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 Complex number - Definition Equality of complex number Algebra of complex number Geometrical representation Conjugate of complex number Properties of modules and arguments Equation involving variables and locus

Complex Numbers Intro Solve x2 + 1 = 0 D = –4(<0)  No real roots Euler Leonhard ( 1707-1783) “i” is the first letter of the latin word ‘imaginarius’

Integral powers of i(iota) Evaluate: Solution Ans: 343i

Illustrative Problem If p,q,r, s are four consecutive integers, then ip + iq + ir + is = 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 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 Note by previous question: u27 = 0

Complex Numbers - Definition z = a + i b a,bR re(z)= a im(z)=b Mathematical notation Re(z) = 4, Im(z) = If a = 0 ?  z is purely real If b = 0 ?  z is purely imaginary z is purely real as well as purely imaginary If a = 0, b = 0 ?

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

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

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): 5) Distributivity: z1(z2 + z3) = z1z2 + z1z3 (z1 + z2)z3 = z1z3 + z2z3

Algebra of Complex Numbers - Division z1 = a1 + ib1, z2 = a2 + ib2 then

Illustrative Problem If one root of the equation is 2 – i then the other root is (a) 2 + i (b) 2 – i (c) i (d) -i Solution:  (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) O X Y P(z) b a Re (z) Representaion of z = a + ib

Modulus and Argument Modulus of z = a + i b Argument of z Y Arg(z) = Amp(z) O X Y Im (z) Re (z) a b P(z) |z| Argument  (-, ] is called principal value of argument

Principal Value of Argument Argument  (-, ] is called principal value of argument 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 (a) 2 – i (b) –2 - i (c) 2 + i (d) -2 + i

Solution Cont. Shortcut. To cancel i, z must have –i and real part must be negative only possible option is b)

Illustrative Problem – Principal argument Step1: Solution Step2: 3rd quadrant ( -,-) Step3:

Conjugate of a Complex Number For z = a + ib, Conjugate of z is Image of z on x – axis b a  P (z) Y X -b - z lies on x -axis

Conjugate of a Complex Number

Properties of modulus (Triangle inequality)

Properties of Argument Arg(purely real) = 0 or  or 2n and vice versa and vice versa Arg(purely imaginary) =

Conjugate Properties

Illustrative Problem (a) (b) (c) (d) Solution:

Square Root of a Complex Number x2 – y2 + 2ixy = a + ib (Squaring) Find x2 , take positive value of x 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 x2 – y2 = a 2xy = b Other root will be – (x+iy)

Illustrative Problem Solution x2 – y2 + 2ixy = 8 –15i x2 – y2 = 8 2xy = -15 x,y are of opposite sign

Illustrative Problem Solution

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 defines shapes as circle , parabola Equation in complex variable (z) defines shapes as circle , parabola etc. Distance between P(x1,y1) and Q(x2,y2) = PQ Distance between P(z1) and Q(z2) = |z1-z2|

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

Illustrative Problem If , then the locus of z is given by Circle with centre on y-axis and radius 5 Circle with centre at the origin and radius 5 A straight line None of these

Solution Let z = x + iy, then As argument is complex number is purely imaginary  x2 + y2 = 25, circle with center (0,0) and radius 5

Illustrative Problem Solution Locus of z As | z+4|  3 -7 -1 (a) 2 (b) 6 (c) 0 (d) -6 Solution -7 -1 Locus of z As | z+4|  3 Least value = ?

Class Exercise

Class Exercise - 1 The modulus and principal argument of –1 – i are respectively Solution: The complex number lies in the third quadrant and principal argument q satisfying is given by q – p.

Solution contd.. arg(z) = is the principal argument. The modulus is = Hence, answer is (d).

Class Exercise - 2 If , then x2 + y2 is equal to (d) None of these

Solution Taking modulus of both sides, Hence, answer is (a).

Class Exercise - 3 If |z – 4| > |z – 2|, then (a) Re z < 3 (b) Re z < 2 (c) Re z > 2 (d) Re z > 3 Solution: If z = x + iy, then |z – 4| > |z – 2| Hence, answer is (a). |x – 4| > |x – 2| x < 3 satisfies the above inequality.

Class Exercise - 4 For x1, x2 , y1, y2 ÎR, if 0< x1 < x2, y1 = y2and z1 = x1 + iy1, z2 = x2 + iy2 and z3 = , then z1, z2 and z3 satisfy (a) |z1| < |z3| < |z2| (b) |z1| > |z3| > |z2| (c) |z1| < |z2| < |z3| (d) |z1| = |z2| = |z3|

Solution y1 = y2 = y (Say) = |z1| < |z3| < |z2| Hence, answer is (a). (As arithmetic mean of numbers)

Class Exercise - 5 If then value of z13 + z23 – 3z1z2 is (a) 1 (b) –1 (c) 3 (d) –3 Solution: We find z1 + z2 = –1. Therefore, 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 = = –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 If z = x + iy and w = , then |w| = 1, in the complex plane z lies on unit circle z lies on imaginary axis z lies on real axis None of these Solution: Putting z = x + iy, we get

Solution contd.. 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 lies on an arc of a circle (b) line joining (1, 0), (–1, 0) (c) pair of lines (d) line joining (0, i) , (0, –i) Solution: If we put z = x + iy, we get By simplifying, we get

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 = 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.. Case 2: x = , then – y2 + = 0 i.e. two solutions given by So in all four solutions. Hence, answer is (c).

Class Exercise - 10 Find the square root of –5 + 12i. Solution: Squaring, x2 – y2 + 2ixy = –5 + 12i x2 – y2 = –5 2xy = 12 xy = 6, Both x and y are of same sign. 2x2 = 8 Þ x = ±2, y = ±3 2 + 3i and –2 – 3i are the values.

Thank you