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 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
Euler Leonhard ( 1707-1783) Complex Numbers Intro Solve x2 + 1 = 0 D = –4(<0) No real roots “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
Mathematical notation im(z)=b re(z)= a Re(z) = 4, Im(z) = Complex Numbers - Definition z = a + i b a,bR 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 ?
Find x and y if 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
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
If one root of the equation is 2 – i then the other root is Illustrative Problem (a) 2 + i (b) 2 – i (c) i (d) -i Solution: (2 – i) + = -2i +2 = -2i +2-2 + i = -i
Im (z) Y P(z) b O X a Re (z) Geometrical Representation Representation of complex numbers as points on x-y plane is called Argand Diagram. Representaion of z = a + ib
Im (z) Y P(z) |z| b O X a Re (z) Modulus and Argument Modulus of z = a + i b Argument of z Arg(z) = Amp(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
Step1: Illustrative Problem – Principal argument Solution Step2: 3rd quadrant ( -,-) Step3:
Conjugate of z is Y P (z) b a - X -b Conjugate of a Complex Number For z = a + ib, Image of z on x – axis z lies on x -axis
Properties of modulus (Triangle inequality)
Properties of Argument Arg(purely real) = 0 or or 2n and vice versa and vice versa Arg(purely imaginary) =
(c)(d) Illustrative Problem (a) (b) 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
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
As argument is complex number is purely imaginary Solution Let z = x + iy, then x2 + y2 = 25, circle with center (0,0) and radius 5
Locus of z -7 -1 Illustrative Problem (a) 2 (b) 6 (c) 0 (d) -6 Solution As | z+4| 3 Least value = ?
The modulus and principal argumentof –1 – i are respectively The complex number lies in the third quadrant andprincipal argument q satisfying is given by q – p. Class Exercise - 1 Solution:
arg(z) = is the principal argument. The modulus is = Solution contd.. Hence, answer is (d).
If , then x2 + y2 is equal to (d) None of these Class Exercise - 2
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.
For x1, x2 , y1, y2ÎR, if 0< x1 < x2, y1 = y2andz1 = x1 + iy1, z2 = x2 + iy2 and z3 = , thenz1, z2 and z3 satisfy (a) |z1| < |z3| < |z2| (b) |z1| > |z3| > |z2| (c) |z1| < |z2| < |z3| (d) |z1| = |z2| = |z3| Class Exercise - 4
(As arithmetic mean of numbers) Solution y1 = y2 = y (Say) = |z1| < |z3| < |z2| Hence, answer is (a).
If then value of z13 + z23 – 3z1z2 is (a) 1 (b) –1 (c) 3 (d) –3 Class Exercise - 5 Solution: We find z1 + z2 = –1. Therefore, Hence, answer is (b).
Sum of the roots = = –2i + 2 Class Exercise - 6 If one root of the equationix2 – 2(1 + i) x + (2 – i) = 0 is2 – i, then the other root is(a) 2 + i (b) 2 – i (c) i (d) –i Solution: One root is 2 – i. Another root = –2i + 2 – (2 – i) = –2i + 2 – 2 + i = –i Hence, answer is (d).
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 Class Exercise - 7 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).
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) If we put z = x + iy, we get By simplifying, we get Class Exercise - 8 Solution: