JEE Main Study Notes for Complex Numbers: Important Concepts, Tricks to Solve Questions & Examples

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JEE Main Study Notes for Complex Number include properties, square roots, cube roots, conjugate, complex number polar form, Euler’s form, logarithm, etc. Complex Number is a very important topic in JEE Main Mathematics Syllabus. The form x+iy is defined as a complex number, where x and y are real numbers and i = √-1. For example, 2+3i, -3+i√2.

JEE Main Mathematics - Complex Number section has a weightage of 2-3% in JEE Main 2022.
JEE Main Question Paper has questions of approximately 8 marks from the topic complex number.

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What are Complex Numbers?

When x, y ∈ R, an ordered pair (x, y) = x + iy is known as a complex number. It is represented as z. Here, x is the real part of Re(z) and y is the imaginary part or Im (z) of the complex number.

(i) Suppose Re(z) = x = 0, it is known as a purely imaginary number

(ii) Suppose Im(z) = y = 0, z is known as a purely real number.

Note: The set of all probably ordered pairs is known as a complex number set and is expressed as C.

Integral Powers of Iota

i (greek letter iota) represents the positive square root of –1, so, It is called an imaginary unit. We have

Thus for any integer k,

i4k = 1, i4k+1 = i, i4k+2 = –1, i4k+3 = –i.

That is if the power of i is m, then divide m by 4 and find the remainder.

If the remainder is zero, then im = 1

If the remainder is one, then im = i

If the remainder is two, then im = –1

If the remainder is three, then im = –i

The sum of four consecutive powers of i is zero, for example, i12 + i13 + i14 + i15 = 0

For any two real numbers a and b, is true only if at least one number is non-negative or zero. If both a and b are negative then

In fact if a > 0 and b > 0 then

Must Read:

Algebraic Operations with Complex Numbers

Addition: (a + ib) + (c + id) = (a + c) + i(b + d)
Subtraction: (a + ib) – (c + id) = (ac) + i(b – d)
Multiplication: (a + ib) (c + id) = (ac – bd) + i(ad + bc)
Reciprocal: If a minimum of one, a or b is non-zero, then the reciprocal of a + bi is represented as 1/(a+ib) = (a – ib)/[(a+ib) (a−ib)] = a/[a2+ b2] – i[b/(a2+ b2)]
Quotient: If a minimum of one, c or d is non-zero, then the quotient of a + bi and c + di is represented as [(a+bi)/(c+di)] = [(a+ib) (c−id)]/[(c+id) (c−id)] = [(ac + bd) + i(bc – ad)]/[a2+d2] = [ac + bd]/[c2+d2] + i[bc−ad]/[c2+d2]

Conjugate of Complex Numbers

Consider = z = a + ib as a complex number. The conjugate of z, shown by z¯ to be the complex number a – ib. that is, if z = a + ib, then z¯ = a – ib.

Properties of Conjugate of Complex Numbers

z1=z2⇔z‾1=z‾2
(zˉ)‾=z
z+z‾=2 Re(z)
z−z‾=2i Im (z)
z=z‾⇔z is fully real
z+z‾=0⇔z is fully imaginary.
zz‾=[Re (z)]2+[Im(z)]2
z1+z2‾=z‾1+z‾2
z1−z2‾=z‾1−z‾2
z1z2‾=z‾1z‾2
(z1z2)‾=z‾1z2, if z2 ≠ 0
If P(z) = a0 + a1 z + a2 z2 + …. + an zn

Where a0, a1, ….. an and z are complex numbers

then, P(z)‾=a‾0+a‾1(z‾)+a‾2(z‾)2+….+a‾n(z‾)n = P‾(z‾)

Where P‾(z)=a‾0+a‾1z+a‾2z2+….+a‾nzn

If R(z) = P(z)Q(z), where P (z) and Q (z) are polynomials in z, and Q(z) ≠ 0, then, R (z)‾=P‾(z‾)Q‾(z‾)

Check:

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Modulus of Complex Numbers

Consider z = a + ib to be a complex number. The modulus or the absolute value of z is the real number √(a2 + b2) and is represented by |z|.

Note that |z| > 0 ∀ z ∈ C

Properties of Modulus

When z is a complex number,

(i) |z| = 0 ⇔ z = 0

(ii) |z| = |z¯| = |-z| = |-z¯|

(iii) – |z| ≤ Re (z) ≤ |z|

(iv) – |z| ≤ Im(z) ≤ |z|

(v) z z¯ = |z|2

Suppose z1, z2 are two complex numbers, then,

(vi) |z1 z2| = |z1|.|z2|

(vii) ∣z1/z2∣ = ∣z1/z2∣, if z2 ≠ 0

(viii) |z1 + z2|2 = |z1|2 + |z2|2 + z¯1 z2 + z1 z–2 = |z1|2 + |z2|2 + 2Re (z1 z¯2)

(ix) |z1+z2|2 + |z1|2 – |z2|2 – z¯1 z2 – z1 z¯2 = |z1|2 + |z2|2 – 2Re (z1 z¯2)

(x) |z1+z2|2 + |z1 – z2|2 = 2(|z1|2 + |z2|2)

(xi) Suppose a and b are real numbers and z1, z2 are complex numbers, then, |az1 + bz2 |2 + |bz1 – az2 |2 = (a2 + b2) (|z1|2+ |z2|2)

(xii) Suppose z1, z2 ≠ 0, then, |z1 + z2|2 = |z1|2 + |z2|2 ⇔z1 z2 is purely imaginary.

(xiii) Suppose z1 and z2 are two complex numbers, then |z1 + z2| < |z1| + |z2|. The equality applies only when z1 z¯2 ≥ 0. This is Triangle Inequality.

Generally, |z1 + z2+…+zn| < |z1| + |z2| +…..+ |zn| and the equality sign is valid only when the ratio of any two non-zero terms is positive.

(xiv) |z1 – z2| ≤ |z1| + |z2| (xv) ||z1| – |z2|| ≤ |z1| + |z2|

(xvi) |z1 – z2| ≥ ||z1| – |z2||

(xvii) If a1, a2, a3, are four complex numbers, then, |z – a1| + |z – a2| + |z – a3| + |z – a4| > max

{|a1−al|+|am−an| : l, m, n are different integers lying in{2, 3, 4}and m

Square Roots of Complex Number

Let z = x + iy and let the square root of z be the complex number a + ib. Then

Equating real and imaginary parts, we get …(1)

and y = 2ab …(2)

Now, …(3)

Solving the equations (1) and (3), we get ;

From (2), we can determine the sign of ab. If ab > 0, then a and b will have the same sign.Thus

abc

If ab < 0, then

abc

Thus square roots of z = a + ib are :

abc for b > 0

and abc for b < 0

For Example:

Square root of i is , as x = 0, y = 1 > 0 and | i | = 1
Square root of -i is , as x = 0, y = -1 < 0 and | i | = 1
Square root of -3 + 4i is ,
or , As x = –3 and y = 4 > 0.
The square root of isor ; As

Check:

Modulus and Argument of Complex Numbers

Suppose z = x + iy = (x, y) for all x, y ∈ R and i = −1

abc

The length OP is known as the modulus of the complex number z and is represented by |z|,

i.e. OP = r = |z| = (x2+y2)

and if (x, y) ≠ (0, 0), then θ is known as the argument or amplitude of z,

That is, θ = tan⁡−1(yx)

[angle made by OP with positive X-axis]

or arg (z) = tan⁡−1(y/x)

Also, the argument of a complex number is not unique. This is because supposed θ is a value of the argument, also it is in 2nπ + θ, where n ∈ 1.

Generally, only that value is taken for which,

0 < θ < 2π. Any two arguments of a complex number differ by 2nπ. The argument of z is θ, π – θ, π + θ and 2π – θ as the point z lies in I, II, III and IV quadrants respectively, where θ = tan⁡−1∣yx∣.

Principal Value of the Argument

The value θ of the argument that fulfills the inequality, −π<θ≤π is known as the principal value of the argument.

If x = x + iy = ( x, y), ∀, x, y ∈ R and i= root of −1, then

Arg(z) = tan⁡−1(y/x) always provides the principal value. It varies according to the quadrant in which the point (x, y) lies.

(i) (x, y) ∈ first quadrant x > 0, y > 0.

The principal value of arg (z) = θ=tan⁡−1(y/x)

This is an acute angle and positive.

(ii) (x, y) ∈ second quadrant x < 0, y > 0.

The principal value of arg (z) = θ = π−tan⁡−1(y∣x∣)

This is an obtuse angle and positive.

abc

(iii) (x, y) ∈ third quadrant x < 0, y < 0.

The principal value of arg (z) = θ = −π+tan⁡−1(y/x)

abc

It is an obtuse angle and negative. (iv) (x, y) ∈ fourth quadrant x > 0, y < 0.

The principal value of arg (z) = θ = −tan⁡−1(∣y∣x)

abc

This is an acute angle and negative.

Also Check:

Polar Form of a Complex Number

Here, z = x + iy

=x2+y2[2x2+y2+ixx2+y2]

= |z| [cosƟ + i sinƟ]

where |z| is the modulus of the complex number, that is, the distance of z from the origin, and Ɵ is the amplitude or argument of the complex number.

Here, the principal value of Ɵ must be taken. The general values of argument z = r[cos(2nπ + Ɵ)] (where n is an integer). It is a polar form of a complex number.

Euler’s form of a Complex Number

eiƟ = cos Ɵ + i sin Ɵ

This representation makes learning complex numbers and its properties easy.

Any complex number can be represented as

z = x + iy (Cartesian form)

= |z| [cos Ɵ + I sin Ɵ] (polar form)

= |z| eiƟ

De Moivre’s Theorem and its Applications

(a) De Moivre’s Theorem for integral index - Suppose n is a integer, then (cos Ɵ + i sin Ɵ)n = cos (nƟ) + I sin (nƟ)

(b) De Moivre’s Theorem for the rational index - Suppose n is a rational number, then the value of or one of the values of (cosƟ + isinƟ)n is cos (nƟ) + i sin (nƟ).

Suppose n = p/q, where p, q ϵ I, q > 0 and p,q have no factors in common, then (cos Ɵ + I sin Ɵ)n has q different values, one of which is cos (nƟ) + i sin (nƟ)

Note: The values of (cos Ɵ + I sin Ɵ)p/q where p, q ϵ I, q ≠ 0, hcf (p,q) = 1 are expressed as cos⁡[pq(2kπ+θ)]+i sin[pq(2kπ+θ)], where k = 0, 1, 2, ….., q -1.

The nth Roots of Unity

The nth root of unity means any complex number z that fulfills the equation zn = 1 (1)

Because an equation of degree n has n roots, there are n values of z that fulfill the equation (1).

To be able to find these n values of z, write 1 = cos (2kπ) + I sin (2kπ)

where k ϵ I and ⇒ z=cos⁡(2kπn)+isin⁡(2kπn)

[using the De Moivre’s Theorem]

where k = 0, 1, 2, …., n -1.

Note: Any n consecutive integral values could be given to k. For example, for 3, you could take -1, 0, and 1 and for 4, you could take – 1, 0, 1, and 2 or -2, -1, 0, and 1.

Notation

ω=cos⁡(2πn)+isin⁡(2πn)

Using De Moivre’s theorem, the nth roots of unity can be written as 1, ω, ω2, …., ωn-1.

The sum of the roots of unity is zero.

Given, 1 + ω + … + ωn – 1 = 1−ωn1−ω But, ωn = 1 as ω is an nth root of unity.

∴ 1+ω+…+ωn−1=0

Note: 1x−1+1x−ω….+1x−ωn−1=nxn−1xn−1

Cube Roots of Unity

Cube roots of unity are represented as 1, ω, ω2, where ω=cos⁡(2π3)+isin⁡(2π3)=−1+3i2and ω2=−1−3i2

Some Results that involve Complex Cube Root of Unity (ω)

(i) ω3 = 1

(ii) 1 + ω + ω2 = 0

(iii) x3 – 1 = (x – 1) ( x – ω) (x – ω2)

(iv) ω and ω2 are roots of x2 + x + 1 = 0

(v) a3 – b3 = (a – b) (a – bω) (a – bω2)

(vi) a2 + b2 + c2 – bc – ca – ab = (a + bω + cω2) (a + bω2 + cω) (

vii) a3 + b3 + c3 – 3abc = (a + b + c) (a + bω + cω2) (a + bω2 + cω)

(viii) x3 + 1 = (x + 1) (x + ω) (x + ω2)

(ix) a3 + b3 = (a + b) (a + bω) (a + bω2)

(x) The cube roots of real number a are a1/3, a1/3ω, a1/3 ω2.

To find the cube roots of a, write x3 = a as y3 = 1 where y = x/a1/3.

The solutions of y3 = 1 are 1, ω, ω2. x = a1/3, a1/3 ω, a1/3 ω2.

Check: JEE Main Study Notes on Differential Calculus

Logarithm of Complex Numbers

Loge(x + iy)

= loge (|z|eiƟ)

= loge |z| + loge eiƟ

= loge |z| + iƟ

= log⁡e(x2+y2)+iarg⁡(z)

∴ log⁡e(z)=loge∣z∣+iarg(z)

Some Useful Identities

x2 + xy + y2 = (x – yω) (x – yω2), in particular,

x2 + x + 1 = (x – ω) (x – ω2)

x2 – xy + y2 = (x + yω) (x + yω2), in particular,

x2 – x + 1 = (x + ω) (x + ω2)

x2 + y2 + z2 – xy – xz – yz = (x + yω + zω2) (x + yω2 + zω)

Geometry of a Complex Number

abc

As in vectors, the position vector of the point helps to represent a point with respect to origin O. Similarly, in the above image, we can see point P is represented by a complex number z, such that length OP = |z| and The point P is called the image of the complex number z and z is said to be an affix or complex coordinate of the point P.

Distance Between Two Points

If two points P and Q have affices z1and z2 respectively then abc

Affix of Q – Affix of P.

Points to Remember for Complex Numbers

||z1| - |z2|| = |z1+z2| and |z1-z2| = |z1| + |z2| if the origins z1, and z2 are collinear and lie between z1 and z2.
|z1 + z2| = |z1| + |z2| and ||z1| - |z2|| = |z1-z2| if the origins, z1 and z2 are collinear and z1 and z2 lie on the same side of origin.
The product of nth roots of any complex number z is given by z(-1)n-1.
amp(zn) = n amp z
The least value of |z - a| + |z - b| is |a - b|

abc

Demoivre's Theorem: It can be stated in two forms:

Case I: Suppose 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 θn + i sin θn) = cos (θ1 + θ2 + θ3 .................. + θn) + i sin (θ1 + θ2 + .............. θn)

Case II: For p and q such that q ≠ 0, we get (cos θ + i sin θ)p/q = cos((2kπ + pq)/q) + isin((2kπ+pq/q) where k = 0,1,2,3,.....,q-1

Solved Examples on Complex Numbers

Question 1: What is the minimum value of |a+bω+cω2|, where a, b and c are all not equal integers and ω≠1 is a cube root of unity?

√3
1/2
1
0

Solution: Assume that z =| a+bω+cω2|

Then, z2 = | a+bω+cω2|2

= (a2 + b2 + c2 – ab– bc – ca)

Or z2 = 1/2 {(a-b)2 + (b-c)2 + (c-a)2} ….. (1)

As a, b and c are all integers but not simultaneously equal,

if a = b, then a ≠ c and b ≠ c.

Because the difference of integers is an integer, (b-c)2 ≥ 1{as the minimum difference of two consecutive integers is ±1}

Also, (c-a)2≥1.

Take a =b so (a-b)2 = 0.

From equation (1), z2≥1/2 (0+1+1) z2 ≥ 1

Therefore, the minimum value of |z| is 1.

Question 2: Suppose a and b are real numbers between 0 and 1 in a way that the points z1 = a + i, z2 = 1+bi, z3 = 0 form an equilateral triangle, what are the values of a and b?

Solution: Given, z1, z2, z3 form an equilateral triangle

z12 + z22+ z32 = z1z2 + z2z3+ z3z1

Hence, (a+i)2 + (1+bi)2 + (0)2 = (a+i)(1+bi) + 0 + 0

a2 – 1 + 2ia + 1 – b2 + 2ib = a + i(ab + 1) – b

Hence, (a2– b2) + 2i(a+b) = (a-b) + i(ab + 1)

Thus, a2 – b2 = a-b and 2(a+b) = ab +1

Hence, a = b or a+b = 1 and 2 (a+b) = ab +1 When a=b, 2(2a) = a2 +1

Hence, a2 – 4a +1 = 0 Thus, a = 2±√3 If a+b = 1, 2 = a(1-a) + 1

Hence, a2 – a + 1 = 0

so, a = (1±√1-4)/2

Because a and b belong to R, the only solution exists when a = b

Therefore, a = b = 2±√3.

Question 3: Suppose iz3 + z2 – z + i = 0, prove that |z| = 1.

Solution: Given, iz3 + z2 – z + i = 0

Hence, iz3–i2z2 – z + i = 0

iz2 (z-i) – 1(z-i) = 0

(iz2 – 1)(z-i) = 0

Thus, either (iz2 – 1) = 0 or (z-i) = 0

So, z = i or z2 = 1/i = -i

When z = i, then |z| = |i| = 1

When z2 = -I, then |z2| = |-i| = 1

Therefore, |z| = 1.

Question 4: Suppose z1, z2, z3 are the vertices of an equilateral triangle ABC in a way that |z1 − i| = |z2 − i| = |z3 − i|, what is the value of |z1 +z2 + z3|?

Solution: |z1 − i| = |z2 − i| = |z3 − i|

Thus, z1,z2, z3 lie on the circle with the center i.

The circumcenter also coincides.

[z1+z2+z3] / 3 = i ⇒|z1+z2+z3|=3

Question 5: The value of λ if the curve y = (λ + 1)x2 + 2 intersects the curve y = λx + 3 at one point exactly is?

Solution: Given, (λ + 1)x2 + 2 = λx + 3 has only one solution,

D = 0 ⇒ λ2 − 4(λ + 1)(−1) = 0

Λ2 + 4λ + 4 = 0

(λ + 2)2 = 0

Therefore, λ = −2

Tricks to Solve Questions from Complex Numbers

Question 1: Suppose ω is an imaginary cube root of unity, then the value of the expression

abc

Trick to solve this question: Find the rth term of the expression. You will get

Now find the value of the expression.

Solution: (a)

Question 2: Find the real part of . abc

The trick to solve this question: Equate x to the given expression. Take logs on both sides. You will get . Now, find the real part.

Solution:

Question 3: Calculate the square root of z=−7−24i.

Trick to solve this question: Assume abc to be a square root. From this, find the equation for abc . Equate the real and imaginary parts. You will get . Solve all equations and you will get the answer.