1. Mathematics

2. 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

3. Session Objectives

4. 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

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

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

7. 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.

8. 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

9. 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 ?

10. 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

11. 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

12. 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

13. 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

14. 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

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

16. 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 i = -i

17. 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

18. 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

19. 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

20. 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

21. Illustrative Problem The complex number which satisfies the equation
(a) 2 – i (b) –2 - i
(c) 2 + i (d) i

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

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

24. 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

25. Conjugate of a Complex Number

26. Properties of modulus
(Triangle inequality)

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

28. Conjugate Properties

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

30. 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)

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

32. Illustrative Problem
Solution

33. 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|

34. 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

35. 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

36. 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

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

38. Class Exercise

39. 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.

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

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

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

43. 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.

44. 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|

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

46. 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).

47. 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).

48. 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

49. Solution contd.. 1 + y2 + x2 + 2y = x2 + y2 – 2y + 1 4y = 0
y = 0 equation of real axis
Hence, answer is (a).

50. 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

51. 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).

52. 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

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

54. 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.

55. Thank you