real coefficients occur in conjugate pairs Proof If z0 is a root of a 0zn + a1zn–1 + ...... + a n–1 z + an = 0, a0, a1, ....... an  R, then a 0 z 0n + a1z 0n 1 + ....... + an–1 z0 + an = 0 By using property (vi) and (vii) we have a 0 z0n + a 1 z0n 1 + ....... + a n 1 z0 + an = 0  Note z0 is also a root. If w = f(z), then w = f( z ) Theorem |z1 ± z2|2 = |z1|2 + |z2|2 ± (z1 z 2 + z1 z2) = |z1|2 + |z2|2 ± 2 Re(z1 z 2 ) = |z1|2 + |z2|2 ± 2 |z1| |z2| cos (1 – 2) Ex. Express the complex number z = – 1 + Sol. z = –1 + i 2 |z|= (1) 2   2 2 = 1 2 = 2 i in polar form. 3  2 Arg z =  – tan–1   =  – tan–1 ( 2) =  (say) 1    z= 3 (cos  + i sin  ) where  =  – tan–1 2 Ex. Express the following complex numbers in polar and exponential form : (i) Sol. 1  3i 1  2i (i) i 1   cos  i sin 3 3 (ii) Let z  1  3i 1  3i 1  2i    1  i 1  2i 1  2i 1  2i | z|  (1)2  1 2  2 tan   1    1  tan     1 44 44  Re(z) < 0 and Im(z) > 0  z lies in second quadrant.   = arg (z) =  –  =  – Hence Polar form is z =  3  4 4 3 3    i sin and exponential form is z  2 e 3  / 4 2  cos 4 4   i 1 i 1 2 (i  1)  =   1 i 3 (1  i 3 ) cos  i sin  3 3 2 2 (ii) Let z   z  Re(z) > 0 and Im(z) > 0   3 1   3 1  | z|      2   2       3 1   3 1   i   z     i   (1  i 3 ) (1  i 3 )  2   2   2 tan   3 1 3 1  tan z lies in first quadrant. 2  1)  2. 4 5 5  12 12 5 5    i sin Hence Polar form is z  2  cos and exponential form is z  2 e 5  / 1 2 12 12   Ex. Find the locus of : (A) |z – 1|2 + |z + 1|2 = 4 Sol. (B) Re(z2) = 0 (A) Let z = x + iy  (|x + iy – 1|)2 + (|x + iy + 1|)2 = 4  (x – 1)2 + y2 + (x + 1)2 + y2 = 4  x2 – 2x + 1 + y2 + x2 + 2x + 1 + y2 = 4  x2 + y2 = 1 Above represents a circle on complex plane with center at origin and radius unity. COMPLEX NUMBER (B) Let z = x + iy  z2 = x2 – y2 + 2xyi  Re(z2) = 0  x2 – y2 = 0  y = ± x Thus Re(z2) = 0 represents a pair of straight lines passing through origin. Ex. Find the minimum value of |1 + z| + |1 – z|. Sol. |1 + z| + |1 – z|  |1 + z + 1 – z| (triangle inequality)  |1 + z | + |1 – z|  2  minimum value of (|1 + z| + |1 – z|) = 2 Geometrically Im P |z + 1| + |1 – z| = |z + 1| + |z – 1| A –1 which represents sum of distances of z from 1 and – 1 z B 1 Re it can be seen easily that minimum (PA + PB) = AB = 2 (A) least positive argument (B) maximum positive argument (C) least modulus (D) maximum modulus The complex numbers z satisfying the condition |z – 25i| 15 are represented by the points inside and on the circle of radius 15 and centre at the point C(0, 25). The complex number having least positive argument and maximum positive arguments in this region are the points of contact of tangents drawn from origin to the circle ge o ri n t fro g in m Sol. Among the complex number z which satisfies |z – 25i|  15, find the complex numbers z having Here  = least positive argument and  = maximum positive argument  In ΔOCP,OP =  OC  –  CP  =  25  – 15  = 20 and sin    2 Ta n Ex. D40i 2 2 2 OP 20 4   OC 25 5 44 44  tan  tan  tan tan 1   33  33 C 25i  Q E O  4 Thus, complex number at P has modulus 20 and argument   tan 1   3  4 3 zp  20  cos   i sin    20   i  5 5  z p  1 2  16 i Similarly zQ = –12 + 16i From the figure, E is the point with least modulus and D is the point with maximum modulus. uuur uuur uuur Hence, z E = OE = OC – EC = 25i –15i = 10i uuur uuur uuur and z D = OD = OC + CD = 25i +15i = 40i P  N Ex. Complex numbers z1, z2, z3 are the vertices A, B, C respectively of an isosceles right angled triangle with right angle at C. Show that (z1 – z2)2 = 2(z1 – z3)(z3 – z2). Sol. In the isosceles triangle ABC, AC = BC and BCAC. It means that AC is rotated through angle /2 to occupy the position BC. Hence we have,  z2  z 3  e  i / 2   i  z2 – z3 = +i(z1 – z3) z1  z3 B(z2) z22  z23  2 z2 z3    z12  z23  2 z1 z3  C(z 3)  A(z1) z12  z22  2z1 z2  2z1 z3  2 z2 z3  2z1 z2  2z23 = 2  z1  z3  z3  z2    z1  z2 2  2  z1  z3  z3  z2  ROTATION Important results (i) arg z =  represnets points (non-zero) on ray eminating from origin making an angle  with positive direction of real axis (ii) arg (z – z 1) =  represents points (z1) on ray eminating from z 1 making z1 an angle  with positive direction of real axis real axis  2 and Arg (z – 3 – 4i) = . 6 3 Ex. Solve for z, which satisfy Arg (z – 3 – 2i) = Sol. From the figure, it is clear that there is no z, which satisfy both ray (3, 4) (3, 2) O real axis COMPLEX NUMBER Ex. Sol.   z 1  If arg  then interpret the locus.  = 3  z 1 z 1m   z 1  arg   = 3  z 1 /3 –2/3 –1 0 Re 1  1 z    arg   = 3  1  z   1  z  represents the angle between lines joining –1 and z, and 1 and z. As this angle is constant, Here arg    1  z  the locus of z will be a larger segment of circle. (angle in a segment is constant). LOGARITHM OF A COMPLEX QUANTITY (i) Loge (+ i ) = (ii) ii 1   Loge (² + ²) + i  2n  tan 1  where n  I. 2    represents a set of positive real numbers given by e    2n  2  , n  I. DEMOIVRE’S THEOREM Case  Statement If n is any integer then (i) (cos  + i sin  )n = cos n + i sin n (ii) (cos 1 + i sin 1) (cos 2 + i sin 2) (cos3 + i sin 2) (cos 3 + i sin 3) .....(cos n + i sin n) = cos (1 + 2 + 3 + ......... n) + i sin (1 + 2 + 3 + ....... + n) Case  Statement If p, q  Z and q  0 then  2k  p   2k  p  (cos  + i sin )p/q = cos   + i sin   q q     where k = 0, 1, 2, 3, ......, q – 1 Continued product of the roots of a complex quantity should be determined using theory of equations. Ex. Sol. If cos + cos + cos = 0 and also sin + sin + sin = 0, then prove that (A) cos2 + cos2 + cos2 = sin2 + sin2 + sin2 = 0 (B) sin3 + sin3 + sin3 = 3sin() (C) cos3 + cos3 + cos3 = 3cos() Let z1 = cos + i sin, z2 = cos+ isin & z3 = cos + isin.  z1 + z2 + z3 = (cos + cos + cos) + i(sin + sin + sin) =0+i.0=0 (A) Also ...... (i) 1 1   cos   isin    cos   i sin  z1 1 1  cos   i sin ,  cos   isin  z2 z3  1 1 1   = (cos + cos + cos) – i(sin + sin + sin) z1 z2 z3 ...... (ii) =0–i.0=0 Now z12  z22  z23   z1  z2  z3   2  z1 z2  z2 z3  z3 z1  2 1 1 1 = 0 – 2z1z2z3     = 0 – 2z1z2z3 . 0 = 0  z3 z1 z2  or  cos   isin  2  (cos   isin  )2   cos   isin  2  0 or cos2 + isin2 + cos2 + isin2 + cos2 + isin2 = 0 + i.0 {using (i) and (ii)} Equating real and imaginary parts on both sides, cos2 + cos2 + cos2 = 0 and sin2 + sin2 + sin2 = 0 (B) If z1 + z2 + z3 = 0 then z13  z23  z33  3z1 z2 z 3  (cos + isin)3 + (cos + isin)3 + (cos + isin)3 = 3(cos + isin) (cos + isin) (cos + isin) or cos3 + isin3 + cos3 + isin3 + cos3 + isin3 = 3{cos() + isin()} Equating imaginary parts on both sides, sin3 + sin3 + sin3 = 3sin() (C) Equating real parts on both sides, cos3 + cos3 + cos3 = 3cos() CUBE ROOT OF UNITY 1  i 3 2 1  i 3 ( ) . ( ) , 2 2 (A) The cube roots of unity are 1 , (B) If  is one of the imaginary cube roots of unity then 1 +  + ² = 0. In general 1 + r + 2r = 0 ; where r  I but is not the multiple of 3 & 1 + r + 2r = 3 if r = 3 ;   I COMPLEX NUMBER (C) In polar form the cube roots of unity are : 1 = cos 0 + i sin 0 ;  = cos (D) 2 2 4 4 + i sin , 2 = cos + i sin 3 3 3 3 The three cube roots of unity when plotted on the argand plane constitute the vertices of an equilateral triangle. (E) The following factorisation should be remembered : (a, b, c  R &  is the cube root of unity) a3  b3 = (a  b) (a  b) (a  ²b) ; a3 + b3 = (a + b) (a + b) (a + 2b) ; x2 + x + 1 = (x  ) (x  2) ; a3 + b3 + c3  3abc = (a + b + c) (a + b + ²c) (a + ²b + c) Ex. Find the value of 192 + 194 Sol. 192 + 194 = 1 + 2 = –  Ex. If  & are imaginary cube roots of unity then n + n is equal to - Sol.   cos 2   i sin 2  3 3   cos 2   i sin 2  3 3 n  n  n   cos 2   isin 2   +  cos 2   i sin 2    3  3 3  3  n  cos 2n   isin 2 n     cos 2n   isin  2n     2n =    3   = 2cos  3  3 3 3 nth ROOTS OF UNITY If 1 , 1 , 2 , 3..... n  1 are the n , nth root of unity then : (A) (B) (C) They are in G.P. with common ratio ei(2/n) Their arguments are in A.P. with common difference 2  n The points represented by n, nth roots of unity are located at the vertices of a regular polygon of n sides inscribed in a unit circle having center at origin, one vertex being on positive real axis. (D) p p 1 p +  1 +  2p +.... + n  1 = 0 if p is not an integral multiple of n = n if p is an integral multiple of n (E) (1  1) (1  2)...... (1  n  1) = n (F) (1 + 1) (1 + 2)....... (1 + n  1) = 0 if n is even and = 1 if n is odd. (G) 1. 1. 2. 3......... n  1 = 1 or 1 according as n is odd or even. Ex. Find the roots of the equation z6 + 64 = 0 where real part is positive. Sol. z6 = – 64 z6 = 26 . e i(2n + 1) n = 0, 1, 2, 3, 4, 5 i(2n 1)  6  z=2 e  z=2 e 6 ,2e  roots with +ve real part are = 2 e 6 , 2 e i  i  2 ,2e i 5 6 , 2e i 7 6 ,2e i 3 2 i Ex. i 11 6 11 6 6 2 k 2 k   Find the value   sin  cos   7 7  k 1  6 Sol. i ,2e   sin k 1 2 k  6  2 k      cos  = 7 7  k 1 6  sin k 1 2 k 6 2 k   cos 1 7 7 k 0 6 =  k 0 (Sum of imaginary part of seven seventh roots of unity) 6 –  (Sum of real part of seven seventh roots of unity) +1 = 0 – 0 + 1 = 1 k 0 THE SUM OF THE FOLLOWING SERIES SHOULD BE REMEMBERED sin  n  / 2   n  1 cos    2  (A) cos  + cos 2  + cos 3  +..... + cos n  = (B) sin  + sin 2  + sin 3  +..... + sin n  =  If  = (2/n) then the sum of the above series vanishes. sin   / 2  sin  n  / 2  sin   / 2   n  1 sin    2  GEOMETRICAL PROPERTIES Section formula If z1 and z2 are affixes of the two points P and Q respectively and point C divides the line segment joining P and Q internally in the ratio m : n then affix z of C is given by z= mz 2  nz1 mn where m, n > 0 If C devides PQ in the ratio m : n externally then z = mz 2  nz1 mn COMPLEX NUMBER If a, b, c are three real numbers such that az1 + bz2 + cz3 = 0 ; where a + b + c = 0 and a,b,c are not all simultaneously zero, then the complex numbers z1, z 2 & z3 are collinear. (1) If the vertices A, B, C of a  are represented by complex numbers z1, z2, z3 respectively and a, b, c are the length of sides then, z1  z 2  z 3 (i) Centroid of the  ABC = (ii) Orthocentre of the  ABC =  a sec A  z1 3 :   b sec B z 2   c secC  z3 or a secA  b sec B  c secC z1 tan A  z2 tan B  z3 tan C tan A  tan B  tanC (iii) Incentre of the  ABC = (az1 + bz2 + cz3)  (a + b + c). (iv) Circumcentre of the  ABC = : (Z1 sin 2A + Z2 sin 2B + Z3 sin 2C)  (sin 2A + sin 2B + sin 2C). (2) amp(z) =  is a ray emanating from the origin inclined at an angle  to the positive x axis. (3) z  a = z  b is the perpendicular bisector of the line joining a to b. (4) The equation of a line joining z1 & z 2 is given by, z = z1 + t (z1  z2) where t is a real parameter. (5) z = z1 (1 + it) where t is a real parameter is a line through the point z 1 & perpendicular to the line joining z 1 to the origin. (6) The equation of a line passing through z 1 & z2 can be expressed in the determinant form as z z z1 z1 1 = 0. This is also the condition for three complex numbers z, z1, z2 to be collinear.. The above z2 1 z2 1 equation on manipulating, takes the form  z   z  r = 0 where r is real and  is a non zero complex constant. If we replace z by zei and z by ze –i then we get equation of a straight line which makes an angle with the (7) given straight line. The equation of circle having centre z 0 & radius  is : z  z0 =  or z z  z0 z  z 0 z + z 0 z0  ² = 0 which is of the form z z   z   z + k = 0, k is real. Centre is  & radius = |  |2  k . Circle will be real if ||2 – k  0. (8) The equation of the circle described on the line segment joining z1 & z2 as diameter isarg (z  z1) ( z  z 2) + (z  z2) ( z  z 1) = 0. z  z2  = ± or z  z1 2 (9) Condition for four given points z1, z2, z3 & z4 to be concyclic is the number z 3  z1 z 4  z 2 . should be real. z 3  z 2 z 4  z1 Hence the equation of a circle through 3 non collinear points z1, z2 & z3 can be taken as  (10)  z  z 2   z3  z1   z  z2   z3  z1  = .  z  z1   z3  z 2   z  z1   z3  z2   z  z1  Arg  z  z  = represent (i) a line segment if  =   2  (ii) Pair of ray if  = 0 (iii) a part of circle, if 0 <  <  z1 1 z2 4i z3 z1 1 (11) Area of triangle formed by the points z 1, z2 & z3 is (12) Perpendicular distance of a point z 0 from the line z   z  r  0 is (13) (i) Complex slope of a line z   z  r  0 is  = – (ii) Complex slope of a line joining by the points z1 & z2 is  = (iii) Complex slope of a line making  angle with real axis  = e 2i (14)  z  z 2   z 3  z1  is real  z  z1   z3  z 2  z2 1 z3 1 | z 0   z0  r | 2||  .  z1  z 2 z1  z2 Dot and cross product Let z1 = x1 + iy1 and z2 = x2 + iy2 be two complex numbers [vectors]. The dot product [also called the scalar product] of z1 and z2 is defined by z1 • z 2 = |z1| |z2| cos  = x1 x2 + y1y2 = Re {z1 z 2 } = 1 { z z  z1 z2 } 2 1 2 where  is the angle between z 1 and z2 which lies between 0 and  If vectors z1, z2 are perpendicular then z 1 • z2 = 0 i.e. z1 z2 + = 0. z1 z2 Sum of complex slopes = 0 The cross product of z1 and z2 is defined by z1 × z2 = |z1| |z2| sin  = x1y2 – y1x2 = m {z1 z 2 } = If vectors z 1, z 2 are parallel then z1 × z2 = 0 i.e.  Complex slopes are equal. 1 {z1 z 2  z1 z 2 } 2i  z1 z2 = . z1 z2 COMPLEX NUMBER 1 & 2 are the compelx slopes of two lines. (i) If lines are parallel then 1 = 2 (ii) If lines are perpendicular then 1 + 2 = 0 (15) If |z – z 1| + |z – z 2| = K > |z 1 – z2| then locus of z is an ellipse whose focii are z 1 & z2 (16) If |z – z0| =  z  z  r then locus of z is parabola whose focus is z 0 and directrix is the line 2||  z +  z + r = 0 (Provided z 0   z0  r  0 ) z  z1 = k  1, 0, then locus of z is circle. z  z2 (17) If (18) If z – z1  – z – z2 = K < z1 – z2 then locus of z is a hyperbola, whose focii are z1 & z 2. Reflection points for a straight line Two given points P & Q are the reflection points for a given straight line if the given line is the right bisector of the segment PQ. Note that the two points denoted by the complex numbers z1 & z2 will be the reflection points for the straight line  z   z  r  0 if and only if ;  z1   z2  r  0 , where r is real and is non zero complex constant. Inverse points w.r.t. a circle Two points P & Q are said to be inverse w.r.t. a circle with centre 'O' and radius , if : (i) The point O, P, Q are collinear and on the same side of O. (ii) OP  OQ = 2. Note that the two points z1 & z2 will be the inverse points w.r.t. the circle z z   z   z  r  0 if and only if z1 z2   z1   z2  r  0 . PTOLEMY’S THEOREM It states that the product of the lengths of the diagonals of a convex quadrilateral inscribed in a circle is equal to the sum of the lengths of the two pairs of its opposite sides. i.e. z1  z3 z2  z4 = z1  z2 z3  z4 + z1  z4 z2  z3. 1. Definition Complex numbers are defined as expressions of the form a+ ib where a, b  R & i = –1 . It is denoted by z i.e. z = a + ib. ‘a’ is called real part of z (Re z ) and ‘b’ is called imaginary part of z (lm z). Every Complex Number Can be Regarded As Purely real if b = 0 Purely imaginary if a = 0 imaginary if b  0 Note (i) The set R of real numbers is a proper subset of the Complex Numbers. Hence the Complex Number system is N  W  I  Q  R  C. (ii) Zero is both purely real as well as purely imaginary but not imaginary. (iii) i = –1 is called the imaginary unit. Also i 2 = – 1 ; i3 = – i ; i4 = 1 (iv) a b = ab only if atleast one of either a or b is non-negative. 2. Conjugate Complex If z = a + ib then its conjugate complex is obtained by chamging the sign of its imaginary part & is denoted by z . i.e. z = a – ib. Note that : (i) z + z = 2 Re (z) (ii) z – z = 2i Im (z) (iv) If z is purely real then z – z = 0 (v) if z is purely imaginary then z + z = 0 3. Representations of a Complex Number in Various Forms (A) Cartesian Form (Geometrical Representation) Every complex nimber z = x + iy can be represented by a point on the cartesian plane know as complex z z = a2+b2 which is real (iii) plane imaginary axis P(x,y) (Argand diagram) by the ordered pair (x, y). r Length OP is called modulus of the complex number denoted by |z| & q is called the argument or amplitude. –1 y e.g. | z |= (angle made by OP with positive + y 2 & θ = tan x x - axis). Geometrically |z| represents the distance of point P from origin. (|z|  0) (B) Trignometric / Polar Representation z = r (cos  + i sin ) where |z| = r ; arg z =  ; z = r(cosθ – isinθ) Note : cos  + i sin  is also written as CiS  Euler’s formula The formula eix = cos x + i sin x is called Euler’s formula. Also cosx = e ix + e –ix eix + e –ix & sinx = are known as Euler’s identities. 2 2i  O Real axis COMPLEX NUMBER (C) Exponential Representation Let z be a complex number such that |z| = r & arg z =  , then z = r.ei 4. 5. IMPORTENT PROPERTIES OF CONJUGATE (A) (z) = z (B) z1 + z 2 = z1 + z 2 (C) z1 - z 2 = z1 - z 2  z  z1 (E)  1  = ; z  0  z2  z2 (F) If f(αα++iβ) iβ)==xx++iy iy f(α f(α ––iβ) iβ) == xx –– iy iy (D) z1 z 2 = z1 . z 2 IMPORTANT PROPERTIES OF MODULUS (A) |z|  0 (B) |z|  Re (z) (C) |z|  lm (z) (E) z z = | z 2 | (F) |z1 z2 | = |z1|.|z2| z1 z1 (G) z = z , z 2  0 2 2 (I) |z1 + z2|2 = |z1|2 + |z2|2 + 2Re (z1 z 2 ) or (D) z = | z | = | –z | = | – z | (H) |zn| = |z|n |z1 + z2|2 = |z1|2 + |z2|2 + 2 |z1| |z2| cos (1 – 2) (J) |z1 + z2|2 + |z1 – z2|2 = 2 [|z1|2 + |z\2|2] (K) | |z1| – |z2| |  | z1 + z2 |  |z1| + |z2| [Triangle Inequality] (L) | |z1| – |z2| |  | z1 – z2 |  |z1| + |z2| [Triangle Inequality] (M) If z + 6. 1 1 a + a2 + 4 = 0(a > 0) , then max | z |= and min | z |= z 2 2  a2 + 4 - a  IMPORTANT PROPERTIES OF AMPLITUDE (A) amp (z1. z2) = amp z1 + amp z2 + 2 k ; k  I z  (B)  1  = amp z1 – amp z 2 + 2kπ; k  I  z2  (C) amp(zn) = n amp (z) + 2k, where proper value of k must be chosen so that RHS lies in (-, ]. (D) log(z) = log(rei) = log r + i = log |z| + i amp (z) 7. DE’MOIVER’S THEOREM The value of (cos + isin)n is cosn + isin if ‘n’ is integer & it is one of the values of (cos + isin)n if n is a rational number of the form p/q, where p & q are co-prime. Note : Continued product of roots of a complex quantity should be determined using theory of equation. 8. CUBE ROOT OF UNITY (A) The cube roots ofb unity are 1, ω = (B) 1 +  + 2 = 0 , 2 = 1, in general –1+ i 3 –1 – i 3 = ei2π/3 & ω2 = = ei4π/3 2 2 0 r is not integral multiple of 3 1+ ω r + ω 2r  3 r is multiple of 3 (C) a2 + b2 + c2 – ab – bc – ca = (a + b + c2) (a + b2 + c) a3 + b3 = (a + b)(a + b2)(a2 + b) a3 – b3 = (a – b)(a – b)(a – b) x2 + x + 1 = (x – ) (x – 2) 9. SQUARE ROOT OF COMPLEX NUMBER  | z | +a | z | –a  | z | –a   | z | +a a + ib = ±  +i –i  for b > 0 & ±   for b > 0 where | z |= 2 2 2   2   C(z ) 10. 2 + b2 . 2 ROTATION z2 – z0 z –z = 1 0 eiθ | z 2 – z 0 | | z1 – z 0 | Take  in anticlockwisev direction.  A(z0 ) 11. RESULT RELATED WITH TRIANGLE (A) Equilateral Triangle z1 – z 2 z3 – z 2 iπ/3 = e l l Also z3 – z 2 z1 – z3 iπ/3 = .e l l A(z1) l .........(i) ......... (ii) B(z1) l /r B(z2) /r l C(z3) from (i) & (ii)  z12 + z 22 + z32 = z1z 2 + z 2 z 3 + z 3 z1 or 1 1 1 + + =0 z1 – z 2 z 2 – z 3 z 3 – z1 (B) Isosceles triangle 4cos2 a (z1 – z2)(z3 – z1) = (z3 – z2)2 (C) 12. z1 1 z2 Area of triangle ABC given by modulus of 4 z3 z1 1 z2 1 z3 1 EQUATION OF LINE THROUGH POINTS Z1 & Z2 z zz11 z 1 zz11 11 == 00  z(z z(z11 –– zz22))++ z(z z(z22 –– zz11)) ++ zz11zz22 –– zz11zz22 == 00 z2 z2 1  z(z1 – z2 )i + z(z 2 – z1 )i + i(z1 z2 – z1z 2 ) = 0 Let (z2 – z1) i = a, then equation of line is [az + az + b = 0] where a  C & b  R. Note : (ii) a a Two lines with slope µ1 & µ2 are parallel or perpendicular if µ1 = µ2 or µ1 + µ2 = 0 (iii) Length of perpendicular from point A() to line az + az + b = 0 is (i) Complex slope of line az + az + b = 0 is – | aα + aα + b | . 2|a | COMPLEX NUMBER 13. EQUATION OF CIRCLE (A) (B) Circle whose centre is z0 & radii = r |z – z0 | r General equation of circle zz + az + az + b = 0 centre ‘–a’ & radii = (C) Diameter form (z – z1 )(z – z2 ) + (z – z 2 )(z – z1 ) = 0 or  z – z1  π arg  =± 2  z – z2  z – z1 = k represent a circle if k 1 and a straight line if k = 1. z – z2 (D) Equation (E) Equation |z – z1|2 + |z – z2|2 = k represent circle if k  (F) 14. | a |2 –b 1 | z1 – z 2 |2 2  z – z1  π arg   = α 0 < α < π,α  z – z 2  2  represent a segment of circle passing through A(z1) & B(z2) STANDARD LOCI (A) (B) |z – z1| + |z – z2| = 2k (a constant) represent (i) if 2k > |z1 – z2|  An ellipse (ii) if 2k = |z1 – z2|  An line segment (iii) if 2k < |z1 – z2|  No solution Equation | |z – z1| – |z – z2 | | = 2 k (a constant) representy (i) if 2k > |z1 – z2|  A hyperbola (ii) if 2k = |z1 – z2|  A line ray (iii) if 2k < |z1 – z2|  No solution