Lecture 19: Complex Numbers 1

Mathematics Session opener What is loge(-1) ? Not Defined It‟s a complex number loge(-1) is defined and is complex no. One of its value is i Session Objectives Session Objective 1. Complex number - Definition 2. Equality of complex number 3. Algebra of complex number 4. Geometrical representation 5. Conjugate of complex number 6. Properties of modules and arguments 7. Equation involving variables and locus Complex Numbers Intro Solve x2 + 1 = 0 D = –4(<0)  No real roots x   -1 -1  i (known as iota ) “i” is the first letter of the latin word „imaginarius‟ Euler Leonhard ( 1707-1783) Integral powers of i(iota) i0  1(as usual) i1  i i2   1 i3  i2 .i   i i4  i3 .i   i.i  1 1 i i1   2   i i i 1 i2  2   1 i 1 1 i3  3   i i i 1 i4  4  1 i Evaluate: 3  17  2 3  i       i    Solution 3 3 8  8 3  16 i .i   i   i  8i    i3   i   Ans: 343i Illustrative Problem If p,q,r, s are four consecutive integers, then ip + iq + ir + is = a)1 b) 2 c) 4 d) None of these Solution: Note q = p + 1, r = p + 2, s = p + 3 Given expression = ip(1 + i + i2 + i3) = ip(1 + i –1 – i) = 0 Remember this. Illustrative Problem If un+1 = i un + 1, where u1 = i + 1, then u27 is a) i b) 1 c) i + 1 d) 0 Solution: u2 = iu1 + 1 = i(i+1) +1 = i2 + i + 1 u3 = iu2 + 1 = i(i2+i+1) +1 = i3 + i2 + i + 1 Hence un = in + in-1 + ….. + i + 1 i28  1 27 26 u27  i  i  .....  i  1  0 i 1 Note by previous question: u27 = 0 Complex Numbers - Definition z=a+ib Mathematical notation re(z) =a z  4  5  4  i 5 If a = 0 ? a,bR im(z) =b Re(z) = 4, Im(z) =  5  z is purely real If b = 0 ?  z is purely imaginary If a = 0, b = 0 ? z is purely real as well as purely imaginary Equality of Complex Numbers If z1 = a1 + ib1 and z2 = a2 + ib2 z1 = z2 if a1 = a2 and b1 = b2 Is 4 + 2i = 2 + i ? No  One of them must be greater than the other?? Order / Inequality (>, <, , ) is not defined for complex numbers Find x and y if 3 5 x  5  i2 5y  2 Illustrative Problem Find x and y if (2x – 3iy)(-2+i)2 = 5(1-i) Hint: simplify and compare real and imaginary parts Solution: (2x – 3iy)(4+i2-4i) = 5 -5i (2x – 3iy)(3 – 4i) = 5 –5i (6x – 12y – i(8x + 9y)) = 5 – 5i 6x – 12y = 5, 8x + 9y = 5 7 1 x ,y  10 15 Algebra of Complex Numbers – Addition (I) Addition of complex numbers z1 = a1 + ib1, z2 = a2 + ib2 then z1 + z2 = a1 + a2 + i(b1 + b2) Properties: 1) Closure: z1 + z2 is a complex number 2) Commutative: z1 + z2 = z2 + z1 3) Associative: z1 + (z2 + z3) = (z1 + z2) + z3 4) Additive identity 0: z + 0 = 0 + z = z 5) Additive inverse -z: z + (-z) = (-z) + z = 0 Algebra of Complex Numbers - Subtraction (II) Subtraction of complex numbers z1 = a1 + ib1, z2 = a2 + ib2 then z1 - z2 = a1 - a2 + i(b1 - b2) Properties: 1) Closure: z1 - z2 is a complex number Algebra of Complex Numbers - Multiplication z1 = a1 + ib1, z2 = a2 + ib2 then z1 . z2 = a1a2 – b1b2 + i(a1b2 + a2b1) Properties: 1) Closure: z1.z2 is a complex number 2) Commutative: z1.z2 = z2.z1 3) Multiplicative identity 1: z.1 = 1.z = z 4) Multiplicative inverse of z = a + ib (0): z 1  1 1 a  ib a  ib  .  2 (remember) 2 a  ib a  ib a  ib a  b 5) Distributivity: z1(z2 + z3) = z1z2 + z1z3 (z1 + z2)z3 = z1z3 + z2z3 Algebra of Complex Numbers - Division z1 = a1 + ib1, z2 = a2 + ib2 then z1 a  ib1 a  ib1 a2  ib2  1  1 . z2 a2  ib2 a2  ib2 a2  ib2  a1a2  b1b2  i(a2b1  a1b2 ) a22  b22 Illustrative Problem If one root of the equation ix2  2(i  1)x  (2  i)  0 is 2 – i then the other root is (a) 2 + i (c) i (b) 2 – i (d) -i Solution: 2(1  i) Sum of the roots is   2i  2 i  (2 – i) +  = -2i +2   = -2i +2-2 + i = -i Geometrical Representation Representation of complex numbers as points on x-y plane is called Argand Diagram. Im (z) Y P(z) b O Representaion of z = a + ib a X Re (z) Modulus and Argument Modulus of z = a + i b OP  z  a2  b2 Argument of z Im (z) Y Arg(z) = Amp(z)   tan1 b a Argument  (-, ] is called principal value of argument P(z) |z| b  O a X Re (z) Principal Value of Argument Argument  (-, ] is called principal value of argument Step1 : Find   tan1 b a   for   0,   2 Step2: Identify in which quadrant (a,b) lies -1+2i ( -,+) ( +,+) ( -,-) ( +,-) -1-2i 1 -2i 1+2i Principal Value of Argument Step3: Use the adjoining diagram to find out the principal value of argument        Based on value of  and quadrant from step 1 and step 2 Illustrative Problem The complex number which satisfies the equation z  2 | z  1 | i  0 is (a) 2 – i (b) –2 - i (c) 2 + i (d) -2 + i Let z  x  i y  z  2 | z  1 | i  0  x  i y  2 (x  1)2  y2  i  0  x  2 (x  1)2  y2  i(1  y)  0 2 2  x  2 (x  1)  y 0 and y  1  0 Solution Cont.  x  2 (x  1)2  y2  0  y1  x  2 (x  1)2  1  0  x2  2(x2  2x  2)  (x  2)2  0  x  2 and y  1  0 Illustrative Problem – Principal argument The principal value of argument in  2  2 3 i a)  3 b)   3 c) Solution   4 3 d) 2 3 Step1:     tan1 2 3   2 3 Step2: 3rd quadrant ( -,-) Step3:     2 3 Conjugate of a Complex Number For z = a + ib,  Conjugate of z is z  a  ib Image of z on x – axis For z  _ 3 i _ , z  ? 2 Y _ z  3 i 2 If z  z  a  ib  a  ib  b  0, z is puerly real _ What if z   z ? z lies on x -axis P (z)  - a b -b Q(z) X Conjugate of a Complex Number a  Re(z)  z  z 2 b  im(z)  z z 2i z  a  ib  a2  b2  z A rg(z)  A rg(a  ib)   tan1 b   A rg(z) a Properties of modulus z1 .z2  z1 . z2 n z z n zz z.z  z 2 Pr oof :z  a  ib, zz  (a  ib)(a  ib)  a2  b2  z 2 z1 z1  z2 z2 z1  z2  z1  z2 z1  z2  z1  z2 z1  z2  z1  z2 z1  z2  z1  z2 2 2  2 z1  z2  z1  z2  2 z1  z2 2  (Triangle inequality) Properties of Argument Arg  z1.z2   Arg  z1   Arg  z2   z1  Arg    Arg  z1   Arg  z2   z2  Arg  z1.z2....zn   Arg  z1   Arg  z2   ....  Arg(zn )  1 Arg z   Arg  z  , Arg     Arg  z  z Arg(purely real) = 0 or  or 2n and vice versa    or  or 2n  1   Arg(purely imaginary) = 2 2 2 and vice versa Conjugate Properties zz Pr oof :z  a  ib, z  a  ib  a  ib  z z1  z2  z1  z2 z1.z2  z1 .z2  z1   z1       provided z2  0  z2   z2  Illustrative Problem 1 Conjugate of is 2i (a) (c) 2i 5 1 2i (b) (d) 5 2i 2i 5 Solution: 1 2  i 2  i z     2  i 2  i 5 2i 2i  z ( )  5 5 Square Root of a Complex Number Find a  ib Let x  iy  a  ib x2 – y2 + 2ixy = a + ib x2 – y2 = a 2 2  x y  x (Squaring) • Find x2 , take positive value of x 2xy = b 2 y 2  2  4x2y2 Other root will be – (x+iy) • Find y2, take value of y which satisfies 2xy = b Note if b > 0 x,y are of same sign, else if b < 0 x,y are of opposite sign • Square root will be x +iy Illustrative Problem Find 8  15i Solution Let x  iy  a  ib x2 – y2 + 2ixy = 8 –15i x2 – y2 = 8 2xy = -15 x,y are of opposite sign  x2  y2  64  225  17 2  2x  8  17  x   5 2 Illustrative Problem Find 8  15i x Solution 2 25 3  y2  17  y   2 2 2 15 5 3 As xy   , for x  y2 2 2 For x  5 5 ,  One of the square root  5 i 2 5 3 Other square root   i 2 2 3 2 Equation involving complex variables and locus Cartesian System – 2D Argand Diagram Point ( x,y) Complex No.( z) Locus of point Locus of complex no. ( point in argand diagram) Equation in x,y Equation in complex variable (z) defines defines shapes as shapes as circle , parabola etc. circle , parabola Distance between Distance between P(z1) and Q(z2) = |z1-z2| P(x1,y1) and Q(x2,y2) = PQ Illustrative Problem If z is a complex number then |z+1| = 2|z-1| represents (a) Circle (b) Hyperbola (c) Ellipse (d) Straight Line Solution Let z  x  iy then (x  1)2  y2  2 (x  1)2  y2  3x2  3y2  10x  3  0 Illustrative Problem  z 5   If arg   , then the locus of z   z 5 2 is given by a) Circle with centre on y-axis and radius 5 b) Circle with centre at the origin and radius 5 c) A straight line d) None of these Solution Let z = x + iy, then  x  iy  5   arg     x  iy  5  2  x2  y2  25  i10y    arg  2 2   2 x  5  y      As argument is complex number 2 is purely imaginary  x2 + y2 = 25, circle with center (0,0) and radius 5 Illustrative Problem If z is a complex no. such that | z  4 | 3 the maximum value of | z  1| is (a) 2 (b) 6 (c) 0 (d) -6 Solution | z  1  3| |z  1 |  3 Locus of z  |z  1 |  3  | z  1  3|  |z  1 |  3  3 As | z+4|  3  |z  1 |  6 Least value = ? -7 -1 Class Exercise Class Exercise - 1 The modulus and principal argument of –1 – i 3 are respectively  (a) 4 and 3 (c) 4 and –  3 2 (b) 2 and 3 2 (d) 2 and – 3 Solution: The complex number lies in the third quadrant and principal argument  satisfying      is given by  – . Solution contd.. arg(z) =   tan1  3   1 3  2 is the principal argument.      3 3 2 The modulus is = 1 Hence, answer is (d).  2 3  2 Class Exercise - 2 a  If 2  1 2  x  iy, then x2 + y2 is equal to 2a  i a  (a)  2 1 4 4a  1 a  (c)  1 2  1 4 4a2  1 2 2 a  (b) 2 2 4a  1 (d) None of these Solution a 2  2 1  x  iy 2a  i Taking modulus of both sides, a 2  1 2  x  iy 2a  i  x y 2 a 2 2  a 2  1 2 4a2  1  1 4 4a2  1 Hence, answer is (a).  x2  y2 Class Exercise - 3 If |z – 4| > |z – 2|, then (a) Re z < 3 (c) Re z > 2 (b) Re z < 2 (d) Re z > 3 Solution: If z = x + iy, then |z – 4| > |z – 2|  x  4   y2  2  x  2  y2 2 |x – 4| > |x – 2| x < 3 satisfies the above inequality. Hence, answer is (a). Class Exercise - 4 For x1, x2 , y1, y2 R, if 0< x1 < x2, y1 = y2and z1 = x1 + iy1, z2 = x2 + iy2 z1  z2 and z3 = , then 2 z1, z2 and z3 satisfy (a) |z1| < |z3| < |z2| (c) |z1| < |z2| < |z3| (b) |z1| > |z3| > |z2| (d) |z1| = |z2| = |z3| Solution y1 = y2 = y (Say) z1  z2  z3  z3  x1  x12  y2 2 x2  y 2 x1  x2 2 iy = |z1| < |z3| < |z2| 2 Hence, answer is (a).  x1  x2  2  y   2   x1  x2 2  x2 (As arithmetic mean of numbers) Class Exercise - 5 If z1  1  3i 1  3i and z2  , 2 2 then value of z13 + z23 – 3z1z2 is (a) 1 (b) –1 (c) 3 (d) –3 Solution: We find z1 + z2 = –1. Therefore, z13  z23  3 z1z2  z13  z23  3z1z2 (z1  z2 ) (z1  z2 )3  (1)3   1. Hence, answer is (b). Class Exercise - 6 If one root of the equation ix2 – 2(1 + i) x + (2 – i) = 0 is 2 – i, then the other root is (a) 2 + i (b) 2 – i (c) i (d) –i Solution: Sum of the roots = 2 1  i i  2i 1  i = –2i + 2 One root is 2 – i. Another root = –2i + 2 – (2 – i) = –2i + 2 – 2 + i = –i Hence, answer is (d). Class Exercise - 7 1  iz If z = x + iy and w = , then |w| = 1, in the complex plane z  i (a) z lies (b) z lies (c) z lies (d) None on unit circle on imaginary axis on real axis of these Solution: w 1  1  iz 1 zi  1  iz  z  i Putting z = x + iy, we get 1i  x  iy   x  iy  i 1  y  ix  x  i (y  1) Solution contd.. 1  y 2  x2  x2  (y  1)2 1 + y2 + x2 + 2y = x2 + y2 – 2y + 1 4y = 0 y = 0 equation of real axis Hence, answer is (a). Class Exercise - 8 The points of z satisfying arg  z  1      lies on (a) an arc of a circle (c) pair of lines  z  1 4 (b) line joining (1, 0), (–1, 0) (d) line joining (0, i) , (0, –i) Solution: z  1  x  1  iy  If we put z = x + iy, we get z  1  x  1  iy z 1  By simplifying, we get z 1   x 2  1  y 2  i  2y   x  12  y 2 Solution contd.. Equation of a circle. Note: But all the points put together would form only a part of the circle. Hence, answer is (a). Class Exercise - 9 The number of solutions of Z2 + 3 z = 0 is (a) 2 (b) 3 (c) 4 (d) 5 Solution: Let z = x + iy (x + iy)2 + 3(x – iy) = 0 x2 – y2 + 2ixy + 3x –3iy = 0 x2 – y2 + 3x + i(2xy – 3y) = 0 x2 – y2 + 3x = 0, 2xy – 3y = 0 Consider y(2x – 3) = 0 Case 1: y = 0, then x2 + 3x = 0, i.e. x = 0 or –3 i.e. two solutions given by 0, –3 Solution contd.. 9 9 3 2 Case 2: x = , then – y + =0 4 2 2 i.e. two solutions given by 3  3 3 i 2 So in all four solutions. Hence, answer is (c). Class Exercise - 10 Find the square root of –5 + 12i. Solution: Let x  iy  5  12i Squaring, x2 – y2 + 2ixy = –5 + 12i x2 – y2 = –5 2xy = 12 xy = 6, Both x and y are of same sign. x y 2 2  x 2 y 2  2  4x2 y2  2x2 = 8  x = ±2, y = ±3 2 + 3i and –2 – 3i are the values. 25  144  13 Thank you