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

    Urvashi Singhal Urvashi Singhal
    Exams Prep Master

    Complex Numbers is considered quite easy to score in JEE Main Mathematics. It forms an integral component of Algebra which is the topic that bears the highest weightage in the Mathematics section. Around 2-3 questions are asked from Complex Numbers that bear a total of about 8 marks. Thus, the weightage of this chapter in JEE Main is around 2-3%. Read JEE Main Mathematics Preparation Tips

    Some of the concepts tested under Complex Numbers are - Modulus of Complex Numbers, Conjugate of Complex Numbers, and Different Forms of Complex Numbers. Candidates are advised to ensure that they do not skip this chapter at any cost while preparing for JEE Main 2020.

    Must Read:

    Read the article below to access the study notes for Complex Numbers that will aid you in preparation for JEE Main Mathematics.

    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 ofRe(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 probable ordered pairs is known as a complex number set and is expressed as C.

    Integral Powers of Iota

    Integral Powers of Iota

    An imaginary number I (iota) is described as √-1 as I = x√-1 and we get i2 = –1 , 13 = –1, i4 = 1

    • To calculate the value of in (n > 4), divide n by 4.

    Let q be the quotient and r the remainder.

    n = 4q + r where o < r < 3

    in = i4q + r = (i4)q , ir = (1)q . ir = ir

    • The sum of four consecutive powers of I is zero.

    In + in+1 + in+2 + in+3 = 0, n ∈ z

    • 1/i = – i
    • (1 + i)2 = 2i and (1 – i)2 = 2i
    • √a . √b = √ab is valid only if a minimum of one of a and b is non negative.
    • When both a and b are negative, then √a × √b = -√(|a|.|b|)
    • √-a × √-b = -a

    Algebraic Operations with Complex Numbers

    Algebraic Operations with Complex Numbers

    1. Addition: (a + ib) + (c + id) = (a + c) + i(b + d)
    2. Subtraction: (a + ib) – (c + id) = (ac) + i(b – d)
    3. Multiplication: (a + ib) (c + id) = (ac – bd) + i(ad + bc)
    4. 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)]
    5. 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)]/[c2 + d2] = [ac + bd]/[c2 + d2] + i[bc−ad]/[c2+d2 ]

    Conjugate of Complex Numbers

    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

    1. z1=z2⇔z‾1=z‾2
    2. (zˉ)‾=z
    3. z+z‾=2 Re(z)
    4. z−z‾=2i Im (z)
    5. z=z‾⇔z is fully real
    6. z+z‾=0⇔z is fully imaginary.
    7. zz‾=[Re (z)]2+[Im(z)]2
    8. z1+z2‾=z‾1+z‾2
    9. z1−z2‾=z‾1−z‾2
    10. z1z2‾=z‾1z‾2
    11. (z1z2)‾=z‾1z2, if z2 ≠ 0
    12. 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

    1. 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‾)

    Modulus of a Complex Numbers

    Modulus of a 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 differentt integers lying in{2, 3, 4}and m

    Square Root of Complex Numbers

    Square Root of Complex Numbers

    Suppose z = x + iy, then

    x+iy={±[∣z∣+x2+i∣z∣−x2]if y > 0±[∣z∣+x2−i∣z∣−x2]if y < 0, Where |x| = x2+y2 

    Note:
    1. x+iy+x−iy=2∣z∣+2x
    2. x+iy−x−iy=i2∣z∣−2x
    3. i=±(1+i2) and −i=±(1−i2)

    Modulus and Argument of Complex Numbers

    Modulus and Argument of Complex Numbers

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

    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, suppose 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 fulfils 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.

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

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

    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)

    This is an acute angle and negative.

    Polar Form of a Complex Number 

    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 the complex number.

    Euler’s 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

    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 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 differentt 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 fulfils the equation zn = 1 (1)

    Because an equation of degree n has n roots, there are n values of z that fulfil 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 the 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 a 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.

    Logarithm of Complex Numbers 

    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)

    Points to Remember for Complex Numbers 

    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|.
    • 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 

    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? 

    1. √3

    2. 1/2

    3. 1

     4. 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 ≠ cand 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

    or

    Λ2 + 4λ + 4 = 0

    or (λ + 2)2 = 0

    Therefore, λ = −2

    Tricks to Solve Questions from Complex Numbers 

    Tricks to Solve Questions from Complex Numbers 

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

    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 .

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

    Solution:

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