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.
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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.
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
In fact if a > 0 and b > 0 then
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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.
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
Check:
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
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
Let z = x + iy and let the square root of z be the complex number a + ib. Then
or
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
If ab < 0, then
Thus, square roots of z = a + ib are :
for b > 0
and for b < 0
For Example:
Check:
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 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∣.
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.
(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.
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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.
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
(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 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.
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 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
Loge(x + iy) = loge (|z|eiƟ)
= loge |z| + loge eiƟ
= loge |z| + iƟ
= loge(x2+y2)+iarg(z)
∴ loge(z)=loge∣z∣+iarg(z)
x2 + x + 1 = (x – ω) (x – ω2)
x2 – x + 1 = (x + ω) (x + ω2)
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.
If two points P and Q have affices z1and z2 respectively then
Affix of Q – Affix of P.
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
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?
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
or
Λ2 + 4λ + 4 = 0
or (λ + 2)2 = 0
Therefore, λ = −2
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
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 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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