Chapter 14 - Complex Numbers PDF

60 30 Complex Numbers 2.1 Introduction. 1 Number system consists of real numbers (5, 7, , 3..............etc.) and imaginary numbers 3 E3 (  5 ,  9....etc.) If we combine these two numbers by some mathematical operations, the resulting number is known as Complex Number i.e., “Complex Number is th...

60 30 Complex Numbers 2.1 Introduction. 1 Number system consists of real numbers (5, 7, , 3..............etc.) and imaginary numbers 3 E3 (  5 ,  9....etc.) If we combine these two numbers by some mathematical operations, the resulting number is known as Complex Number i.e., “Complex Number is the combination of real and imaginary numbers”. (1) Basic concepts of complex number complex number so the quantity ID (i) General definition : A number of the form x  iy, where x , y  R and i   1 is called a  1 is denoted by 'i' called iota thus i   1. i.e., U A complex number is usually denoted by z and the set of complex number is denoted by c c  {x  i y : x  R, y  R, i   1} Note D YG For example, 5  3 i,  1  i, 0  4 i, 4  0 i etc. are complex numbers. :  Euler was the first mathematician to introduce the symbol i (iota) for the square root of – 1 with property i 2  1. He also called this symbol as the imaginary unit.  Iota (i) is neither 0, nor greater than 0, nor less than 0.  The square root of a negative real number is called an imaginary unit.  For any positive real number a, we have  a  1a  1 a  i a  i  a   a. U  The property a b  ab is valid only if at least one of a and b is non-negative. If a and b are both negative then ST  If a  0 then a b   ab. a  | a| i. (2) Integral powers of iota (i) : Since i   1 hence we have i2  1 , i3  i and i 4  1. To find the value of i n (n  4 ), first divide n by 4. Let q be the quotient and r be the remainder. i.e., n  4 q  r where 0  r  3 i n  i 4 q r  (i 4 )q. (i)r  (1)q. (i)r  i r In general we have the following results i 4 n  1, i 4 n 1  i, i 4 n  2  1, i 4 n  3  i , where n is any integer. In other words, i n  (1)n / 2 if n is even integer and i n  (1)n 1 / 2 i if n is odd integer. The value of the negative integral powers of i are found as given below : Complex Numbers 31 i 1  i i 1 i3 1 1 1 1 1  4  i 3  i, i 2  2   1, i 3  3  4   i, i 4  4   1 i i 1 1 1 i i i i Important Tips The sum of four consecutive powers of i is always zero i.e., in  in 1  in  2  in  3  0, n  I.  i n  1, i,  1,  i, where n is any integer.  (1  i) 2  2i, (1  i) 2  2i  1i 1i 2i  i,  i, 1i 1i 1i i1 E3 60  200 Example: 1 If i 2  1, then the value of i n is n 1 (b) – 50 200 i n i (1  1) i (1  i )  0. (since G.P.)  1i 1i  i  i2  i3 .....  i200  n 1 Example: 2 If i   1 and n is a positive integer, than in  in 1  in  2  in  3  (a) 1 Solution: (d) n 1 i i n (b) i n 2 i n3 i (d) 100 (c) i [Rajasthan PET 2001; Karnataka CET 1994] n (d) 0 U Solution: (c) (c) 0 200 ID (a) 50 [MP PET 1996]  i (1  i  i  i )  i (1  i  1  i)  o. n 2 3 n Trick: Since the sum of four consecutive powers of i is always zero. D YG  in  in 1  in  2  in  3  0, n  I. x Example: 3 1  i If    1 then 1  i [AIEEE 2003; Rajasthan PET 2003] (a) x = 4n, where n is any positive integer (b) x = 2n, where n is any positive integer (c) x = 4n +1, where n is any positive integer (d) x = 2n +1, where n is any positive integer x 1i 1i 1i 1i x  .     1  i  1  x  4 n, n  I. 1  i 1i 1i 1i   Example: 4 1  i2  i4  i6 .....  i2 n is U Solution: (a) (a) Positive Example: 5 (d) Can not be determined i 2k  1 or – 1 (which is depend upon the value of n). k 0 If x  3  i, then x 3  3 x 2  8 x  15  (a) 6 Solution: (d) (c) Zero n 1  i2  i4  i6 .....  i2 n  ST Solution: (d) (b) Negative [EAMCET 1980] [UPSEAT 2003] (b) 10 (c) – 18 (d) – 15 Given that; x  3  i  (x  3)  i  x  6 x  10  0 2 2 2 Now, x 3  3 x 2  8 x  15  x (x 2  6 x  10 )  3(x 2  6 x  10 )  15  0  0  15  15. Example: 6 The complex number (a) 0 Solution: (d) 2n (1  i)2 n  , (n  Z ) is equal to 2n (1  i) 2n (b) 2 (1  i)2n  ((1  i)2 )n  (1  i2  2i)n  (1  1  2i)n  2n in (1  i)2n  ((1  i)2 )n  (1  i2  2i)n  (1  1  2i)n  (2)n in (c) [1  (1)n ].in (d) None of these 32 Complex Numbers  2n (1  i)2 n 2n 2n in 1 1  (1)n i2 n 1  (1)n (i2 )n    n   in   n n n n n n (1  i)2 2 (2) i 2 (1) i (1)n in (1)n in  1  (1)n (1)n 1  (1)2 n 1 1 2   . (1)n in (1)n in (1)n in (1)n in 2.2 Real and Imaginary Parts of a Complex Number. 60 If x and y are two real numbers, then a number of the form z  x  iy is called a complex number. Here ‘x’ is called the real part of z and ‘y’ is known as the imaginary part of z. The real part of z is denoted by Re(z) and the imaginary part by Im(z). Note E3 If z = 3 – 4i, then Re(z) = 3 and Im(z) = – 4. :  A complex number z is purely real if its imaginary part is zero i.e., Im(z) = 0 and purely imaginary if its real part is zero i.e., Re(z) = 0.  i can be denoted by the ordered pair (0,1). ID  The complex number (a, b) can also be split as (a, 0) + (0, 1) (b, 0). Important Tips A complex number is an imaginary number if and only if its imaginary part is non-zero. Here real part may or may not be zero.  All purely imaginary numbers except zero are imaginary numbers but an imaginary number may or not be purely imaginary.  A real number can be written as a + i.0, therefore every real number can be considered as a complex number whose imaginary part is zero. Thus the set of real number (R) is a proper subset of the complex number (C) i.e., R  C.  Complex number as an ordered pair : A complex number may also be defined as an ordered pair of real numbers and may be denoted by the symbol (a,b). For a complex number to be uniquely specified, we need two real numbers in particular order. D YG U  2.3 Algebraic Operations with Complex Numbers. U Let two complex numbers z 1  a  ib and z 2  c  id Addition : (a  ib)  (c  id)  (a  c)  i(b  d ) : (a  ib)  (c  id)  (a  c)  i(b  d ) Multiplication : (a  ib)(c  id)  (ac  bd )  i(ad  bc) ST Subtraction Division : a  ib c  id (when at least one of c and d is non-zero) a  ib (a  ib) (c  id) . c  id (c  id) (c  id) a  ib (ac  bd ) i(bc  ad)  2  2. c  id c d2 c d2 (Rationalization) Complex Numbers 33 Properties of algebraic operations with complex numbers : Let z 1 , z 2 and z 3 are any complex numbers then their algebraic operation satisfy following operations: (i) Addition of complex numbers satisfies the commutative and associative properties i.e., z 1  z 2  z 2  z 1 and (z 1  z 2 )  z 3  z 1  (z 2  z 3 ). 60 (ii) Multiplication of complex number satisfies the commutative and associative properties. i.e., z 1 z 2  z 2 z 1 and (z 1 z 2 )z 3  z 1 (z 2 z 3 ). (iii) Multiplication of complex numbers is distributive over addition Note E3 i.e., z 1 (z 2  z 3 )  z 1 z 2  z 1 z 3 and (z 2  z 3 )z 1  z 2 z 1  z 3 z 1. 0  0  0 i is the identity element for addition. :  1  1  0 i is the identity element for multiplication. ID  The additive inverse of a complex number z  a  ib is  z (i.e. – a – ib). 1  2i 4  i   2  i 3  2i (a) 24 10  i 13 13 (b) 24 10  i 13 13 (c) D YG Example: 7 U  For every non-zero complex number z, the multiplicative inverse of z is [Rajasthan PET 1987] 10 24  i 13 13 Solution: (d) 1  2i 4  i 50  120 i 10 24 (1  2i)(3  2i)  (4  i)(2  i)       i. 2  i 3  2i 65 13 13 (2  i)(3  2i) Example: 8 3   3  4i   1      1  2i 1  i   2  4 i  (a) (b) (d) 10 24  i 13 13 [Roorkee 1979; Rajasthan PET 1999] 1 9  i 2 2 (c) 1 9  i 4 4 (d) 1 9  i 4 4 3  3 i   6  16  12 i  8 i  3   3  4i   1  2i  1  2  4 i  15  15 i   1  2i   2       2    2 10 1  1 2   22  4 2  1  2i 1  i   2  4 i    2  1  2  U Solution: (d) 1 9  i 2 2 1. z (17  11i)(1  2i) 5  45 i 1 9    i. 20 20 4 4 ST  Example: 9 The real value of  for which the expression (a) n   / 2 Solution: (c) Given that 1  i cos  1  2i cos  1  i cos  is a real number, is 1  2i cos  (b) n   / 2  (c) n   / 2 (d) None of these  (1  2 cos 2  )   3 cos   (1  i cos  )(1  2i cos  )   i  2 2 (1  2i cos  )(1  2i cos  )  (1  4 cos  )   1  4 cos   Since Im (z)  0 , then 3 cos   0    n   / 2. 2.4 Equality of Two Complex Numbers. Two complex numbers z 1  x 1  iy1 and z 2  x 2  iy 2 are said to be equal if and only if their real parts and imaginary parts are separately equal. 34 Complex Numbers i.e., z 1  z 2  x 1  iy1  x 2  iy 2  x 1  x 2 and y 1  y 2. Thus , one complex equation is equivalent to two real equations. Note : A complex number z  x  iy  0 iff x  0, y  0.  The complex number do not possess the property of order i.e., (a  ib)  (or)  (c  id) is Example: 11 6 i  3i 1 If 4 3 i  1  x  i y , then 20 3 i (a) x = 3, y=1 Solution: (d) (d) None of these E3 Solution: (d) Which of the following is correct (a) 6  i  8  i (b) 6  i  4  i (c) 6  i  4  2i Because, inequality is not applicable for a complex number. [MP PET 2000; IIT 1998] (b) x =1, y=3 (c) x = 0, y=3 6i  3i 1 4 3 i  1 Applying C2 C 2  3 i C 3 20 3 i (d) x = 0, y=0 ID Example: 10 60 not defined. For example, the statement 9  6 i  3  2i makes no sense. The real values of x and y for which the equation (x 4  2 xi)  (3 x 2  yi)  (3  5i)  (1  2yi) is satisfied, are [Roorkee (a) x  2, y  3 Solution: (c) 1 3 D YG Example: 12 U 6i 0 1 4 0  1 = 0 = 0+ 0 i, Equating real and imaginary parts x = 0, y = 0 20 0 i (b) x  2, y  (c) Both (a) and (b) (d) None of these Given equation (x 4  2 xi)  (3 x 2  yi)  (3  5i)  (1  2yi)  (x 4  3 x 2 )  i(2 x  3 y)  4  5i Equating real and imaginary parts, we get x4  3x2  4.....(i) and 2 x  3 y  5 Form (i) and (ii), we get x   2 and y  3, U Trick: Put x  2, y  3 and then x  2, y .....(ii) 1. 3 1 , we see that they both satisfy the given equation. 3 2.5 Conjugate of a Complex Number. ST (1) Conjugate complex number : If there exists a complex number z = a  i b, (a , b)  R, then its conjugate is defined as z  a  i b. zz zz and Im (z ) . Geometrically, the 2i 2 conjugate of z is the reflection or point image of z in the real axis. (2) Properties of conjugate : If z , z1 and z 2 are existing complex Imaginary axis Hence, we have Re(z )  Y O P(z )  – Real axis X Q (z ) numbers, then we have the following results: (i) (z )  z (iii) z1  z 2  z1  z 2 (ii) z1  z 2  z1  z 2 (iv) z1 z 2  z1 z 2 , In general z1.z 2.z 3..... z n  z1.z 2.z 3..... z n Complex Numbers 35 z (v)  1  z2  z1   , z 2  0  z2 (vi) (z )n  (z n ) (vii) z  z  2 Re( z )  2 Re( z )  purely real (viii) z  z  2i Im(z )  purely imaginary (x) z1 z 2  z1 z 2  2 Re( z1 z 2 )  2 Re( z1 z 2 ) (ix) z z  purely real 60 (xi) z  z  0 i.e., z  z  z is purely real i.e., Im(z )  0 (xii) z  z  0 i.e., z  z  either z  0 or z is purely imaginary i.e., Re(z )  0 (xiv) z  0  z  0 (xiii) z 1  z 2  z 1  z 2 (xv) zz  0  z  0 rei  re i E3 (xvi) If w  f (z ) then w  f (z ) (xvii) Important Tips Complex conjugate is obtained by just changing the sign of i.  Conjugate of i  i  Conjugate of iz  iz  (z 1  z 2 ) and (z 1. z 2 ) real  z 1  z 2 or z 2  z 1  z1 z 2  z1 z 2 U ID  Example: 13 D YG (3) Reciprocal of a complex number : For an existing non-zero complex number z  a  ib , the a  ib Re(z ) i[ Im(z )] 1 z z 1 reciprocal is given by z 1   i.e., z 1   2 =.   2 2 2 2 z | z| a  ib | z|2 a b | z| | z| If the conjugate of (x  i y) (1  2 i) be 1+i, then (a) x  1 5 (b) y  3 5 [MP PET 1996] (c) x  i y  Given that (x  i y) (1  2 i)  1  i  x  i y  Example: 14 For the complex number z, one from z  z and z z is U Example: 15 1  (b) x   n    2  (c) x  0 [IIT 1988] (d) No value of x sin x  i cos 2 x and cos x  i sin 2 x are conjugate to each other if sin x  cos x and cos 2 x  sin 2 x  5 9  5 9  5 9 , ,....... (ii) There , , , ,........ (i) and tan 2 x  1  2 x  , ,........ or x  , 8 8 8 4 4 4 4 4 4 exists no value of x common in (i) and (ii). Therefore there is no value of x for which the given complex numbers are conjugate. 2  3i The conjugate of complex number is [MP PET 2004] 4 i or tan x  1  x  Example: 16 (b) An imaginary number (d) Both are imaginary numbers The complex numbers sin x  i cos 2 x and cos x  i sin 2 x are conjugate to each other for (a) x  n  Solution: (d) 1i 1  2i Here z  z  (x  i y)  (x  i y)  2 x (Real) and zz  (x  i y )(x  iy)  x 2  y 2 (Real). ST Solution: (c) (d) x  i y  1i 1i  x iy  1 2i 1  2i Solution: (c) (a) A real number (c) Both are real numbers 1i 1  2i (a) 3i 4 (b) 11  10 i 17 (c) 11  10 i 17 (d) 2  3i 4i 36 Complex Numbers 11  10 i 2  3i (2  3i)(4  i) 8  3  12 i  2i 11  10 i  Conjugate .    17 4 i (4  i)(4  i) 16  1 17 The real part of (1  cos   2i sin  )1 is Example: 17 1 3  5 cos  (a) (b) 1 5  3 cos  (1  cos  )  i. 2 sin  1  2 sin 2  Solution: (c)  [Karnataka CET 2001; IIT 1978, 86] 2  i. 4 sin 1 3  5 cos  (c)  cos 2  1  2   =  2 sin  2   1 (d)    sin  i. 2 cos  2 2   1 5  3 cos  1 60 Solution: (b)   Hence, real part  2 sin   2    1  3 cos 2  2 2 The reciprocal of 3  7 i is 3 7  i 4 4 (a) 1 Solution: (c) 3  7i  (b) 3  7 i (c) 3 7  i 16 16 (d) 7  3i 1 3  7i 3  7i 3  7i 3 7.     i. 9 7 16 16 16 3  7i 3  7i D YG 2.6 Modulus of a Complex Number. U Example: 18 1 1 . 5  3 cos     2 1  3 cos 2  2  ID sin E3 1 sin  i.2 cos sin  i.2 cos  1  2 2  2 2   2 sin .    2  2  sin   i.2 cos  sin   i.2 cos   2 sin  sin  4 cos 2  2 2 2 2 2 2 2 Modulus of a complex number z  a  ib is defined by a positive real number given by | z |  a2  b 2 , where a, b real numbers. Geometrically |z| represents the distance of point P (represented by z) from the origin, i.e. |z| = OP. Y P(z) If |z| = 0, then z is known as zero modular complex number and is used to represent the origin of reference plane. U If |z| = 1 the corresponding complex number is known as ST unimodular complex number. Clearly z lies on a circle of unit radius having centre (0, 0). Note M O X :  In the set C of all complex numbers, the order relation is not defined. As such z 1  z 2  or z 1  z 2 has no meaning. But | z 1 | | z 2 | or | z1 | | z 2 | has got its meaning since | z1 | and | z 2 | are real numbers. Properties of modulus (i) z  0  z  0 iff z  0 and |z| 0 iff z  0. (ii)  z  Re (z )  z and  z  Im (z )  z (iii) z  z   z   z | zi | Complex Numbers 37 2 (iv) z z  z | z | 2 (v) z 1 z 2  z 1 z 2. In general z 1 z 2 z 3...... z n  z 1 z2 z 3.... z n (vii) | z n | | z | n , n  N 2  (z 1  z 2 )(z 1  z 2 )  z 1  z1 2  z1  z 2 (xi) az1  bz 2 2  z2 2 2    2 z1  bz 1  az 2 2 2 2  z2  (z 1 z 2  z 1 z 2 ) or | z1 | 2  | z 2 | 2  2 Re( z1 z 2 ) z z1 is purely imaginary or Re 1 z2  z2 2  z2 2  (a2  b 2 )     0  E3 2 (ix) z 1  z 2 (x) z 1  z 2 2 (Law of parallelogram) z 2  z2 2  , where a, b  R. ID (viii) z 1  z 2 60 z1 z1  , (z 2  0) z2 z2 (vi) 1 Important Tips Modulus of every complex number is a non-negative real number.  | z | | Re(z)|  Re(z) and | z | | Im(z)|  Im(z) z 1 z  z is always a unimodular complex number if z  0 z D YG    | z |  0 iff z  0 i.e., Re( z )  Im(z )  0 U   | z|  1 z   1 z z is always a unimodular complex number if z  0 | z|  | Re( z)|  | Im(z)|  2 | z | || z 1 |  | z 2 || | z 1  z 2 | | z 1 |  | z 2 | Thus | z 1 |  | z 2 | is the greatest possible value of | z 1  z 2 | and || z 1 |  | z 2 || is the least possible value of | z 1  z 2 | 1  a  a2  4 a  a2  4  a, the greatest and least values of | z | are respectively and z 2 2 If z   | z 1  z 12  z 22 |  | z 1  z 12  z 22 | | z 1  z 2 |  | z 1  z 2 | U  (1  i) (2  i)  (3  i) [MP PET 1995, 99; Rajasthan PET ST Example: 19 1998] Solution (c) (a) – 1/2 (b) 1/2 (1  i)(2  i) 1  3i 3  i 3  4 i z     | z|  1 (3  i) 3 i 3 i 5 Trick : | z|  Example: 20 2. 5 (d) – 1 1 10 If  and  are different complex numbers with |  |  1, then (a) 0 Solution (c) | z1 | | z 2 |  | z3| (c) 1 (b) 1/2   1   (c) 1       1   1   ,   1 1   |  | (   ) |  |      (   ) is equal to (d) 2 {| z| | z| } 38 Complex Numbers Example: 21 For any complex number z, maximum value of | z| | z  1| is Solution (b) (a) 0 (b) 1 We know that | z 1  z 2 | | z 1 | | z 2 | (c) 3/2 (d) None of these | z| | z  1| | z  (z  1)| or | z| | z  1|  1 ,  Maximum value of | z| | z  1| is 1. If z  x  iy and iz 2  z  0 , then | z | is equal to (a) 1 [Bihar CEE 1998] (b) 0 or 1 (c) 1 or 2 (d) 2 60 Example 22 Solution: (b) iz  z | iz | | z | | z| | z| | z| (| z |  1)  0  | z|  0 or | z | = 1 Example: 23 For x1, x 2 , y1, y2  R, if 0  x 1  x 2 , y 1  y 2 and z 1  x 1  iy1 , z 2  x 2  iy 2 , and z 3  2 2 2 z 3 satisfy (b) | z 1 | | z 2 | | z 3 | E3 [Roorkee 1991] (a) | z 1 | | z 2 | | z 3 | Solution: (d) 1 (z 1  z 2 ), then z 1 , z 2 and 2 (d) | z 1 | | z 3 | | z 2 | (c) | z 1 | | z 2 | | z 3 | 0  x 1  x 2 , y 1  y 2 (Given) | z1  z 2 |  x  x2   1 2 2  2   y  y2    1 2      2 ID | z 1 |  x 12  y 12 , | z 2 |  x 22  y 22 | z 2 | | z 1 | | z 3 |  2  x1  x 2     y12 | z 2 | | z1 |. Hence, | z 1 | | z 3 | | z 2 |  2  U 2.7 Argument of a Complex Number. Let z  a  ib be any complex number. If this complex number is represented geometrically D YG by a point P, then the angle made by the line OP with real axis is known as argument or amplitude of z and is expressed as b arg (z )    tan 1  ,   POM. Also, argument of a complex a number is not unique, since if  be a value of the argument, so also is 2n   , where n I. – (–,+)   X' (–, – –) + Y (+,+) P(z) -  – M O Y' X (+,–) – (1) Principal value of arg (z) : The value  of the argument, which satisfies the inequality       is called the principal value of argument. Principal values of argument z will be U  ,   ,     and   according as the point z lies in the 1st , 2nd , 3rd and 4th quadrants ST respectively, where   tan 1 b   (acute angle). Principal value of a argument of any complex number lies between      . b (i) a, b  First quadrant a  0, b  0. arg (z )    tan 1  . It is a an acute angle and positive. Y (a, b) X'  a O Y' b X Y – X' (– ,+) (–,–) – ( – )  (+,+) O (+,–) – Y' X Complex Numbers 39  b  . It is an obtuse angle and (ii) (a, b ) Second quadrant, a  0, b  0, arg (z )      tan 1  | a |  y (a, b) b x'  a x O 60 positive. y' negative. a x' b O x  y' ST U D YG U (a, b) ID y E3 b  (iii) (a, b)  Third quadrant a  0, b  0, arg (z )      tan 1  . It is an obtuse angle and a Complex Numbers 39 39 | b |  (iv) (a, b ) Fourth quadrant a  0, b  0, arg (z )     tan  . It is an acute angle and  a  1 negative. a O x'  60 y x b x y arg(z) Interval of  I + +  0    / 2 II – +    / 2    III – – (   )       / 2 IV + –    / 2   0 ID Quadrant U Note E3 y' (a, b)  Argument of the complex number 0 is not defined. : D YG  Principal value of argument of a purely real number is 0 if the real number is positive and is  if the real number is negative.  Principal value of argument of a purely imaginary number is  / 2 if the imaginary part is positive and is   / 2 if the imaginary part is negative. (2) Properties of arguments (i) arg (z1 z 2 )  arg (z1 )  arg (z 2 )  2k  , (k  0 or 1 or – 1) U In general arg (z 1 z 2 z 3......... z n )  arg(z 1 )  arg (z 2 )  arg (z 3 ) ..........  arg (z n )  2k  , (k  0 or 1 or  1) z (iii) arg 1  z2 ST (ii) arg (z 1 z 2 )  arg (z 1 )  arg (z 2 )    arg z 1  arg z 2  2k  , (k  0 or 1  or – 1) z (iv) arg   2arg z  2k  , (k  0 or 1 or – 1) z z (vi) If arg  2  z1 z     , then arg  1  z2  (vii) arg z  arg z  arg (ix) arg ( z )  arg (z )   (xi) arg (z )  arg (z )   1 z (v) arg(z n )  n arg z  2k  , (k  0 or 1 or – 1)    2k    , where k  I  (viii) arg (z  z )   / 2 (x) arg (z )  arg (z )  0 or arg (z )  arg (z ) 40 Complex Numbers (xii) z 1 z 2  z 1 z 2  2 | z 1 | | z 2 | cos ( 1   2 ), where  1  arg (z 1 ) and  2  arg(z 2 ) Note :  Proper value of k must be chosen so that R.H.S. of (i), (ii), (iii) and (iv) lies in ( ,  )  The property of argument is same as the property of logarithm. 60 If arg (z) lies between   and  ( inclusive), then this value itself is the principal value of arg (z). If not, see whether arg (z)   or   . If arg(z )   , go on subtracting 2 until it lies between   and  (  The general value of a rg (z ) is 2n   arg (z ). E3 obtained will be the principal value of arg (z). inclusive). The value thus Important Tips  If z1  z 2  z1  z 2 and arg z 1 = arg z2. z1  z 2  z1 |  | z 2  arg (z1 )  arg (z 2 ) i.e., z1 and z2 are parallel.  z1  z 2  z1 |  | z 2  arg (z1 )  arg (z 2 )  2n , where n is some integer.   | z1  z 2 |  | | z1 |  | z 2 ||  arg(z1 )  arg(z 2 )  2n , where n is some integer. z 1  z 2  z 1  z 2  arg (z 1 ) – arg (z 2 )   / 2. If | z 1 | 1,| z 2 | 1 then (i) z1  z 2   z1  z 2 2  (arg (z1 )  arg (z 2 ))2 (ii) z1  z 2 2   z1  z 2 2  D YG arg (z1 )  arg (z 2 )2 2 U  ID   z1  z 2 2  z1  z 2  2| z1 | | z 2 | cos(1   2 ).  z1  z 2 2  z1  z 2  2| z1 | | z 2 | cos(1   2 ). 2 2 2 2  If | z 1 | | z 2 | and amp (z 1 )  amp (z 2 )  0, then z 1  z 2 are conjugate complex numbers of each other.  z  0, amp (z  z )  0 or  ; amp (zz )  0; amp (z  z )   / 2.  arg (1)  0, arg (1)   ; arg (i)   / 2, arg (i)   / 2.  arg (z )   Amplitude of complex number in I and II quadrant is always positive and in III rd and IVth quadrant is always negative.  If a complex number multiplied by i (Iota) its amplitude will be increased by  / 2 and will be decreased by  / 2 ,   if multiplied by –i, i.e. arg(iz)   arg(z ) and arg(iz)  arg(z ) . 2 2   Re (z )  Im(z ). ST U 4 Complex number +ve Re (z) –ve Re (z) +ve Im (z) Value of argument 0 –ve Im (z)   /2 3 / 2 or   / 2 – (z) |    |, if  is  ve and  ve respectively (iz)     arg (z ) 2  –(iz)   arg (z )   2  Complex Numbers 41 41 (z n ) n. arg (z) (z 1.z 2 ) arg (z1) + arg (z2)  z1   z2 arg (z1) – arg (z2)    1  i Amplitude of   is 1  i [Rajasthan 1996]  2 (a) 1i 1i  i 1i 1i   1     tan 1     0 2   Example: 25 (c) If z 1  z 2 and amp z 1  amp z 2  0 , then (b) z 1  z 2 (a) z 1  z 2 Let z 1  OP, z 2  OQ (c) z 1  z 2  0 Y D YG Q is point image of P U Since amp (z1 ) =   amp (z 2 ) = –   (d) O  z1  z 2 Trick : arg z  arg z  0, Example: 26 [MP PET (d) z 1  z 2 P  X – z1 must be equal to z 2. Q Let z, w be complex numbers such that z  iw  0 and arg zw   , then arg z equals [AIEEE 2004] (b)  / 2 (a) 5 / 4 Solution: (d)  6 (Since z lies on negative imaginary axis) 1999] Solution: (b)  4 E3 z  2 ID Solution: (a) (b) PET 60 Example: 24 (c) 3 / 4 (d)  / 4 z  iw  0 U  z  iw    ( / 2   )   ,     / 4. Example: 27 The amplitude of sin  5    i  1  cos  5  ST (a)  / 5 Solution: (c) sin  5  i (1  cos  5 [Karnataka CET 2003] (c)  / 10 (b) 2 / 5 )  2 sin  10 sin. cos  10  i 2 sin 2  10  2 sin (d)  / 15       i sin  cos  10  10 10   10  tan     . For amplitude, tan    10 10 cos 10 Example: 28 If z  4 and arg z  5 , then z  6 [MP 1987] (a) 2 3  2i Solution: [c] z  4 and arg z  (b) 2 3  2i 5  150 º , 6 (c)  2 3  2i (d)  3  i PET 42 Complex Numbers Let z  x  iy , then z  r  x 2  y 2  4 and   5  150 º 6  x  r cos   4 cos 150 º   2 3 and y  r sin   4 sin 150 º  4  1  2. 2  z  x  i y  2 3  2i. 5  150 º , here the complex number must lie in second quadrant, so (a) and (b) 6 60 Trick: Since arg z  rejected. Also z  4 , which satisfies (c) only. equal to [AIEEE 2003] (a) 1 (b) – 1 From equation (i) and (ii), | z | |  |  1 and (c) i z z z  1 i .....(i) and arg         2 | z ||  |  1 z   z   0; z  z   0  z    z  , then z  is (d) – i.....(ii) z     z   i |  | 2  i. U 2.8 Square Root of a Complex Number. 2 ID Solution: (d)  E3 If z and  are to non-zero complex numbers such that | z |  1 and arg (z) – arg ( )  Example: 29 Let a  ib be a complex number such that D YG Then a  ib  x  iy, where x and y are real numbers. a  ib  x  iy  a  ib  (x  iy)2  a  ib  (x 2  y 2 )  2ixy  x 2  y2  a.....(i) and 2 xy  b.....(ii) [On equating real and imaginary parts]  a  ib       a2  b 2  a    i   2   ST  U  a2  b 2  a   and y   Solving, x      2   Therefore Note :  a2  b 2  a      2    a2  b 2  a      2     | z | a  | z | a | z | a  | z | a  a  ib    i i  for b>0     for b1) U Then point z describes a hyperbola having foci at z 1 and z 2 and a  R  Example: 38 S(z1) O S'(z2 ) If in the adjoining diagram, A and B represent complex number z 1 and z 2 respectively, then C represents (a) z 1  z 2 ST Y (b) z 1  z 2 C B (c) z 1.z 2 A (d) z 1 / z 2 O X Solution: (a) It is a fundamental concept. Example: 39 If centre of a regular hexagon is at origin and one of the vertex on argand diagram is 1  2 i, then its perimeter is [Rajasthan PET 1999; Himachal CET 2002] (a) 2 5 Solution: (d) (b) 6 2 (c) 4 5 Let the vertices be z 0 , z 1 ,....., z 5 w.r.t. centre O and | z 0 |  5 (d) 6 5 Complex Numbers 51  A 0 A1 | z 1  z 0 | | z 0 e i  z 0 | | z 0 | | cos   i sin  1 |  5 (cos   1) 2  sin 2    5 2(1  cos  )  5. 2 sin( / 2) 2       A0 A1  5. 2 sin( / 6)  5    6 3  .....(i) A5 A0 Hence, the perimeter of regular polygon is  A0 A1  A1 A2  A2 A3  A3 A4  A4 A5  A5 A0  6 5. (a) Of area zero z1  z 3 1  i 3 are the vertices of a triangle which is  z2  z3 2 (b) Right-angled isosceles Taking mod of both sides of given relation (c) z1  z 3 1 3  i  z2  z3 2 2  z  z3    = tan 1 ( 3 )   So, | z 1  z 3 | | z 2  z 3 |. Also, amp  1  3 z  z 3   2 (d) 1 3  1. 4 4 z z or amp  2 3  z1  z 3     3 or z 2 z 3 z 1  60    The triangle has two sides equal and the angle between the equal sides  60 . So it is equilateral. Let the complex numbers z 1 , z 2 and z 3 be the vertices of an equilateral triangle. Let z 0 be the U Example: 41 [IIT Screening2001] Equilateral ID Solution: (b) A1 E3 The complex numbers z 1 , z 2 and z 3 satisfying A2 O (0,0 ) 60 Similarly, A1 A 2  A 2 A3  A3 A4  A4 A5  A5 A0  5 Example: 40 (1+2i) A3 A4 circumcentre of the triangle, then z 12  z 22  z 32  Solution: (c) (b)  z 02 D YG (a) z 02 (c) 3z 02 (d)  3z 02 Let r be the circum-radius of the equilateral triangle and  the cube root of unity. Let ABC be the equilateral triangle with z 1 , z 2 and z 3 as its Y A'(z1 – r  z0) O 2 3 vertices A, B and C respectively with circumcentre O' (z 0 ). The vectors O' A, O' B, O' C are equal and parallel to O A, O B, OC ' respectively. Y Then the vectors OA'  z1  z 0  rei 2 3  OB'  z 2  z 0  rei(  2 / 3)  r e i X C (z3 – z0 ) U Y  OC '  z 3  z 0  rei(  4  / 3)  r 2e i A(Z1)  z1  z 0  rei , z 2  z 0  r e i , z 3  z 0  r 2 e i ST 2 3 Squaring and adding, we get, O'(Z0) z12  z 22  z 32  3 z 02  2(1     2 )z 0 re i  (1   2   4 ) r 2 e i 2  3 z 02 , since 1     2  0  1   2   4. Example: 42 B(Z2) C(Z3) The points z 1 , z 2 , z 3 , z 4 in the complex plane are the vertices of a parallelogram taken in order, if and only if [IIT 1981,83] (a) z1  z 4  z 2  z 3 Solution: (b) (b) z 1  z 3  z 2  z 4 (c) z 1  z 2  z 3  z 4 (d) None of these Diagonals of a parallelogram ABCD are bisected each other at a point i.e., z1  z 3 z  z4  2 2 2 z1  z 3  z 2  z 4. Example: 43 If the complex number z 1 , z 2 and the origin form an equilateral triangle then z 12  z 22   52 Complex Numbers (a) z 1 z 2 Solution: (a) (b) z 1 z 2 (d) | z 1 | 2 | z 2 | 2 (c) z 2 z 1 Let OA, OB be the sides of an equilateral  OAB and OA, OB represent the complex numbers or vectors z 1 , z 2 respectively. Y From the equilateral  OAB, AB  Z 2  Z1  z    arg(z 2  z 1 )  argz 2  and arg 2  3   z1     arg(z 2 )  arg(z 1 )   3  z  z1 z Also, 2  1  2 , since triangle is equilateral. z2 z1 z 2  z1 z2 vectors are equal, that is and Z 2– Z 1  Z1 3 A(Z1 ) O X z2 have same modulus and same argument, which implies that the z1 E3 Thus the vectors Z2 60  z  z1  arg 2  z2 B(Z2 ) z 2  z1 z  2  z 1 z 2  z 12  z 22  z 12  z 22  z 1 z 2. z2 z1 2.13 Rotation Theorem. ID Rotational theorem i.e., angle between two intersecting lines. This is also known as coni method. Let z 1 , z 2 and z 3 be the affixes of three points A, B and C respectively taken on argand U plane. Then we have AC  z 3  z 1 and AB = z 2  z 1 Y C(z3) and let arg AC  arg (z 3  z 1 )   and AB  arg (z 2  z 1 )   D YG  Let CAB   ,  CAB       B(z2) A(z1)   z  z1   = arg AC  arg AB = arg (z 3  z 1 )  arg (z 2  z 1 ) = arg  3  z 2  z1   O X  affix of C  affix of A   angle between AC and AB = arg   affix of B  affix of A  or U For any complex number z we have z | z | e i(argz) ST  z  z1    Similarly,  3  z 2  z1  Note :  z 3  z1   z 3  z 1  i  a rg z 2  z1  z  z1 | z 3  z1 | i(CAB ) AC i   e or 3  e  e z 2  z1 | z 2  z1 | AB  z 2  z1  Here only principal values of the arguments are considered.  z  z2   z  z2     , if AB coincides with CD, then arg  1   0 or   , so that  arg  1  z3  z4   z3  z4  z1  z 2 z  z2 is real. It follows that if 1 is real, then z3  z4 z3  z4 the points A, B, C, D are collinear.  If AB is perpendicular to P(z1) S(z4) A CD, then  R(z3) arg C Q(z2) D B Complex Numbers 53  z1  z 2   z3  z4     / 2 ,  so z1  z 2 is purely imaginary. It follows that if z 1  z 2 = z3  z4  k z 3  z 4  , where k purely imaginary number, then AB and CD are perpendicular to each other. (1) Complex number as a rotating arrow in the argand plane : Let z  rcos   i sin    re i plane. Then OP | z |  r and POX   Q(zei) Y   X' P(z) X O E3 Now consider complex number z1  ze i or z1  re i.e i  re i   60 r. e i be a complex number representing a point P in the argand..…(i) Y' {from (i)} Clearly the complex number z 1 represents a point Q in the argand plane, when OQ  r and ID QOX    . Clearly multiplication of z with e i through angle  rotates the vector OP in U anticlockwise sense. Similarly multiplication of z with e  i will rotate the vector OP in clockwise sense. : If z 1 , z 2 and z 3 are the affixes of the points A,B and C such that AC  AB and D YG Note CAB  . Therefore, AB  z 2  z1 , AC  z 3  z1. C(z3) Then AC will be obtained by rotating AB through an angle  in anticlockwise sense, and therefore, AC  AB e i or (z 3  z1 )  (z 2  z1 )e i or B(z2)  z 3  z1  e i z 2  z1 A(z1) U  If A, B and C are three points in argand plane such that AC  AB and CAB   then use the rotation about A to find e i , but if AC  AB use coni method. ST  Let z 1 and z 2 be two complex numbers represented by point P and Q in the argand plane such that  POQ  . Then, z1e i is a vector of magnitude | z 1 |  OP along OQ and z1e i | z1 | is a unit vector along OQ. Consequently, | z 2 |. magnitude | z 2 |  OQ along OQ i.e., z 2  z1 e i | z1 | is a vector of | z2 | z.z1e i  z 2  2. | z1 | z1 (2) Condition for four points to be concyclic : If points A,B,C and D are concyclic  ABD   ACD Using rotation theorem A(z1)  B(z2) D(z4)  C(z3) 54 Complex Numbers In  ABD (z1  z 2 ) z 4  z 2 i  e z1  z 2 z4  z2 In  ACD (z1  z 3 ) z 4  z 3 i  e z1  z 3 z4  z3.....(i).....(ii) 60 From (i) and (ii) (z1  z 2 ) (z 4  z 3 ) (z1  z 2 ) (z 4  z 3 ) =Real  z1  z 3 z 4  z 2 (z1  z 3 )(z 4  z 2 ) if z 1 , z 2 , z 3 and z 4 are such that (z1  z 2 ) (z 4  z 3 ) is real, then these four points are (z1  z 3 ) (z 4  z 2 ) E3 So concyclic. Example: 44 If complex numbers z1 , z 2 and z 3 represent the vertices A, B and C respectively of an isosceles ID triangle ABC of which C is right angle, then correct statement is (a) z 12  z 22  z 32  z 1 z 2 z 3 (b) (z 3  z 1 )2  z 3  z 2 (d) (z1  z 2 )2  2(z1  z 3 ) (z 3  z 2 ) Solution: (d) U (c) (z 1  z 2 )2  (z 1  z 3 ) (z 3  z 2 ) BC  AC and C   / 2 A(z1) By rotation about C in anticlockwise sense CB  CAe i / 2 D YG  (z 2  z 3 )  (z 1  z 3 )e i / 2  i(z 1  z 3 )  (z 2  z 3 )2  (z1  z 3 )2  z 22  z 32  2 z 2 z 3   z12  z 32  2 z1 z 3 90°  z 12  z 22  2 z 1 z 2  2 z 1 z 3  2 z 2 z 3  2 z 32  2 z 1 z 2  (z 1  z 2 )  2[(z 1 z 3  2 Example: 45 C(z3)  z 2 z 3 )]  (z 1  z 2 )  2(z 1  z 3 )(z 3  z 2 ). 2 In the argand diagram, if O,P and Q represents respectively the origin, the complex numbers z and z  iz, then the angle OPQ is (a)  4 (b)  3 (c)  2 (d) 2 3 It is a fundamental concept. The centre of a regular polygon of n sides is located at the point z  0 and one of its vertex z 1 is U Solution: (c) Example: 46 z 32 )  (z 1 z 2 B(z2) known. If z 2 be the vertex adjacent to z 1 , then z 2 is equal to ST 2  2      i sin (a) z 1  cos  (b) z 1  cos  i sin  n n n  n   Solution: (a)      i sin (c) z 1  cos  (d) None of these 2n 2n   Let A be the vertex with affix z 1. There are two possibilities of z 2 i.e., z 2 can be obtained by rotating z 1 through 2 either in n B(z2) 2/n 2/n clockwise or in anticlockwise direction.  z2 z  2 e z1 z1 i 2 2  z 2  z1 e 2 2    i sin  z 2  z1  cos  n n   i 2 2 (| z 2 | | z 1 |) A(z1) B(z2) Complex Numbers 55 Let z 1 , z 2 , z 3 be three vertices of an equilateral triangle circumscribing the circle | z |  z1  If 1 3i  and z 1 , z 2 , z 3 are in anticlockwise sense then z 2 is 2 2 (b) 1  3 i (a) 1  3 i Solution: (d) 1. 2 1 3 z 2  z 1 e i 2 / 3    2 2  (c) 1 1   3  2 2   i sin i i   cos       3 3  2  2  (d) – 1  1 3  3 1   i    1.  2  4 4 2   60 Example: 47 2.14 Triangle Inequalities. (3) | z 1  z 2 | | | z 1 |  | z 2 || : (4) | z 1  z 2 | || z 1 |  | z 2 || In a complex plane | z 1  z 2 | is the distance between the points z 1 and z 2. ID Note E3 In any triangle, sum of any two sides is greater than the third side and difference of any two side is less than the third side. By applying this basic concept to the set of complex numbers we are having the following results. (1) | z 1  z 2 | | z 1 |  | z 2 | (2) | z 1  z 2 | | z 1 |  | z 2 |  The equality | z 1  z 2 | | z 1 |  | z 2 | holds only when arg (z 1 ) = arg (z 2 ) i.e., z 1 and z 2 z 2 are antiparallel. U are parallel.  The equality | z 1  z 2 | || z 1 |  | z 2 || holds only when arg (z 1 ) – arg (z 2 ) =  i.e., z 1 and D YG  In any parallelogram sum of the squares of its sides is equal to the sum of the squares of its diagonals i.e. | z1  z 2 | 2  | z1  z 2 | 2  2 (| z1 | 2  | z 2 | 2 )  Law of polygon i.e., | z 1  z 2 ....  z n | | z 1 |  | z 2 | .... | z n | Important Tips 1 | z|2. 2 The area of the triangle whose vertices are z, iz and z + iz is  If z 1 , z 2 , z 3 be the vertices of a triangle then the area of the triangle is  Area of the triangle with vertices z, wz and z  wz is  If z 1 , z 2 , z 3 be the vertices of an equilateral triangle and z o be the circumcentre, then z 12  z 22  z 32   3z 02.  If z 1 , z 2 , z 3..... z n be the vertices of a regular polygon of n sides and z 0 be its centroid, then z 12  z 22 .....  z n2  nz 02.  If z 1 , z 2 , z 3 be the vertices of a triangle, then the triangle is equilateral iff (z 1  z 2 ) 2  (z 2  z 3 ) 2  (z 3  z 1 ) 2  0 or ST U  z 12  z 22  z 32  z 1 z 2  z 2 z 3  z 3 z 1 or  (z 2  z 3 )| z1 | 2 4 iz1. 3 | z2 |. 4 1 1 1   0. z1  z 2 z 2  z 3 z 3  z1  If z 1 , z 2 z 3 are the vertices of an isosceles triangle, right angled at z 2 then z 12  z 22  z 32  2 z 2 (z 1  z 3 ).  If z 1 , z 2 , z 3 are the vertices of right-angled isosceles triangle, then (z 1  z 2 ) 2  2z 1  z 3 (z 3  z 2 ).  If one of the vertices of the triangle is at the origin i.e., z 3  0, then the triangle is equilateral iff z 12  z 22  z 1 z 2  0. 56 Complex Numbers z 1 z 2 z 3  z1 If z 1 , z 2 , z 3 and z 1 , z 2 , z 3 are the vertices of a similar triangle, then z 2 z3  If z 1 , z 2 , z 3 be the affixes of the vertices A, B, C respectively of a triangle ABC, then its orthocentre is 1 1 0. 1 Solution: (b) The points 1  3i, 5  i and 3  2i in the complex plane are (a) Vertices of a right angled triangle (b) Collinear (c) Vertices of an obtuse angled triangle (d) Vertices of an equilateral triangle E3 Example: 48 Let z1  1  3i, z 2  5  i and z 3  3  2i. Then area of triangle A  z1 , z 2 and z 3 are collinear. x1 1 x2 2 x3 y1 y2 y3 1 1 1  1 3 1 1 5 1 1  0 , Hence 2 3 2 1 If z  x  iy , then area of the triangle whose vertices are points z, iz and z  iz is ID Example: 49 60 a(sec A)z 1  b(sec B)z 2  (c sec C )z 3. a sec A  b sec B  c sec C [IIT 1986; MP PET 1997, 2001; DCE 1997; AMU 2000; UPSEAT 2002] 1 (b) | z | 2 2 (a) 2 | z | 2 (d) 3 | z| 2 2 Let z  x  iy , z  iz  x  y   ix  y  and iz   y  ix U Solution: (b) (c) | z | 2 If A denotes the area of the triangle formed by z , z  iz and iz , then x y 1 x y x y 2 y x 1 We get A  2 Example: 50 1 1 1 (Applying transformation R2 R2  R1  R3 ) D YG A x 0 y y 0 x 1 1 1  1 = (x 2  y 2 )  | z | 2. 2 2 0 | z1  z 2 | | z1 |  | z 2 | is possible if (b) z 2  1 z1 (c) arg (z1 )  arg( z 2 ) (d) | z1 | | z 2 | U (a) z 2  z1 [MP PET 1999] Solution: (c) Squaring both sides, we get | z1 | 2  | z 2 | 2  2 | z1 || z 2 | cos(1   2 ) | z1 | 2  | z 2 | 2 2 | z1 || z 2 | ST  2| z1 || z 2 | cos(1   2 )  2 | z1 || z 2 |  cos(1   2 )  1  1   2  0 o  1   2 Hence arg (z1 )  arg (z 2 ) Trick: Let z1 and z2 are the two sides of a triangle. By applying triangle inequality (z 1  z 2 ) is the third side. Equality holds only when 1   2 i.e., z1 and z 2 are parallel. 2.15 Standard Loci in the Argand Plane. (1) If z is a variable point in the argand plane such that arg (z )   , then locus of z is a straight line (excluding origin) through the origin inclined at an angle  with x–axis. (2) If z is a variable point and z 1 is a fixed point in the argand plane such that arg (z  z 1 )   , then locus of z is a straight line passing through the point representing z 1 and inclined at an angle  with x-axis. Note that the point z 1 is excluded from the locus. Complex Numbers 57 (3) If z is a variable point and z 1 , z 2 are two fixed points in the argand plane, then  (i) | z  z 1 | | z  z 2 | Locus of z is the perpendicular bisector of the line segment joining z 1 and z 2 z1  z 2 |   Locus of z is an ellipse 60 | (ii) | z  z 1 |  | z  z 2 | = constant  (iii) | z  z 1 |  | z  z 2 | | z 1  z 2 | Locus of z is the line segment joining z 1 and z 2 but z does not lie between z 1 and z 2.  Locus of z is a hyperbola. (vi) | z  z 1 | 2  | z  z 2 | 2 | z 1  z 2 |  Locus of z is a circle with z 1 and z 2 as the (vii) | z  z 1 |  k | z  z 2 | k  1 ID | z1  z |  Locus of z is a straight line joining z 1 and z 2 E3  (iv) | z  z 1 |  | z  z 2 | | z 1  z 2 | (v) | z  z1 | | z  z 2 |  constant 2 extremities of diameter.  Locus of z is a segment of circle.   =   / 2   Locus of z is a circle with z 1 and z 2 as the D YG  z  z1 (ix) arg   z  z2 vertices of     (fixed )  Locus of z is a circle. U  z  z1 (viii) arg   z  z2   z  z1   = 0 or  (x) arg    z  z2  and z 2.  diameter. Locus z is a straight line passing through z 1 U (xi) The equation of the line joining complex numbers z 1 and z 2 is given by z  z1 z  z1  z 2  z1 z 2  z 1 ST z z 1 or z1 z1 1  0 z2 z2 1 Example: 51  z 1  The locus of the points z which satisfy the condition arg  is   z 1 3 (a) A straight line (b) A circle z  1 x  iy  1 (x  y  1)  2iy   z  1 x  iy  1 (x  1)2  y 2 2 Solution:(c) We have  arg z 1 2y  tan 1 2 z 1 x  y2  1 Hence tan 1 2y x2  y2  1   3 2 (c) A parabola [Rajasthan PET 2000,2002; MP PET 200 (d) None of these 58 Complex Numbers 2y  2  tan  3  3  x2  y2 1  y  x 2  y2  2 3 3 y  1  0 , which is obviously a circle. z 2  1 | z | 2 1, then z lies on If (a) An ellipse Solution: (b) 2 [AIEEE 2004] (b) The imaginary axis (c) A circle 2  | z  1 | 2| z  1 | 2  (zz  1)2  (z  1) (z  1) (z  1) (z  1)  (zz  1)2  z  z  0  z lies on imaginary axis.  z 1  The locus of the point z satisfying arg    k. (where k is non-zero) is  z 1 E3 Example: 53 (a) Circle with centre on y–axis (b) (b) A straight line parallel to x–axis Circle with centre on x–axis (d) A straight line making an angle 60° with x–axis  (x  1)  iy   z 1  arg   k  arg   k  arg[(x  1)  iy]  arg[(x  1)  iy]  k  z 1   (x  1)  iy  ID Solution: (a) (d) The real axis | z  1 | | z | 1 2 60 Example: 52 x  y 1 2  y y     y  2y 1  y  1 x  1 x  1   k  tan k  y(x  1)  y(x  1)   tan    tan    k  tan  2 x 2  y2 1 x2  y2 1 y    x 1   x 1  1 2  x 1    U 1  2y 2y  x 2  y2 1  x 2  y2  1  0 tan k tan k Example: 54 D YG It is an equation of circle whose centre is (g,  f )  (0, cot k ) on y–axis. The locus of z satisfying the inequality log1 / 3 | z  1|  log1 / 3 | z  1| is (a) R(z )  0 Solution: (a) (b) R(z )  0 (c) I(z )  0 (d) None of these log1 / 3 | z  1|  log1 / 3 | z  1|  | z  1| | z  1|  x 2  1  2 x  y 2  x 2  1  2 x  y 2  x  0  Re( z )  0. Example: 55 If   i  tan 1 (z), z  x  iy and  is constant, the locus of 'z' is (a) x 2  y 2  2 x cot 2  1 (b) cot 2(x 2  y 2 )  1  x (d) x 2  y 2  2 x sin 2  1 tan(  i )  x  iy U Solution: (a) (c) x 2  y 2  2y tan 2  1 [EAMCET 1995; KCET 1996]  tan(  i )  x  iy (conjugate),  is a constant and  is known to be eliminated ST tan 2  tan(  i    i )  tan 2  x  iy  x  iy 1  (x 2  y 2 )  1  (x 2  y 2 )  2 x cot 2  x 2  y 2  2 x cot 2  1. 2.16 De' Moivre's Theorem. (1) If n is any rational number, then (cos   i sin  )n  cos n  i sin n. (2) If z  (cos  1  i sin  1 )(cos  2  i sin  2 )(cos  3  i sin  3 ).....(cos  n  i sin  n ) then z  cos( 1   2   3 .....   n )  i sin( 1   2   3 .....   n ) , where 1 ,  2,  3..... n  R.   2k      2k    (3) If z  r(cos   i sin  ) and n is a positive integer, then z 1 / n  r1 / n cos    i sin  n n       ,  Complex Numbers 59 where k  0, 1, 2, 3,.....(n  1).  2k   p    2k   p    ,   i sin  (4) If p, q  z and q  0, then (cos   i sin  ) p / q  cos  q q     where k  0, 1, 2, 3.....(q  1). 60 Deductions: If n  Q, then (ii) (cos   i sin  )n  cos n  i sin n (i) (cos   i sin  )n  cos n  i sin n     (iv) (sin   i cos  )n  cos n      i sin n    2  2  E3 (iii) (cos   i sin  )n  cos n  i sin n Applications (i) In finding the expansions of trigonometric functions i.e. cos n  cos n   n C 2 cos n  2  sin 2  ID  n C 4 cos n  4  sin 4  –...... sin n  nC1 cos n 1  sin  n C 3 cos n  3  sin 3   n C 5 cos n 5  sin 5  ....... (ii) In finding the roots of complex numbers. Note : U (iii) In finding the complex solution of algebraic equations. This theorem is not valid when n is not a rational number or the complex D YG number is not in the form of cos   i sin . Powers of complex numbers : Let z  x  iy  r(cos   i sin  )  z n  r n (cos   i sin  )n  r n (cos n  i sin n ) Number x + iy form Standard complex General form cos 0  i sin 0 cos 2n  i sin 2n –1 – 1+i0 cos   i sin  cos( 2n  1)  i sin(2n  1) i 0 +i(1) ST U 1+i0 1 –i Example: 56 1  i If   1  i cos 0 +i(–1) cos  2  2  i sin  i sin  2  2 cos(4 n  1)  cos(4 n  1) 2  2  i sin(4 n  1)  2  i sin(4 n  1)  100  a  ib , then (a) a = 2, b = –1 (b) a = 1, b = 0 (c) a = 0, b = 1 (d) a = –1, b = 2 2 60 Complex Numbers Solution: (b) 1i 1i       i  cos     i sin    (i)100  cos( 50  )  i sin(50  ) = 1  i (0)  a  1, b  0 1i 1i  2  2 Example: 57     If x r  cos  r  + i sin  r  , then x1. x 2.x 3........... is 2  2  [Rajasthan PET 1990, 2000; Karnataka CET 2000; UPSEAT 1990; Haryana CEE 1998; BIT Ranchi 1996] (c) –1        x 1. x 2. x 3..... upto    cos  i sin   cos 2  i sin 2  ……………….. 2 2  2 2            = cos   2 ...   i sin  2 ....  = cos  2 1  2 2  2 2  1  2  r n 2  i sin r n2 , where r  1, 2,3,....., n, then lim z1 z 2 z 3..... z n is equal to (a) cos   i sin  Solution: (c) z r  cos z 2  cos r  i sin n2 2  i sin n2 n n2 r  z 1  cos 2  i sin ;……………….. n2  i sin  n2 n n2  n2 (c) e i / 2 (d) 3 [UPSEAT 2001] e i ;       lim (z 1 , z 2 , z 3 ,..... z n )  lim cos  2 (1  2  3 .....  n)  i sin  2 (1  2  3 .....  n) n  n  n n      D YG  z n  cos n  (b) cos( / 2)  i sin( / 2) n2    = cos   i sin   1    ID If z r  cos        i sin  2 1   1   2   U Example: 58 (d) 0 E3 Solution: (c) (b) –2 60 (a) –3  n(n  1)  n(n  1)      lim cos  i sin  cos  i sin  e i / 2.  2 2 n   2 2 2n 2n  n  1  sin   i cos       1  sin   i cos    n   n   n    i sin  n  (b) cos   2   2   n   n   n    i cos   n  (c) sin 2 2      n   n   n    i sin n  n  (d) cos n 2 2     n  1  cos   i sin    1  sin   i cos       1 sin  cos    i  1  cos   i sin     ST Solution: (a) [Kerala (Engg.) 2002]  n   n   n    i sin  n  (a) cos   2   2  U Example: 59      2 cos 2  2i sin cos 2 2 2     2   2i sin. cos  2 cos 2 2 2  n n       cos  i sin 2 2         cos  i sin 2 2       where      2   n n      n  cis     2    cis      cis(n  )            2 2    cis       2     n   n   n   cis n     cis      cos   n    i sin  n  . 2 2 2 2         2.17 Roots of a Complex Number. (1) nth roots of complex number (z1/n) : Let z  r(cos  i sin  ) be a complex number. To find the roots of a complex number, first we express it in polar form with the general value of its Complex Numbers 61 amplitude and use the De' moivre's theorem. By using De'moivre's theorem nth roots having n distinct values of such a complex number are given by 2m    2m      z 1 / n  r 1 / n cos  i sin , where m  0, 1, 2,....., (n  1). n n   60 Properties of the roots of z1/n : (i) All roots of z1/n are in geometrical progression with common ratio e 2 i / n. (ii) Sum of all roots of z1/n is always equal to zero. E3 (iii) Product of all roots of z 1 / n  (1)n1 z. (iv) Modulus of all roots of z1/n are equal and each equal to r 1 / n or | z | 1 / n. (v) Amplitude of all the roots of z1/n are in A.P. with common difference 2. n ID (vi) All roots of z1/n lies on the circumference of a circle whose centre is origin and radius equal to | z | 1 / n. Also these roots divides the circle into n equal parts and forms a polygon of n sides. U (2) The nth roots of unity : The nth roots of unity are given by the solution set of the equation x n  1  cos 0  i sin 0  cos 2k   i sin 2k  x  cos D YG x  [cos 2k  i sin 2k ]1 / n 2k  2k   i sin , where k  0, 1, 2,....., (n  1). n n Properties of nth roots of unity (i) Let   cos 2 2  i sin  e i(2 / n) , the nth roots of unity can be expressed in the form of a n n U series i.e., 1, ,  2 ,..... n1. Clearly the series is G.P. with common difference  i.e., e i(2 / n). (ii) The sum of all n roots of unity is zero i.e., 1     2 .....   n 1  0. ST (iii) Product of all n roots of unity is (1)n1. (iv) Sum of pth power of n roots of unity 0, when p is not multiple of n 1   p   2 p .....   (n 1) p   n, when p is a multiple of n (v) The n, nth roots of unity if represented on a complex plane locate their positions at the vertices of a regular plane polygon of n sides inscribed in a unit circle having centre at origin, one vertex on positive real axis. Note : x n  1  (x  1)(x n1  x n2 .....  x  1)  (sin   i cos  )  i2 sin   i cos   i(cos   i sin  ) 62 Complex Numbers (3) Cube roots of unity : Cube roots of unity are the solution set of the equation x 3  1  0  2k   2k    i sin  x  (1)1 / 3  x  (cos 0  i sin 0)1 / 3  x  cos  , where k  0,1,2 3  3  Therefore roots are 1, cos 2 2 4 4 or 1, e 2 i / 3 , e 4  i / 3.  i sin , cos  i sin 3 3 3 3 x  1, 60 Alternative : x  (1)1 / 3  x 3  1  0  (x  1)(x 2  x  1)  0 1  i 3 1  i 3 , 2 2 E3 If one of the complex roots is  , then other root will be  2 or vice-versa. Properties of cube roots of unity (i) 1     2  0 0, if r not a multiple of 3 (iii) 1   r   2 r   3, if r is a multiple of 3 (iv)    2 and ( )2   and .   3. U (v) Cube roots of unity from a G.P. ID (ii)  3  1 (vi) Imaginary cube roots of unity are square of each other i.e., D YG ( 2 )2   3.   . 1 (vii) Imaginary cube roots of unity are reciprocal to each other i.e.,  ()2   2   2 and 1 2 and . (viii) The cube roots of unity by, when represented on complex plane, lie on vertices of an equilateral triangle inscribed in a unit circle having centre at origin, one vertex being on positive real axis. (ix) A complex number a  ib, for which | a : b |  1 : 3 or 1  i 3  e 2 i / 3 , 2 U of i, ,  2. Note :  If then 2  3 : 1, can always be expressed in terms 1  i 3  e  4  i / 3  e  2 i / 3 2 or vice-versa ST .   3.  a  b   c 2  0  a  b  c, if a, b, c are real.  Cube root of – 1 are  1,, 2. Important Tips   x 2  x  1  x    x   2   x  x  1  (x   )(x   )  x 2  xy  y 2  x  y   x  y  2  x  y  (x  iy)(x  iy)  x  y  (x  y ) (x  y  ) (x  y  )  x 3  y 3  (x  y ) (x  y  ) (x  y  2 )  x 2  y 2  z 2  xy  yz  zx  (x  y   z 2 )(x  y  2  z)  x 3  y 3  z 3  3 xyz  (x  y  z ) (x   y   2 z ) (x   2 y   z ) 2  2 2  2   x 2  xy  y 2  x  y   x  y  2 3 3  2 Complex Numbers 63 Fourth roots of unity : The four, fourth roots of unity are given by the solution set of the equation x 4  1  0.  (x 2  1)(x 2  1)  0  x   1,  i Note : Sum of roots = 0 and product of roots =–1. Continued product of the roots 60  Fourth roots of unity are vertices of a square which lies on coordinate axes. If z  r(cos   i sin  ) i.e., | z |  r and amp (z )   then continued product of roots of z 1 / n is n 1 2m     (n  1)  . n m 0  E3  r(cos   i sin  ) , where   z , if n is odd Thus continued product of roots of z 1 / n  r[cos{(n  1)   }  i sin{( n  1)   }]    z , if n is even ID m  z , if n is odd Similarly, the continued product of values of z m / n is   m  (-z) , if n is even Important Tips  If n be a positive integer then , (1  i)n  (1  i)n  2 2  If z is a complex number, then e z is periodic.  nth root of –1 are the solution of the equation z n  1  0 U  1 1 1 1 1 If x   2 cos  or x   2i sin  then x  cos   i sin  ,  cos   i sin  , x n  n  2 cos n  , x n  n  2i sin n . x x x x x 1 cos n. 4 D YG n z n  1  (z  1)(z   )(z   2 ).....( z   n 1 ), where   n th root of unity (n  2 ) / 2 z n  1  (z  1)(z  1)  (z 2  2 z cos r 1 2r  1), if n is even. n  (n  2) / 2   (2r  1)   2    1, if n is even.  z  2 z cos  n      r 0  n z 1   (n - 3)/2  2   (2r  1)     1, if n is odd.  z  2 z cos  (z  1) n    r 0    If x  cos   i sin , y  cos   i sin  , z  cos   i sin  and given, x  y  z  0, then ST  U  (i) 1 1 1    0 (ii) yz  zx  xy  0 (iii) x 2  y 2  z 2  0 (iv) x 3  y 3  z 3  3 xyz x y z then, putting, values if x, y, z in these results cos(    )  cos(   )  cos(   )  0 x  y  z  0  cos   cos   cos   0  sin   sin   sin   yz  zx  xy  0   sin(   )  sin(   )  sin(   )  0  cos 2  0 the summation consists 3 terms  sin 2  0,   x2  y2  z2  0     x 3  y 3  z 3  3 xyz , gives similarly cos 3  3 cos(     )   sin 3  3 sin(     ) 64 Complex Numbers If the condition given be x  y  z  xyz , then Example: 60 cos   cos(     ) etc. If the cube roots of unity be 1, ,  2 , then the roots of the equation (x  1)3  8  0 are [DCE 2000; IIT 1979; UPSEAT 1986] (a)  1,1  2,1  2 2 (b)  1,1  2,1  2 2 (c) 1,1,1  x  1  2,2,2 (d) None of these  x  1, 1  2 , 1  2 Solution: (c) (x  1)  8  x  1  (8) Example: 61  is an imaginary cube root of unity. If (1   )  (1   4 )m , then least positive integral value of m is 1/3 2 2 m 2 60 3 [IIT Screening 2004] (a) 6 (b) 5 (c) 4 (d) 3 The given equation reduces to ( )m  ( 2 )m   m  1  m  3. Example: 62 If  is the cube root of unity, then (3  5  3 2 ) 2  (3  3  5 2 ) 2 = (a) 4 Solution: (c) (b) 0 (c) – 4 (3  5  3 )  (3  3  5 ) 2 2 2 2   is equal to [IIT 1999] (d)  i 3 (c) i 3 334  1 3   3   i  2 2   365 U  1 3  Given equation is 4  5    i  2 2   365  1  i 3   i 3  4  5 334  3 365  4  5  3 2 1  2  1  2   2   Let  is an imaginary cube root of unity then the value of D YG Example: 64 2   1 i 3   3    2 2   (b)  1  i 3 (a) 1  i 3 Solution: (c) 2 334 2 2 (1     2  0,  3  1) 4   If i   1 , then 4  5   1  i 3  2 ID Example: 63 2 2 (d) None of these  (3  3  3  2 )  (3  3  3  2 ) 2 2 2 = 2   (2 )  4  4  4(1)   4 2 2(  1)( 2  1)  3(2  1)(2 2  1) .....  (n  1)(n   1)(n  2  1) is 2  n(n  1)  (a)   n  2  Solution: (a)  n(n  1)  (b)    2  n U   [Orissa JEE 2002] 2  n(n  1)  (c)   n  2  (d) None of these n (r  1)(r  1)(r 2  1) r 1 n n n (r  1)(r 2 3  r r 2  1)  r 1 (r  1)(r 2  r  1)   (r 3  r 2  r  r 2  r  1)   r 1 r 1 r 1 n (r 3 )   ST [MP PET 1986] (c) 1,1,  , (b) 1,1, i,i 2 Given equation x  1  0  (x  1)(x  1)  0  x  1 and x  1 4 2 2 2  n(n  1)  (1)     n.  2  r 1 The roots of the equation x 4  1  0 , are (a) 1,1, i,i Solution: (b) 2 2(  1)( 2  1)  3(2  1)(2 2  1) .....  (n  1)(n   1)(n  2  1)   Example: 65 [MP PET 1999] E3 Solution: (d) 2 2 (d) None of these  x  1,i 2.18 Shifting the Origin in Case of Complex Numbers. Let O be the origin and P be a point with affix z 0. Let a point Q Y Y Q has affix z with respect to the co-ordinate system passing through O. x When origin is shifted to the point P(z 0 ) then the new affix Z of the point Q with respect to new origin P is given by Z  z  z 0 i.e., to shift the origin at z 0 we should replace z by Z  z 0. O P (z0) M X X Complex Numbers 65 Example: 66 If z 1 , z 2 , z 3 are the vertices of an equilateral triangle with z 0 as its circumcentre then changing origin to z 0 , then (where z1 , z 2 , z 3 are new complex numbers of the vertices) (a) z12  z 22  z 32  0 (c) Both (a) and (b) (d) None of these In an equilateral triangle the circumcentre and the centroid are the same point. So, z0  z1  z 2  z 3  z 1  z 2  z 3  3z 0 3..... (i) 60 Solution: (a) (b) z1z 2  z 2 z 3  z3 z1  0 To shift the origin at z 0 , we have to replace z 1 , z 2 , z 3 and z 0 by z1  z0 , z 2  z0 , z 3  z0 and 0  z 0 then equation (i) becomes (z1  z0 )  (z 2  z0 )  (z 3  z0 )  3(0  z0 )  z1  z 2  z 3  0 On squaring z12  z 22  z 32  2(z1 z 2  z 2 z 3  z 3 z1 )  0 E3..... (ii) But triangle with vertices z1 , z 2 and z 3 is equilateral, then z12  z 22  z 32  z1 z 2  z 2 z 3  z 3 z1.....(iii) From (ii) and (iii) we get, 3 (z12  z 22  z 32 )  0. Therefore, z12  z 22  z 32  0. ID 2.19 Inverse Points. U (1) Inverse points with respect to a line : Two points P and Q are said to be the inverse points with respect to the line RS. If Q is the image of P in RS, i.e., if the line RS is the right bisector of PQ. D YG P R S Q (2) Inverse points with respect to a circle : If C is the centre of the circle and P,Q are the inverse points with respect to the circle then three points C,P,Q are collinear, and also CP. CQ C P Q ST U  r 2 , where r is the radius of the circle. Example: 67 z 1 , z 2 , are the inverse points with respect to the line z a  a z  b if (a) z1 a  z 2 a  b Solution: (b) (b) z 1 a  a z 2  b (c) z 1 a  a z 2  b Let RS be the line represented by the equation z a  a z  b (d) None of these.....(i) P  z1 Let P and Q are the inverse points with respect to the line RS. The point Q is the reflection (inverse) of the point P in the line RS if the line RS is the right bisector of PQ. Take any point z in the line RS, then lines joining z to P and z to Q are equal. R A(z ) Q  z2 S 66 Complex Numbers i.e., | z  z 1 | | z  z 2 | or | z  z 1 | 2 | z  z 2 | 2 i.e., (z  z 1 )(z  z 1 )  (z  z 2 )(z  z 2 )  z (z 2  z 1 )  z (z 2  z 1 )  (z 1 z 1  z 2 z 2 )  0.....(ii) Hence, equations (i) and (ii) are identical, therefore comparing coefficients, we get z 2  z1  z1a az 2 z a  az 2  b a b b     1 So that, z 2  z1 0 z1 z1  z 2 z 2 z 1 (z 2  z 1 ) z 2 (z 2  z 1 ) z 1 z 1  z 2 z 2 60 a (By ratio and proportion rule) Hence, z 1 a  a z 2  b = 0 or z 1 a  a z 2  b. Inverse of a point a with respect to the circle | z  c |  R (a and c are complex numbers, centre C and radius R) is the point c  (a) c  (b) c  R2 a c (c) c  R c a (d) None of these Let a' be the inverse point of a with respect to the circle | z  c |  R, then by definition the points c, a, ID Solution: (a) R2 a c R2 a c E3 Example: 68 a' are collinear. We have, arg(a'c)  arg(a  c)  arg(a  c) ( argz  argz)  arg(a'c)  arg(a  c)  0  arg{(a'c)(a  c )}  0 C By definition | a'c | | a  c |  R a' a |z – c| = R ( CP. CQ  r ) 2 (| z | | z |) 2 D YG  | a'c | | a  c |  R 2 U  (a'c)(a  c ) is purely real and positive.  | (a'c)(a  c )|  R 2  (a'c)(a  c )  R 2 { (a'c)(a  c) is purely real and positive}  a'  c  R2 R2. Therefore, the inverse point a' of a point a, a'  c . a c a c 2.20 Dot and Cross Product. Let z 1  a1  ib1  (a1 , b1 ) and z 2  a 2  ib 2  (a 2 , b 2 ) be two complex numbers. z 2  0 | z 2 | i  e z1  0 | z1 | Z2(a2, b 2) z 2 z 1 | z 2 | i z z | z |  e  2 12  2 e i  z 2 z 1 | z 1 || z 2 | e i z1 z1 | z1 | | z1 | | z1 | ST  U If  POQ   then from coni method  O  z 2 z 1 | z 1 || z 2 | (cos   i sin  )  Re( z 2 z 1 ) | z 1 || z 2 | cos .....(i) Z1(a1, b1 ) and Im(z 2 z 1 ) | z 1 || z 2 | sin .....(ii) The dot product z 1 and z 2 is defined by z 1 o z 2 | z 1 | | z 2 | cos   Re( z 1 z 2 )  a1 a 2  b1 b 2 (From(i)) Cross product of z 1 and z 2 is defined by z 1  z 2 | z 1 | | z 2 | sin   Im(z 1 z 2 )  a1 b 2  a 2 b1 (From(ii)) Hence, z 1 oz 2  a1 a 2  b1 b 2  Re( z 1 z 2 ) and z 1  z 2  a1 b 2  a 2 b1  Im(z 1 z 2 ) Important Tips  If z 1 and z 2 are perpendicular then z 1 o z 2  0  If z 1 and z 2 are parallel then z 1  z 2  0 Complex Numbers 67  Projection of z 2 on z 1  (z 1 o z 2 ) / | z 1 |  Projection of z 1 on z 2  (z 1 o z 2 ) / | z 2 |  Area of triangle if two sides represented by z 1 and z 2 is 1 | z1  z 2 | 2  Area of a parallelogram having sides z 1 and z 2 is | z 1  z 2 | Example: 69 If z 1  2  5i, z 2  3  i then projection of z 1 on z 2 is (a) 1/10 Solution: (b) (c) 7 / 10 (b) 1 / 10 Projection of z 1 on z 2  z1o z 2 | z2 |  a1 a 2  b 1 b 2 a 22 (d) None of these  b 22  1 10 ST U D YG U ID *** 60 1 | z1  z 2 | 2 Area of parallelogram if diagonals represents by z 1 and z 2 is. E3 