Complex Number Notes

Complex Number PRELIMINARY (i) The numbers of the form x + iy are known as complex numbers, where x, y R and i = Ex.1 1 . i is an imaginary unit and it is known as iota. (ii) Complex numbers are denoted by z . Let z = x + iy, then Re (z) = x and Im(z) = y Sol. 2 3 4 (iii) i = –1, i = –i, i = 1, i4n = 1, i4n+1 = i,i4n+2 = i2 = –1, i4n+3 = i3 = –i If n is a positive integer, then which of the following relations is false (A) i4n = 1 (B) i4n–1 = i (C) i4n+1 = i (D) i–4n = 1 We know that i2 = –1  (i2)2 = (–1)2 = 1 i4n = 1n and therefore i4n–1 = –i Ans.(B)  (iv) The sum of four consecutive powers of i is always zero, i. e. i4n + i4n+1 + i4n+2 + i4n+3 = 0 200 Ex.2 If i2 = –1, then the value of Note : is n 1 (i) i2 = i × i =  1 ×  1  1 (ii) n i (A) 50 (B) –50 (C) 0 (D) 100  x   y  xy 200 So for two real numbers x and y, x  y  x  y Sol. i n  i  i2  i3  ......  i200  n 1 possible if both x, y are non-negative. (iii) ' i ' is neither positive, zero nor negative, Due to this reason order relations are not defined for imaginary numbers. i(1  i 200 ) 1 i (since G.P.)  i (1  1) 0 1 i Ans.(C) 1 Complex Number i592 + i590 + i588 + i586 + i584 The value of i582 + i580 + i578 + i576 + i574 –1 is Ex.3 (A) 2 (B) 2 (D) 1 (C) 1 Given expression Sol.  i10 i582  i580  i578  i576  i574 i582  i580  i578  i576  i574  i10  1  i2 5 1 5  1  1  1  1  1  2 Ans.(C) Ex.4 Find the value of [i]198 Sol. [i]198 = [i2]99 = [–1]99 = – 1 Ex.5 Find the value of in + in+1+ in+2 + in+3 Sol. in + in+1+ in+2 + in+3 = in [1 + i + i2 + i3] = in [1 + i – 1 – i] = in [0] =0 Note : (i) Inequalities in complex number are not defined because 'i' is neither positive, zero nor negative so 4 + 3i < 1 + 2i or i< 0 or i > 0 is meaning less. (ii) If two complex numbers are equal, then their real and imaginary parts are separately equal. Thus if a + ib = c + id  a = c and b = d so if z = 0x + iy =0  x =0 and y = 0 The student must note that x, y  R and x, y  0. Then if x + y = 0  x = y is correct but x + i y = 0  x =– iy is incorrect Hence a real number cannot be equal to the imaginary number, unless both are zero. (iii) The complex number 0 is purely real and purely imaginary both. ALGEBRAIC OPERATIONS WITH COMPLEX NUMBERS Ex.6 The sum of series terms is (A) 0 (C) n Sol. i2 + i4 + i6 + .......(2n + 1) (i) Addition : (a + ib) +(c + id) = (a + c) + i (b + d) (ii) Subtraction : (a + ib) – (c + id) = (a – c) + i (b – d) (B) 1 (D) – 1 (iii) Multiplication : (a + ib) (c + id) = (ac – bd) + i (ad + bc) Given series is a G.P. So, Sum of a G. P. is = = 1 i2 1 1 1 2 = =–1 1 1 i 4n 1 1 2 Ans.[D] COMPLEX NUMBER A number of the form z = x + iy where x, y R and i = 1 is called a complex number where x is called as real part and y is called imaginary part of complex number and they are expressed as Re ( z ) = x, m ( z ) = y Here if x = 0 the complex number is purely Imaginary and if y = 0 the complex number is purely Real. A complex number may also be defined as an ordered pair of real numbers any may be denoted by the symbol (a, b). If we write z = (a, b) then a is called the real part and b the imaginary part of the complex number z. (iv) Division : a  ib ac  bd i(bc  ad) = + 2 2 (c + id  0) c d c  id c 2  d 2 Properties of Algebraic operations with Complex Number Let z, z1, z2 and z3 are any complex number then their algebraic operation satisfy following propertiesCommutativity : z1 + z2 = z2 + z1 & z1 z2 = z2z1 Associativity : (z1+ z2) + z3 = z1 + (z2 + z3) and (z1 z2) z3 = z1(z2 z3) Identity element : If O = (0, 0) and 1 = (1, 0) then z + 0 = 0 + z = z and z.1 = 1. z = z. Thus 0 and 1 are the identity elements for addition and multiplication respectively. Inverse element : Additive inverse of z is – z and multiplicative inverse of z is 1 . z 2 Complex Number Cancellation Law : UV W and Sol.  (4 + 2i) x + (9 – 7i) y – 3i – 3 = 10i Ans.(B) Equating real and imaginary parts, we get 2x – 7y = 13 and 4x + 9y = 3. Hence x = 3 and y = –1. UV W z1  0 Distributivity : z1 (z2 + z3) = z1 z2 + z1 z3 and (z2 + z3) z1 = z2 z1 + z3 z1 MULTIPLICATIVE INVERSE OF A Ex.9 NON-ZERO COMPLEX NUMBER Multiplicative inverse of a non-zero complex number z = x + iy is Sol. z 1  1 1 1 x  iy x  iy     z x  iy x  iy x  iy x 2  y 2 x y = x2  y 2  i x2  y 2 Re(z) (1  i)x  2i ( 2  3i) y  i  i 3i 3i Ex.10 If a + ib = c + id, then (A) a – c = i (b – d) (B) a – ib = c – id (C) a = d, b = c (D) none of these If a + ib = c + id. Equating real and imaginary parts, we get a = c and b = d  –b = –d. Therefore a – ib = c – id Ans.(B) 3  2i sin  1  2i sin  will be purely imaginary, if  = (A) 2n   3 Im(z) 1 i.e. z  | z |2  i | z |2 (B) n   Ex.7 (C) n  3 If z = – 3 + 2i, then 1/z is equal to (A) (C) 1 (3 + 2i) 13 1 13 (3 + 2i) 1 (B) – (3 + 2i) 13 (D)  1 13 Sol. 3  2i sin  1  2i sin  (D) none of these will be purely imaginary, if the real part vanishes, i.e., (3 + 2i)  3 3  4 sin 2  1  4 sin 2  0  3 – 4 sin2  = 0 (only if  be real) Sol. z 1 3 2 1  i   (3  2i) 13 13 13 Ans.(B) EQUALITY OF COMPLEX NUMBERS Two complex numbers are said to be equal if and only if their real parts and imaginary parts are separately equal i.e. If a + ib = c + id, then a = c & b = d Ex.8 The values of x and y satisfying the equation (1  i)x  2i ( 2  3i)y  i  i 3i 3i (A) x = –1, y =3 (C) x = 0, y = 1      = n + (–1)n   3  = n ± . 3   Ans.(C) Ex.11 If(x + iy) (2 – 3i) = 4 + i, then(A) x = –14/13, y = 5/13 (B) x = 5/13, y = 14/13 (C) x = 14/13, y = 5/13 (D) x = 5/13, y = –14/13 are (B) x = 3, y = –1 (D) x = 1, y = 0 3 = sin 2  sin  = ± Sol. x + iy = 4 i i = i 13  x = 5/13, y = 14/13. i = i 13 Ans.[B] 3 Complex Number z 1 Ex.12 If Complex Number z 1 is purely imaginary then locus of z is (A) a circle (C) a parabola Sol. (B) a straight line (D) None of these 1 = 1 1 1 = ( = ( = = z1 – z2 (vii) rei = re–i FG z IJ H K z1 1 bg ) ( n (x) z1 z2 = z1 z 2 (xi) z + z = 0 or z = – z  z = 0 or z is purely imaginary ) (xii) z = z  z is purely real 1) 1 1 Ex.13 The conjugate of If it is purely Imaginary i (A) (3 – 4i) 1 ( (vi) (ix) zn = z 1 1 1 1 × = z1 + z 2 (viii) z = z 2 2 Let z = x + iy then z 1 = z 1 (v) = 0 1) (C)  x2 + y2 – 1 = 0  x2 + y2 Sol. =1 which is the equation of a circle. Ans.[A] 1 (B) 1 (3 – 4i) 25 = ( i conjugate of is 1 (3 + 4i) 25 (D) None of these i i)( FG H i) 1 IJ iK = = 1 (3 – 4i) 25 1 (3 + 4i). 25 Ans. [B] CONJUGATE COMPLEX NUMBER The complex numbers z = (a, b) = a + ib and 2 2 z = (a, – b) = a – ib where b  0 are said to be Ex.14 If z is a complex number such that z = ( z ) , complex conjugate of each other (Here the then complex conjugate is obtained by just changing (A) z is purely real the sign of i) e.g.conjugate of z = – 3 + 4i is (B) z is purely imaginary z = – 3 – 4i. (C) Either z is purely real or purely imaginary Note : Image of any complex number in x-axis is (D) None of these called its conjugate. Sol. Let z = x + iy, then its conjugate z  x  iy Properties of Conjugate Complex Number Let z = a + ib and z = a – ib then (i) b zg = z (ii) z + z = 2a = 2 Re (z) = purely real (iii) z – z = 2ib = 2i m (z) = purely imaginary Given that z2 = ( z )2  x2 – y2 + 2ixy = x2 – y2 – 2ixy  4ixy = 0. If x  0 then y = 0 and if y  0 then x = 0. Ans.(C) (iv) z z = a2 + b2 = |z|2 4 MODULUS OF A COMPLEX NUMBER Modulus of a complex number z = x + iy is denoted as mod (z) or | z |, is defined as |z| = x 2  y 2 , where x = Re(z), y = Im(z). Complex Number Ex.16 If for any complex number z, |z – 4| < |z – 2|, then (A) R(z) > 2 (C) R(z) > 0 Sol. (B) R(z) < 0 (D) R(z) > 3 Let z = x + iy, then Sometimes, |z| is called absolute value of z. Note that | z |  0. |z – 4| < |z – 2|  (x – 4)2 + y2 < (x – 2)2 + y2 For example, if z = 3 + 2i, then |z| =  – 4x < – 12  R(z) > 3 3 2  2 2  13 . Properties of modulus x > 3 Ans.[D] 3 = 1 then the locus of z is 3 (i) |z|  0 and |z| = 0 if and only if z = 0, i.e., x = 0, y = Ex.17 If 0 (A) x axis (ii) |z| = | z | = |– z| = |– z |. (B) x – y = 0 (iii) z z = |z|2 (iv) – |z|  Re (z)  | z| and – |z|  Im(z)  |z| (C) Circle passing through origin (D) y axis Sol. (v) |z1z2| = |z1| |z2| (vi) 3 3 z1 | z1 | z 2 = | z 2 | (provide z2  0) (vii) |z1 ± z2|  |z1| + |z2|  (viii) |z1 – z2| | | z1| - |z2| | 2 2 n  n (ix) |z | = |z| or |z | = |z| also |z1z2 .... zn| = |z1| |z2|.... |zn| 2 2 2 2 (x) |z1 + z2| + |z1 – z2| = 2(|z1| + |z2| ) 2 2 2 (xi) |z1 + z2| = |z1| + |z2| + 2 Re (z1 z2 ) (xii) |z1 – z2|2 = |z1|2 + |z2|2 – 2 Re (z1 z2 ) Ex.15 The modulus of z= (A) e 1 1 z + 3i| |x + iy – 3i| = |x + iy + 3i| ( 3) = ( 3)  12 y = 0  y = 0, which is equation of x - axis Ans.[A] (A) (a2 + b2) (|z1| + |z2|) (B) (a2 + b2) (|z1|2 + |z2|2) 1 (C) (a2 + b2) (|z1|2 – |z2|2) 3 (D) none of these Sol. |(az1 – bz2)|2 + |(bz1 + az2)|2 = a2 |z1|2 + b2 |z2|2 – 2ab Re (z1z2 ) + b2 |z1|2 + a2 |z| = 2 1 = |z – 3i| = | Ex.18 For any two complex numbers z1 and z2 and any real numbers a and b; is- (D) 1 2 =1 |(az1 – bz2)|2 + |(bz1 + az2)|2 = j (B) 3 2 (C) Sol. Let z = x + iy then |z2|2 + 2ab Re (z1z2 ) 2 2 2 = 1 2 Ans.[C] = (a2 + b2) (|z1|2 + |z2|2) Ans.(B) 5 Complex Number Ex.20 The polar form of – 1 + i is- REPRESENTATION OF A COMPLEX NUMBER (a) Cartesian Representation : The complex number z = x + iy = (x, y) is represented by a point P whose coordinates are refered to rectangular axis xox´ and yoy´, which are called real and imaginary axes respectively. Thus a complex number z is represented by a Sol. point in a plane, and corresponding to every point in this plane there exists a complex number such a plane is called Argand plane or Argand diagram or complex plane or gussian plane. Ex.21 Note : (i) Distance of any complex number from the origin is called the modulus of complex number and is denoted by |z|. Thus, |z| = . (ii) Angle of any complex number with positive direction of x-axis is called amplitude or argument y of z. Thus, amp (z) = arg (z) =  = tan–1 . x (b) Polar Representation : If z = x + iy is a complex number then z = r (cos  + i sin ) is a polar form of complex number z where x = r cos , y = r sin  and r = = |z|. Sol. (c) Exponential Form : If z = x + iy is a complex number then its exponential form is z = r ei where r is modulas and  is amplitude of complex number. (d) Vector Representation : If z = x + iy is a complex (A) 2 (cos / 4 + i sin / 4) (B) 2 (cos 5/ 4 + i sin 5/ 4) (C) 2 (cos 3/ 4 + i sin 3/4) (D) 2 (cos / 4 – i sin / 4)  i = 2 , amp (– 1 + i) = – /4 = 3/4 1 –1 + i = 2 (cos 3/ 4 + i sin 3/ 4) Ans. [C] The polar form of is 3 3   (A) 2  cos 4  i sin 4        (B) 2  cos 4  i sin 4    3 3   (C)  cos 4  i sin 4    (D) none of these 1  7i (1  7i) (3  4i) 25  25i    1  i 2 (3  4i) (3  4i) 25 ( 2  i) Let z = x + iy = – 1 + i  r cos  = –1 and r sin  = 1 = vector representation is z = OP Thus 3 and r = 4 2 1  7i 3 3    2 cos  i sin  4 4 ( 2  i)  2 Ex.19 If A  1 + 2i, B  –3 + i, C  – 2 – 3i and D  Ans.(A) 2 – 2i are vertices of a quadrilateral, then it is a (A) rectangle (B) parallelogram (C) square (D) rhombus ARGUMENT OF A COMPLEX NUMBER Sol.  A  (1, 2); B  (–3, 1); C  (–2, –3); D  (2, –2) The amplitude or argument of a complex number z  AB2 = 16 + 1 = 17, BC2 = 1 + 16 = 17 is the inclination of the directed line segment CD2 = 16 + 1 = 17, AC2 = 9 + 25 = 34 representing z, with real axis. If z = x + iy then BD2 = 25 + 9 = 34. y Now since AB = BC = CD and AC = BD amp ( z )  tan 1  x  ABCD is square. Ans.[C] 6 The argument of any complex number is not unique. Sol. 2n +  (n integer) is also argument of z for various values of n. The value of  satisfying the inequality –  <   is called the principle value of the argument. For finding the principle argument of any complex number first check that the complex number is in Ex.23 which quadrant and then find the angle  and amplitude using the adjacent figure. Note : (i) If a complex number is multiplied by iota(i) its amplitude will be be increased by /2 and will be Sol. decreased by /2, if is multiplied by –i. (ii) Amplitude of complex number in I and II quadrant is always positive and in III and IV is always negative. (iii) Argument of zero is not defined. (iv) Let z1,z 2 ,z3 be the affixes of P, Q, R respectively in the Argand Plane. Then from the figure the angle between PQ and PR is   z -z    2  1 = arg PR - arg PQ = arg  3 1   z 2 - z1  The amplitude of the complex number z = sin  + i (1 – cos ) is (A) 2 sin  2 (B) (C)   2 (D) None of these z = sin  + i (1 – cos )  1  cos    tan 1 arg(z)=tan–1  sin     = tan–1 tan  2  =   2     2 sin2  2    2 sin  cos     2 2  Ans.(B)     Ex.24 The amplitude of sin 5  i 1  cos 5    Properties of Arguments (i) arg (z1z2) = arg(z1) + arg(z2) (ii) arg(z1/z2) = arg(z1) – arg(z2) (iii) arg(zn) = n arg(z) (iv) If arg (z) = 0 z is a positive real number (v) arg(z) + arg( z ) = 0 Sol. (vi) arg(z – z ) = ± /2 (vii) arg(z) =  z is a negative real number (viii) arg( z ) = – arg (z) = arg(1/z) (ix) arg(–z) = arg(z) ±  (x) arg(i y) = /2 if y > 0 = – /2 if y < 0 (where y  R) Ex.22 Let z1 and z2 be two complex numbers with  and  as their principal arguments such that  +  > , then principal arg (z1 z2) is given by (A)  +  +  (B)  +  –  (C)  +  –2 (D)  +  Complex Number We know that principal argument of a complex number lie between – and , but  +  > , therefore principal arg (z1z2) = arg z1 + arg z2 =  + , is given by  +  – 2. Ans.(C) 2  (A) 5 (B) 5   (C) 10 (D) 15  10  tan  For amplitude, tan  =  10 cos 10 sin    = 10 . Ans.(C) 7 Complex Number Ex.25 The amplitude of F (A ) tan–1 G H Sol. DIFFERENT WAYS OF WRITING A is equal to- I JK F (B) tan–1 GH 2ab I JK F I F 2ab I (C) tan–1 GH JK (D) tan–1 GH 2ab JK F IJ = amp (a + ib) – amp (a – ib) amp GH K F bI F bI = tan–1 GH a JK – tan–1 GH a JK L 2(b / a) O F 2ab I = tan–1 M 1 (b / a ) P = tan–1 GH JK MN PQ 2 Ex.26 If z = z = a + ib (Algebraic form) z = r (cos  + i sin ) (Polar form) where r is modulus of complex number and  is it's argument. z = r ei ei = cos  + i sin  e–i = cos  – i sin  cos  = (A)  (B) – (C) 0 (D) 1 = × i i = i arg ( z ) = 3 i 2 Sol. = i (D) – i i or eiz = e(–r sin  + ir cos ) = e–r sin  eri cos  or |eiz| = |e–r sin| |eri cos| = e–r sin [cos2 (r cos ) + sin2 (r cos )]1/2 = e–r sin  2 i z1  z0 z z z z i  = 2 0 e i(2  ) = 2 0 e | z1  z0 | | z  z | | z 2  z0 | 2 0 Ex29 If the points z 1, z2, z3 are the vertices of an equilateral triangle in the complex plane, then the value of z12 + z22 + z32 is equal to i (iz) = 2i, z which is the positive Imaginary quantity 2 Ans.(B) z 2  z0 z1  z0 i e = | z 2  z0 | | z1  z0 | or z = 2  arg (iz) = (D) er cos ROTATION THEOREM 3 i i (C) e–r cos Ans.[D] Ex.27 If z = 2 i + 2 i then arg (zi) is(A) – (B)  2 3 i 3 i z= 2 i + 2 i (B) e–r sin  iz = ir (cos  + i sin ) = –r sin  + ir cos  2 i =–i 1 2 (C) (A) er sin If z = r ei = r (cos  + i sin ) Sol.  z = i, which is the positive Imaginary quantity  e i   e i  ei  e i , sin   2 2i Ex.28 If z = rei, then |eiz| = 1 then arg ( z ) is i 1 z = i (Euler's formula) 2 Ans.[B] Sol. COMPLEX NUMBER z z 1 2 3 (A) z  z  z 2 3 1 (B) z1z2 + z2z3 + z3z1 (C) z1z2 – z2z3 – z3z1 1 2 3 (D)  z  z  z 2 3 1 Ans.[D] z z z 8 Complex Number AC  AB ei / 3 Sol. By rotating /3 in clockwise sense  (z3 – z1) = (z2 – z1)ei/3 Also (z1 – z2) = (z3 – z2) ei/3 DIviding (i) by (ii) we get Ex.30 The square roots of 7 + 24i is. ....(i) ....(ii) z 3  z1 z 2  z1  z1  z 2 z 3  z 2  Sol. (A) ± (4 + 3i) (B) ±(3 + 4i) (C) ± (2 + 3i) (D) ± (4 – 3i) Here | z | = 25, x = 7, Hence square root =  z12 + z22 + z32 = z1z2 + z2z3 + z3z1 Ans.(B) 1/ 2  25  7 1/ 2  25  7     i   = ± (4 + 3i)    2    2  SQUARE ROOT OF A COMPLEX NUMBER Ans.(A) If z  x  iy suppose MISCELLANEOUS RESULTS z = x + iy = a + ib (i) If ABC is an equilateral triangle having vertices z1, z2, z3 then z12 + z22 + z32 = z1z2 + z2z3 + z3z1 or  x  iy  a 2  b 2  2 iab On comparing the real and imaginary parts x  a b , 2 Now, y  2ab 2 2 2 2 a b  2 x y  z .....(i) z x 2 , b 1 + 1 = 0. (iii) Let z1 = x1 + iy1 and z2 = x2 + iy2 be two complex numbers represented by points P and Q respectively in Argand Plane then - From Equation (i) and (ii) a + (ii) If z 1, z 2, z3, z 4 are vertices of parallelogram then z1 + z3 = z2 + z4. .....(ii) a 2  b2  x 1 z x PQ = 2 2 2 Solving these two equations we shall get the required = | (x2 – x1) + i (y2 –y1) | = |z2 – z1| square roots as follows : (iv) If a point P divides AB in the ratio of m : n,    z x    z x 2 2 i i z x  if y  0 2   and z x   if y  0 2   Note : FG 1 i IJ H 2 K (Here b = 1) F 1 iI (ii) The square root of – i is ± GH JK (Here b = –1) 2 (i) The square root of i is ± 2 (iii)The square root of  is ± (iv) The square root of 2 is ±  then z = where z1, z2 and z represents the point A, B and P respectively. (v) |z – z1| = |z – z2| represents a perpendicular bisector of the line segmentjoining the points z1 and z2. (vi) Let P be any point on a circle whose centre C and radius r, let the affixes of P and C be z and z0 then |z – z0| = r. (a) Again if |z – z0 | < r represent interior of the circle of radius r. (b)|z – z0| > r represent exterior of the circle of radius r. 9 Complex Number (vii) Let z1, z2, z3 be the affixes of P, Q, R respectively in Ex.32 If the complex numbers, z1, z2, z3 represent the the Argand Plane. Then from the figure the angle vertices of an equilateral triangle such that between PQ and PR is. z1  z 2  z3 , then z1 + z2 + z3 =  = 2 – 1 (A) 0 (B) 1 = arg. PR – arg PQ (C) –1 (D) None of these = arg FG H IJ K Sol. (a) If z1, z2, z3 are collinear, thus = 0 therefore is purely real. Therefore |z1| = |z2| = |z3|  OA = OB = OC (b) If z1, z2, z3 are such that PR PQ, = / 2 So Let the complex number z1, z2, z3 denote the vertices A, B, C of an equilateral triangle ABC. Then, if O be the origin we have OA = z 1 , OB = z 2 , OC = z3 i.e., O is then circumcentre of ABC is purely imaginary. Hence z1 + z2 + z3 = 0. Ans.(A) Ex.33 If z  2 , then the points representing the Ex.31 The points represented by the complex numbers 1 + i, – 2 + 3i, 5 3 i on the Argand diagram are (A) Vertices of an equilateral triangle (B) Vertices of an isosceles triangle (C) Collinear (D) None of these Sol. complex numbers –1 + 5z will lie on a Sol. (A) Circle (B) Straight line (C) Parabola (D) None of these Let  = –1 + 5z, then  + 1 = 5z  | + 1| = 5 |z| = 5 × 2 = 10 ( |z| = 2, given value) Thus  lies on a circle. 5 Let z1 = 1 + i, z2 = –2 + 3i and z3 = 0 + 3 i Ex.34 The equation Then x1 x2 y1 1 1 y2 1   2 x3 y3 1 0 1 3 1 1 5/3 1 4 10 4  6  10 2  0 3 3 3 zz  (2  3i) z  (2  3i) z  4  0 represents a circle of radius 5    10   1  3    1( 2)  1   3    3   Ans.(A) Sol. Ans.(C) (A) 2 (B) 3 (C) 4 (D) 6 Here a = 2 – 3i, a  2  3i and b = 4. Hence radius = a a  b = (2  3i)( 2  3i)  4  3 10