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Complex Numbers

Complex numbers is an important topic of IIT JEE Mathematics as it fetches several questions in the exam. Locus based approach of complex numbers is one of the several mathematical tools which can be usefully employed in solving problems of coordinate geometry. This chapter discusses the operations of complex numbers and the application of properties such as multiplication of two complex numbers. The concept of rotation has been dealt with adequate number of tricks of different levels. This has been expressed in detail with ample problems and by illustrations. The application of properties of modulus and conjugate in solving problems has been dealt with in a unique way and discussed by illustrations in depth.

What exactly do we mean by complex numbers?

A complex number, generally represented by the number z is the number of the form z = a + ib, where a and b are real numbers and ‘i’ is the imaginary number iota satisfying the condition i2 = -1.

Example: 3 – 4i is a complex number where 3 is called the real part of the complex number while 4 is the imaginary part.

The chapter has been divided into several parts like:

Representation of complex number

Algebraic Operations of complex numbers

Geometrical representation of algebraic operations

Distance, Triangle Inequality

Conjugate and argument of complex numbers

Cube roots of Unity

Demoivre’s Theorem

Rotation

We shall give a brief outline of all the topics here. The heads have been discussed in detail with illustrative examples in the coming sections:

Every complex number can be classified as:

Two complex numbers say z1 and z2 are said to be equal if their corresponding real and imaginary parts are equal. For example:

If z1 = a + ib and z2 = c + id then z1 = z2 implies that a = c and b = d.

If we have a complex number z = a + ib, then its conjugate is denoted by or z* and is equal to a – ib. In fact, for any complex number z, its conjugate is given by

z* = Re(z) – Im(z).

View the video on complex numbers

The absolute value of a complex number z = a + ib is defined by |z| = √x2 + y2. The absolute value basically signifies the magnitude of the complex number.

Argument of a complex number is a quite tricky but an important concept of complex numbers. The argument of the complex number is the magnitude of the angle which it makes with the positive direction of x-axis. Given the complex number z = x + iy, the below list shows the value of argument according to the positive or negative real and imaginary parts:

Note:

If z is purely real positive complex number then amp (z) = 0.

If z is purely imaginary positive complex number then amp (z) = π/2.

If z is purely real negative complex number then amp (z) = π.

If z is purely imaginary negative complex number then amp (z) = -π/2.

In order to add two complex numbers, the corresponding real and imaginary parts get added to each other.

(a + bi) + (c + di) = (a + c) + (b + d)i

Example: (3 + 2i) + (1 + 7i) = (4 + 9i)

(B) Subtraction of Complex Numbers::

z1 - z2 = (x1 + iy1) - ( x2 + iy2) = (x1 - x2) + i(y1 - y2) ∈ C

Re(z1 - z2) = Re(z1) - Re(z2) and

Im(z1 - z2) = Im(z1) - Im(z2)

(C) Multiplication of Complex Numbers:

The multiplication of two complex numbers is defined by the formula:

(a + ib)(c + id) = (ac – bd) + i(ad + bc), using the value of i2 = -1.

(D) Division of Complex Numbers:

The division of two complex numbers is performed by first multiplying the term by the conjugate of denominator and then simplifying it further. For example: if we wish to perform the following division (a + ib)/(c + id), then we first multiply both the numerator and denominator by the (c – id) i.e. the conjugate of the denominator and then simplify.

Case-1:

Statement :

If n is any integer then

(i) (cos θ + i sin θ)n = cos nθ + i sin nθ

(ii) (cos θ1 + i sin θ1)(cos θ2 + i sin θ2)(cos θ3 + i sin θ2)(cos θ3 + i sin θ3) …… (cos θn + i sin θn) = cos (θ1 + θ2 + θ3 + . .  . . θn) + i sin (θ1 + θ2 + θ3 + . .  . . θn)

Case–2:

Statement :

If p, q ∈ Z and q ≠ 0 then



where k = 0, 1, 2, ….. , q - 1.

There are three cube roots of unity which are given as shown:

z3 – 1 = 0 gives z = (1)1/3

= (cos 0 + i sin 0)1/3

= (cos 2mπ + i sin 2mπ)1/3

= cos 2mπ/3 + i sin 2mπ/3 , m = 0 , 1, 2.

Now, for m = 0, z1 = cos 0 + i sin 0 = 1.

m = 1, z2 = cos 2π/3 + i sin 2π/3 = – 1/2 + √3i /2 = ω

m = 2, z3 = cos 4π/3 + i sin 4π/3 = – 1/2 - √3i /2 = ω2

Let us have a look at the types of questions asked in the exam from this topic:

Illustration 1: Let a and b be roots of the equation x2 + x + 1 = 0. Then find the equation whose roots are a19 and b7.

Solution: Given that x2 + x + 1 = 0.

This means that either x = ω or x = ω2.

Hence, a19 = ω19 and b7 = ω14 = ω2

Hence, the equation becomes x2 – (ω + ω2)x + ω ω2 = 0.

Hence the required equation is x2 + x + 1 = 0.

Illustration 2: Dividing f(z) by z - i, we obtain the remainder i and dividing it by z + i, we get remainder 1 + i. Find the remainder upon the division of f(z) by z2 + 1.

Solution: z – i = 0 gives z = i.

Remainder when f(z) is divided by (z – i) = f(i).

Similarly, the remainder when f(z) is divided by (z + i) = f(- i)   ….. (1)

According to question f(i) = i

and f( -i) = 1 + i.                                                                  ..... (2)

Since, z2 + 1 is a quadratic expression, therefore remainder when f(z) is divided by z2 + 1 will be in general a linear expression.

Let g(z) be the quotient and az + b the remainder when g(z) is divided by z2 + 1.

Then f(z) = g(z) (z2 + 1) + az + b                                          ..... (3)

So, f(i) = g(i) (i2 + 1) + ai + b = ai + b                                     .… (4)

and f(-i) = g(-i) (i2 + 1) – ai + b = -ai + b                                  .… (5)

From (1) and (4), we have b + ai = i                                        .… (6)

from (2) and (5) we have b – ai = 1 + i                                     …. (7)

Solving (6) and (7), we have b = ½ + i and a = i/2.

Hence, the required remainder  = az + b = ½ iz + ½ + i.

Illustration 3: Find all complex numbers z for which arg [(3z-6-3i)/(2z-8-6i)] = π/4 and |z-3+4i| = 3.

Solution: We have



= 

= 

Since arg(a + ib) = π/4, so tan π/4 = b/a which gives a = b

So, 6x2 + 6y2 – 36x – 24y + 66 = 12x – 12y -12

So, x2 + y2 – 8x – 2y + 13 = 0 . … … (1)

Again, |z – 3 + i| = 3 gives |x + iy - 3 + i| = 3

So, (x-3)2 + (y+1)2 = 9

This yields x2 + y2 - 6x + 2y +1 = 0 …. (2)

Subtracting (2) form (1), we have

-2x – 4y + 12 = 0

This gives x = – 2y + 6 … (3)

Putting the value of x in (2), we get

(-2y + 6)2 + y2 – 6 (-2y + 6) + 2y + 1 = 0

So, 5y2 – 10 y +1 = 0

So, y = (-10 ± 4√5) /10

= 1 ± 2/√5

So, x = – 2y + 6 = 4 ∓ (- 4)/√5

so, z = x + iy = 4 ∓ 4/√5 + i (1 ± 2/√5)

Q1. Any integral power of ‘i’ (iota) can be expressed as

(a) ± 1

(b) ± i

(c) ± 1 or ± i

(d) none of these

Q2. If z is purely real negative complex number then

(a) amp (z) = π

(b) amp (z) = – π

(c) amp (z) = π/2

(d) amp (z) = π/4

Q3. arg (z1z2) =

(a) arg(z1) + arg(z2) + 2π

(b) arg(z1) + arg(z2)

(c) arg(z1) . arg(z2)

(d) arg(z1) + arg(z2) + 2kπ, k ∈ I

Q4. 1 + ωr + ω2r =

(a) 0

(b) 3

(c) 0 if r is not a multiple of 3 and 3 if r is a multiple of 3.

(d) 0

Q5. If ω1 = ω2 are the complex slopes of two lines, then

(a) If ω1 = ω2 then the lines are parallel.

(b) If ω1 + ω2 = 0 then the lines are parallel.

(c) If ω1 = ω2 then the lines are not parallel.

(d) If ω1 = ω2 then the lines are perpendicular.

Q1.

Q2.

Q3.

Q4.

Q5.

(c)

(a)

(d)

(c)

(a)

Related resources:

Look into the Previous Year Papers with Solutions to get a hint of the kinds of questions asked in the exam.

You can get the knowledge of Recommended Books of Mathematics here.

To read more, Buy study materials of Complex Numbers comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.

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