Detailed Notes: Complex Numbers | Mathematics (Maths) for JEE Main & Advanced PDF Download

What is a Complex Number? 

• Complex Numbers are the numbers which along with the real part also have the imaginary part included with it.
• It is defined as the combination of real part and imaginary part. Either of the parts can be zero.
• If ‘a’ is the real part and ‘b’ represents the imaginary part, then the complex number is represented as 
•  z = a + ib where i, stands for iota which itself is a square root of negative unity.
• Examples

Thus, we can also write z = Re(z) + i Im(z). This form of representation is also called as the Cartesian or algebraic form of representation.

If z = -2 + j4, then Re(z) = -2 and Im(z) = 4.

Similarly, for z = 3+j5, Re(z) = 3 and Im(z) = (5).

Representation of Complex Number 

1.  Cartesian , algebraic or rectangular form
2.  Trigonometric or polar form
3.  Exponential form
4.  Vector form

Can we take the square root of a negative number?

• Yes of course, but to understand this question, let’s go into more deep of complex numbers,
• Consider the equation x²+1 = 0, 
  If we try to get its solution, we would stuck at x = √(-1) so in Complex Number we assume that √(-1) =i or i² =-1 which means i can be assumed as the solution of this equation. i is called an iota in Complex Numbers.
• We can further formulate as,
  i² = -1
  i³ = i² x i = -i
  i⁴ =i² x i² =1
• So, we can say now, i⁴ⁿ = 1 where n is any positive integer.
  Also, note that i + i² + i³ + i⁴ = 0 or in + i²ⁿ + i³ⁿ + i⁴ⁿ= 0
  This means the sum of the consecutive four powers of iota leads the result to zero.

Set of complex Numbers: The set of all complex numbers is denoted by C.
i.e. C = {a + ib | a, b ∈ ℝ}
Equality of Complex Numbers: Two complex numbers z₁ = a₁ + ib₁ and z₂ = a₂ + ib₂ are equal if a₁ = a₂ and b₁ = b₂, i.e. Re(z₁) = Re(z₂) and Im(z₁) = Im(z₂).

[Question: 1571380]​

What is the form a + ib?

• We know from the above discussion that, Complex Numbers can be represented in four different ways. Out of which, algebraic or rectangular form is one of the forms.
• Z = a + ib is the algebraic form in which ‘a’ represents the real part and ‘b’ represents the imaginary part. Two mutually perpendicular axes locate any complex point on the plane. The horizontal axis represents the real part while the vertical axis represents the imaginary part.

How do we locate any Complex Number on the plane?  

Let us take a few examples to understand, how can we locate any point on the complex or argand plane.
Example 1:
Consider a complex number z = 6 +j4 (‘i’ and ‘j’, both can be used for representing imaginary part), if we compare this number with z = a + jb form. Then we can easily equate the two and get a = 6 and b = 4. Since both a and b are positive, which means the number will be lying in the first quadrant. ‘z’ will be 6 units on the right and 4 units upwards from the origin. You can see the same point in the figure below.
Example 2:
Now consider a point in the second quadrant that is. z = -7 + j6, Here since a= -7 and b = 6 and thus will be lying in the second quadrant.
Point z is 7 units in the left and 6 units upwards from the origin. Refer the figure to understand it pictorially.
Example 3:
Now let’s consider a point in the third quadrant as z = -2 – j3. Since in third quadrant both a and b are negative and thus a = -2 and b = -3 in our example. This point will be lying 2 units in the left and 3 units downwards from the origin.
Example 4:
Let now take the fourth (of the fourth quadrant) and the last case where z = 5 – j6. Here, a = 5 and b = - 6 i.e. a positive and b negative. This point will be lying 5 units in the right and 6 units downwards.

Are all Real Numbers  Complex Numbers?

• A complex number has two parts, the real part and the imaginary part that is. z = a + ib
• if b = 0, z = a which is called the Purely Real Number
  and if a = 0, z = ib which is called the Purely Imaginary Number.
• Thus we can say that all real numbers are also complex numbers with imaginary part zero.

[Question: 1571381]

Complex Equations 

For example: x = (2+3i) (3+4i), In this example, x is a multiple of two complex numbers. On multiplying these two complex number we can get the value of x.

z² + 2z + 3 = 0 is also an example of complex equation whose solution can be any complex number.

Solved Examples

Illustration 1: Let a and b be roots of the equation x² + x + 1 = 0. Then find the equation whose roots are a = 19 and b = 7.

Sol: Given that x² + x + 1 = 0, its roots are the non-real cube roots of unity, namely ω and ω². This means that a = ω and b = ω².
Since ω³ = 1, we can reduce the exponents modulo 3.
For a¹⁹: 19 mod 3 = 1, so a¹⁹ = ω¹⁹ = ω¹⁸ · ω = (ω³)⁶ · ω = 1⁶ · ω = ω.
For b⁷: 7 mod 3 = 1, so b⁷ = (ω²)⁷ = ω¹⁴ = ω¹² · ω² = (ω³)⁴ · ω² = 1⁴ · ω² = ω².
Thus, the new roots are ω and ω², just as in the original equation.
Now, the quadratic equation with roots ω and ω² is formed by:   x² – (ω + ω²)x + (ω · ω²) = 0.
It is known that 1 + ω + ω² = 0, which means ω + ω² = –1. Also, ω · ω² = ω³ = 1.
Substitute these into the equation:   x² – (–1)x + 1 = x² + x + 1 = 0.
Therefore, 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.

Sol: Since z² + 1 is a quadratic polynomial, the remainder when f(z) is divided by (z² + 1) will be a linear polynomial. Let this remainder be a·z + b.
Then we can write: f(z) = g(z) · (z² + 1) + a·z + b

Substituting z = i: f(i) = g(i) · (i² + 1) + a·i + b Since i² + 1 = –1 + 1 = 0, we have:   
f(i) = a·i + b But f(i) = i, so   a·i + b = i       (1)
Similarly, substituting z = –i:   f(–i) = g(–i) · ((–i)² + 1) + a·(–i) + b Again, (–i)² + 1 = –1 + 1 = 0, so:   f(–i) = –a·i + b And we are given f(–i) = 1 + i, hence:   –a·i + b = 1 + i   (2)
Now, we have two equations:   (1) a·i + b = i   (2) –a·i + b = 1 + i
Subtract equation (2) from equation (1):   (a·i + b) – (–a·i + b) = i – (1 + i)   2a·i = –1   a = –1/(2i) Recall that 1/i = –i, so:   a = (i)/2
Next, substitute a = i/2 into equation (1): (i/2)·i + b = i   
i²/2 + b = i Since i² = –1, this becomes:  
(–1)/2 + b = i   b = i + 1/2
Thus, the required remainder is:   
a·z + b = (i/2)·z + (1/2 + 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: arg[(3z – 6 – 3i)/(2z – 8 – 6i)] = π/4
and write z = x + it.
Step 1. Rewrite the numerator and denominator in a convenient form. Notice that
3z – 6 – 3i = 3(x + iy) – 6 – 3i = 3(x – 2) + i·3(y – 1)
2z – 8 – 6i = 2(x + iy) – 8 – 6i = 2(x – 4) + i·2(y – 3)
Thus the quotient becomes
(3/2)·[(x – 2) + i(y – 1)]/[(x – 4) + i(y – 3)]
Since (3/2) is a positive real number, its argument is 0. Therefore the given condition reduces to
arg[(x – 2) + i(y – 1)] – arg[(x – 4) + i(y – 3)] = π/4
When a complex number has an argument equal to π/4 its imaginary part equals its real part. Thus if we write the quotient in standard form as A + iB, we must have A = B. Carrying out the division (by multiplying numerator and denominator by the complex conjugate of the denominator) leads to
A = ( (x – 2)(x – 4) + (y – 1)(y – 3) )/( (x – 4)² + (y – 3)² )
B = ( (y – 1)(x – 4) – (x – 2)(y – 3) )/( (x – 4)² + (y – 3)² )
Setting A = B and multiplying through by the common denominator gives
(x – 2)(x – 4) + (y – 1)(y – 3) = (y – 1)(x – 4) – (x – 2)(y – 3)
Expanding both sides and simplifying yields
x² + y² – 8x – 2y + 13 = 0                    (1)
Step 2. The second condition is
|z – 3 + i| = 3
Since z = x + iy this becomes
| (x – 3) + i(y + 1) | = 3
which leads to the circle
(x – 3)² + (y + 1)² = 9                   (2)
Expanding (2) we get
x² + y² – 6x + 2y + 1 = 0
Step 3. Subtract equation (2) from equation (1):
[ x² + y² – 8x – 2y + 13 ] – [ x² + y² – 6x + 2y + 1 ] = 0
Simplify term‐by‐term:
(–8x + 6x) + (–2y – 2y) + (13 – 1) = 0
which gives
–2x – 4y + 12 = 0
Divide both sides by –2:
x + 2y – 6 = 0
Thus
x = 6 – 2y                   (3)
Step 4. Substitute (3) into equation (2). Replace x by (6 – 2y) in
(6 – 2y – 3)² + (y + 1)² = 9
That is,
(3 – 2y)² + (y + 1)² = 9
Expanding:
(9 – 12y + 4y²) + (y² + 2y + 1) = 9
Combine like terms:
5y² – 10y + 10 = 9
Subtract 9 from both sides:
5y² – 10y + 1 = 0
Solve this quadratic in y using the quadratic formula:
y = [10 ± √(100 – 20)]/(2·5) = [10 ± √80]/10 = [10 ± 4√5]/10
Simplify:
  y = 1 ± (2/√5)

Step 5. Now, substitute these y-values back into (3) to find x:
x = 6 – 2y = 6 – 2[1 ± (2/√5)] = 6 – 2 ∓ (4/√5) = 4 ∓ (4/√5)
Thus the solutions for z are:
z = x + iy = [4 ∓ (4/√5)] + i[1 ± (2/√5)]
In summary, the complex numbers z that satisfy the given conditions are

z = 4 ∓ 4/√5 + i(1 ± 2/√5)

Fundamental Operations on Complex Numbers

1. Addition 

Let z₁ = a₁ + ib₁ and z₂ = a₂ + ib₂ be two complex numbers. 
Then their sum z₁ + z₂ is defined as the complex number (a₁ + a₂) + i (b₁ + b₂).

Properties of addition of complex numbers

(i) Addition is commutative: For any two complex numbers z₁ and z₂, we have
z₁ + z₂ = z₂ + z₁
(ii) Addition is associative: For any three complex numbers z₁, z₂, z₃, we have
(z₁ + z₂) + z₃ = z₁ + (z₂ + z₃)
(iii) Existence of additive identity: The complex number 0 = 0 + i0 is the identity element for addition, i.e.,
z + 0 = z = 0 + z for all z ∈ C
(iv) Existence of additive inverse: For every complex number z, there exists -z such that
z + (-z) = 0 = (-z) + z
The complex number -z is called the additive inverse of z.

2. Subtraction

Let z₁ = a₁ + ib₁ and z₂ = a₂ + ib₂ be two complex numbers. Then the subtraction of z₂ from z₁ is denoted by z₁ - z₂ and is defined as the addition of z₁ and -z₂.
Thus,
z₁ - z₂ = (a₁ - a₂) + i (b₁ - b₂)

3. Multiplication

Let z₁ = a₁ + ib₁ and z₂ = a₂ + ib₂ be two complex numbers. Then, the multiplication of z₁ with z₂ is denoted by z₁z₂ and is defined as the complex number.
(a₁a₂ - b₁b₂) + i (a₁b₂ + a₂b₁)

Properties of Multiplication

(i) Multiplication is commutative: For any two complex numbers z₁ and z₂, we have
z₁ z₂ = z₂ z₁
(ii) Multiplication is associative: For any three complex numbers z₁, z₂, z₃, we have
(z₁ z₂) z₃ = z₁ (z₂ z₃)
(iii) Existence of identity element for multiplication: The complex number 1 = 1 + i0 is the identity element for multiplication, i.e., for every complex number z, we have
z ⋅ 1 = z
(iv) Existence of multiplicative inverse: Corresponding to every non-zero complex number z = a + ib, there exists a complex number z₁ = x + iy such that
z ⋅ z₁ = 1 ⇒ z₁ = 1 / z
The complex number z₁ is called the multiplicative inverse or reciprocal of z and is given by:
z₁ = (a / (a² + b²)) + i(-b / (a² + b²))

Did You Kmow

Multiplication of Complex Numbers is Distributive Over Addition
For any three complex numbers z₁, z₂, z₃, we have:
(i) z₁(z₂ + z₃) = z₁z₂ + z₁z₃ (Left distributivity)
(ii) (z₂ + z₃)z₁ = z₂z₁ + z₃z₁ (Right distributivity)

4. Division

The division of a complex number z₁ by a non-zero complex number z₂ is defined as the multiplication of z₁ by the multiplicative inverse of z₂ and is denoted by (z₁ / z₂).
Thus,
z₁ / z₂ = z₁ z₂^-1 = z₁ (1 / z₂)

Conjugate , Modulus and Argument of a Complex Number

1. Conjugate 

Let z = a + ib be a complex number. Then the conjugate of z is denoted by z̅ and is equal to a - ib.

Thus, z = a + ib ⇒ z̅ = a - ib

Properties of Conjugate

If z, z₁, z₂ are complex numbers, then:

• z + z̅ = 2 Re(z)
• z - z̅ = 2 Im(z)
• z = z̅ ⇔ z is purely real
• z + z̅ = 0 ⇒ z is purely imaginary
• z z̅ = {Re(z)}² + {Im(z)}²
•  z̅₁ + z̅₂ = z̅₁ + z̅₂
• z̅₁ - z̅₂ = z̅₁ - z̅₂
•  z̅₁ z₂ = z̅₁ z̅₂
• (z₁ / z₂) z̅ = z̅₁ / z̅₂, z₂ ≠ 0
• z̅(z) = z

2. Modulus

• The modulus of a complex number z = a + i b is denoted by |z| and is defined as
• 
  
• The multiplicative inverse of a non-zero complex number z is same as its reciprocal and is given by
  
• If b is positive then 
  
• If b is negative then 
  

Properties of Modulus of z

• |z₁ + z₂|² = |z₁|² + |z₂|² + 2|z₁||z₂| cos(θ₁ - θ₂)
  or
  |z₁ + z₂|² = |z₁|² + |z₂|² + 2 Re(z₁z̅₂)
• |z₁ - z₂|² = |z₁|² + |z₂|² - 2 |z₁||z₂| cos(θ₁ - θ₂)
  or
  |z₁ - z₂|² = |z₁|² + |z₂|² - 2 Re(z₁z̅₂)
• |z₁ + z₂|² + |z₁ - z₂|² = 2 (|z₁|² + |z₂|²)
• |z₁ + z₂| = |z₁ - z₂| ⇒ arg(z₁) - arg(z₂) = π/2
• |z₁ + z₂| = |z₁| + |z₂| ⇔ arg(z₁) = arg(z₂)
• |z₁ + z₂|² = |z₁|² + |z₂|² ⇔ (z₁ / z₂) is purely imaginary.
• |z₁ + z₂| ≤ |z₁| + |z₂|
• |z₁ - z₂| ≤ |z₁| + |z₂|
• |z₁ + z₂| ≥ ||z₁| - |z₂||
• |z₁ - z₂| ≥ ||z₁| - |z₂||
• |z₁ z₂| = |z₁| |z₂|
• |zⁿ| = |z|ⁿ
• |z|² = z z̅
• |z| = |z̅| = |-z| = |-z̅|

3. Argument

The argument of a complex number is defined as the angle inclined from the real axis in the direction of the complex number represented on the complex plane. 
It is denoted by “θ” or “φ”. It is measured in the standard unit called “radians”.

Argument or (amplitude) of a Complex Number

• If x and y both are positive, then the argument of z = x + iy is the acute angle given by: tan^-1 |y/x|
• If x < 0 and y > 0, then the argument of z = x + iy is π - α, where α is the acute angle given by: tan^-1 |y/x|
• If x < 0 and y < 0, then the argument of z = x + iy is α - π, where α is the acute angle given by: tan α = |y/x|
• If x > 0 and y < 0, then the argument of z = x + iy is -α, where α is the acute angle given by: tan α = |y/x|

Properties of Argument of z

• arg(z̅) = -arg(z)
• arg(z₁z₂) = arg(z₁) + arg(z₂)
• arg(z₁z̅₂) = arg(z₁) - arg(z₂)
• arg(z₁ / z₂) = arg(z₁) - arg(z₂)
• arg(zⁿ) = n arg(z)
• |z₁ + z₂| = |z₁ - z₂| ⇒ arg z₁ - arg z₂ = π/2
• |z₁ + z₂| = |z₁| + |z₂| ⇒ arg z₁ = arg z₂
• If arg z = 0, then z is purely real
• If arg z = ± π/2, then z is purely imaginary

Polar Form of a Complex Number

• Let z = x + iy be a complex number represented by a point P (x, y) in the Argand plane. Then, by the geometrical representation of z = x + iy, we have 
  ⇒ z = r(cosθ + i sinθ), where r = |z| and θ = arg(z) 
• This form of z is called a polar form of z.
• The Euclearian form of a complex number is e^iθ = cosθ + i sinθ and e^-iθ = cosθ - i sinθ

1. Distance Between Two Points

If z₁ and z₂ are the affixes of points P and Q respectively in the Argand plane, then:
PQ = |z₂ - z₁|

2. Section Formula

Let z₁ and z₂ be the affixes of two points P and Q respectively in the Argand plane. Then, the affix of a point R dividing PQ internally in the ratio m : n is:
(m z₂ + n z₁) / (m + n)
but if R is an external point, then the affix of R is:
(m z₂ - n z₁) / (m - n)

3. Mid Point Formula

• If R be the mid-point, then the affix of R is: (z₁ + z₂) / 2
• If z₁, z₂, z₃ are affixes of the vertices of a triangle, then the affix of its centroid is:
  (z₁ + z₂ + z₃) / 3
• The equation of the perpendicular bisector of the line segment joining points having affixes z₁ and z₂ is:
  z(z̅₁ - z̅₂) + z̅(z₁ - z₂) = |z₁|² - |z₂|²
• The equation of a circle whose centre is at point having affix z₀ and radius R is:
  |z - z₀| = R

Note

• If the centre of the circle is at the origin and radius R, then its equation is |z| = R.
• General Equation of a circle is:
  z z̅ + a z̅ + az + b = 0, where b ∈ R and a is a complex number.
  This represents a circle having centre at '-a' and radius:
  √(|a|² - b) = √(a z̅ - b)

Complex Number as a Rotating Arrow in the Argand Plane

To obtain the point representing z e^iα, we rotate OP through angle α in an anticlockwise sense. Thus, multiplication by e^iα to z rotates the vector OP in an anticlockwise sense through an angle α.

Let z₁ and z₂ be two complex numbers represented by points P and Q in the Argand plane such that ∠POQ = θ.

Then, z₁ e^iθ is a vector of magnitude |z₁| = OP along OQ and (z₁ e^iθ) / |z₁| is a unit vector along OQ.

Some Important Results

1. I f z₁, z₂, z₃ are the affixes o f the points A, B and C in the Argand plane, then
(i) ∠BAC = arg ( (z₃ - z₁) / (z₂ - z₁) )
(ii) ∠BAC = arg ( (z₃ - z₁) / (z₂ - z₁) ) = |(z₃ - z₁) / (z₂ - z₁)| (cos α + i sin α), where α = ∠BAC.
If z₁, z₂, z₃, and z₄ are the affixes of the points A, B, C, and D respectively in the Argand plane, then AB is inclined to CD at the angle: arg ( (z₂ - z₁) / (z₄ - z₃) )
(iii) The equation of the circle having z₁ and z₂ as the endpoints of a diameter is:
(z - z₁)(z̅ - z̅₁) + (z̅ - z̅₁)(z - z₂) = 0

De-Moivre’s Theorem

Statement
(i) If n ∈ Z (the set of integers), then:
(cosθ + i sinθ)ⁿ = cos(nθ) + i sin(nθ)
(ii) If n ∈ Q (the set of rational numbers), then cos(nθ) + i sin(nθ) is one of the values of (cosθ + i sinθ)ⁿ.
(iii) ¹ / _{(cosθ + i sinθ)} = cosθ - i sinθ
(iv) (cosθ₁ + i sinθ₂)(cosθ₂ + i sinθ₂) = cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)

n^th Roots of Unity

n^th roots of unity are: α⁰ = 1, α, α², α³, ...... α^n-1 where α = e^i2π/n = cos(2π/n) + i sin(2π/n)

Properties of n^th Roots of Unity

• Property 1: n^th roots of unity form a G.P. with common ratio e^i2π/n
• Property 2: Sum of the n^th roots of unity is always zero.
• Property 3: Sum of p^th powers of n^th roots of unity is zero, if p is not a multiple of n.
• Property 4: Sum of p^th powers of n^th roots of unity is n, if p is a multiple of n.
• Property 5: Product of n^th roots of unity is (-1)^n-1
• Property 6: n^th roots of unity lie on the unit circle |z| = 1 and divide its circumference into n equal parts.

Properties of Cube Roots of Unity and Some Useful Results Related To Them

(i) Cube roots of unity are 1, ω, ω² where:
ω = -1/2 ± i(√3/2), ω² = -1/2 - i(√3/2)
(ii) arg(ω) = 2π/3
(iii) Cube roots of -1 are -1, -ω, -ω²
(iv) 1 + ω + ω² = 0
(v) ω³ = 1

• Four fourth roots of unity are -1, 1, -i, i
• log(α + iβ) = (1/2) log(α² + β²) + i tan^-1(β/α)

Condition for points A(z₁), B(z₂), C(z₃), D(z₄) to be concyclic:
α + β

Condition(s) for four points A(z₁), B(z₂), C(z₃), and D(z₄) to represent vertices of a

1. Parallelogram:
(i) The diagonals AC and BD must bisect each other
⇔ (1/2) (z₁ + z₃) = (1/2) (z₂ + z₄)

(ii) Rhombus:
(a) The diagonals AC and BD bisect each other.
⇔ z₁ + z₃ = z₂ + z₄
(b) A pair of two adjacent sides are equal, i.e., AD = AB.
⇔ |z₄ - z₁| = |z₂ - z₁|
(iii) Square:
(a) The diagonals AC and BD bisect each other.
⇔ z₁ + z₃ = z₂ + z₄
(b) A pair of adjacent sides are equal, AD = AB.
⇔ |z₄ - z₁| = |z₂ - z₁|
(c) The two diagonals are equal, AC = BD.
⇔ |z₃ - z₁| = |z₄ - z₂|
(iv) Rectangle:
(a) The diagonals AC and BD bisect each other.
z₁ + z₃ = z₂ + z₄
(b) The diagonals AC and BD are equal.
⇔ |z₃ - z₁| = |z₄ - z₂|

(v) Incentre: I (z) of the ΔABC is given by
(vi) Circumcentre (z) of the ΔABC is given by
(v) Orthocentre (z) of the ΔABC is given by

(vi) Area of triangle ABC with vertices A(z₁), B(z₂), C(z₃) is given by
(vi) Equation of line passing through A(z₁) and B(z₂) is

(vii) General equation of a line is:
a̅z + az̅ + b = 0, where a is a complex number and b is a real number.
(viii) Complex slope of a line joining points A(z₁) and B(z₂) is given by:
w = (z₁ - z₂) / (z̅₁ - z̅₂)
(ix) Two lines with complex slopes w₁ and w₂ are parallel if w₁ = w₂ and perpendicular if w₁ + w₂ = w

Length of Perpendicular from a Point to a Line

Length of perpendicular from a point A(ω) to the line a̅z + az̅ + b = 0 is:
p = |aω + a̅ω̅ + b| / (2 |a|)

Recognizing Some Loci by Inspection

(i) If z₁ and z₂ are two fixed points, then:
|z - z₁| = |z - z₂| represents the perpendicular bisector of the line segment joining A(z₁) and B(z₂).
(ii) If z₁ and z₂ are two fixed points and k > 0, k ≠ 1 is a real number, then:
|z - z₁| / |z - z₂| = k represents a circle. For k = 1, it represents the perpendicular bisector of the segment joining A(z₁) and B(z₂).
(iii) Let z₁ and z₂ be two fixed points and k be a positive real number.
(a) If k > |z₁ - z₂|, then:
|z - z₁| + |z - z₂| = k represents an ellipse with foci at A(z₁) and B(z₂) and length of major axis = k = CD.
(b) If k = |z₁ - z₂|, it represents the line segment joining z₁ and z₂.
(c) If k < |z₁ - z₂|, then |z - z₁| + |z - z₂| = k does not represent any curve in the Argand plane.

(iv) Loci Related to Fixed Points
(a) If k < |z₁ - z₂|, then |z - z₁| - |z - z₂| = k represents a hyperbola with foci at A(z₁) and B(z₂).
(b) If k = (z₁ - z₂), then:
(z - z₁) - (z - z₂) = k
represents the straight line joining A(z₁) and B(z₂) excluding the segment AB.
(v) Special Circle Property
If z₁ and z₂ are two fixed points, then:
|z - z₁|² + |z - z₂|² = |z₁ - z₂|²
represents a circle with z₁ and z₂ as extremities of a diameter.
(vi) Argument Condition for a Circle Segment
Let z₁ and z₂ be two fixed points and α be a real number such that 0 ≤ α ≤ π, then:
(a) If 0 < α < π and α ≠ π/2, then: arg((z - z₁) / (z - z₂)) = α represents a segment of the circle passing through A(z₁) and B(z₂).

(b) If α₂ = π/2, then: arg((z - z₁) / (z - z₂)) = α = π/2 represents a circle with diameter as the segment joining A(z₁) and B(z₂).
(c) If α = π, then: arg((z - z₁) / (z - z₂)) = α represents the straight line joining A(z₁) and B(z₂) but excluding the segment AB.

(d) If α = 0, then: arg((z - z₁) / (z - z₂)) - α ( = 0 ) 
represents the segment joining A(z₁) and B(z₂).