Last modified on February 22nd, 2024

chapter outline

 

Complex Number

In mathematics, a complex number is the sum of real and imaginary numbers. It is expressed as:

Complex Number Venn Diagram

It is the standard or rectangular form of representing complex numbers.

Real numbers are all positive, negative, rational, and irrational numbers. Some examples are -2, 5.8, ${\sqrt{3}}$, and ${\dfrac{7}{2}}$.

Imaginary numbers are the numbers whose squares give a negative value. ${\sqrt{-5}}$, ${\sqrt{-9}}$, and ${\sqrt{-8}}$ are few examples.

Complex Number

Here are a few examples of complex numbers with their real and imaginary parts:

Complex NumbersReal PartImaginary Part
2 + 3i23i
${\sqrt{3}+\sqrt{5}i}$${\sqrt{3}}$${\sqrt{5}i}$
-6i (purely imaginary)0-6i
7 (purely real)70i
0 (zero complex number)00i

Express z = ${-7+\sqrt{-8}}$ in the standard form and identify its real and imaginary part.

Solution:

z = ${-7+\sqrt{-8}}$
= ${-7+\sqrt{8}\sqrt{-1}}$
= ${-7+\sqrt{8}i}$, which is in the standard form.
The real and the imaginary parts are -7 and ${\sqrt{8}i}$ respectively.

Representing Graphically

The complex numbers are often plotted in the Euclidean plane where the real and the imaginary parts are expressed as (Re(z), Im(z)). This plane with the complex number is called the argand plane (or complex plane).

Plotting the complex number z = 5 + 4i (a + bi), we get

Graphical Representation of Complex Number

In the graph, ‘a’ represents the real part along the x-axis, and ‘ib’ is the imaginary part along the y-axis. The number 5 + 4i refers to a point P (5, 4) at 5 units along the x-axis and 4 units along the y-axis.

We further analyze the point P using some important terms in the argand plane: modulus and argument.

Modulus (Absolute Value)

The point P forms the linear distance from the origin (0, 0), called the modulus (or the absolute value) of the complex number z.

By using the Pythagorean theorem, the distance (denoted by r) = |z|

= |5 + 4i| = ${\sqrt{5^{2}+4^{2}}}$ = ${\sqrt{41}}$, where base = 5 and perpendicular height = 4.

Thus, for any complex number z, the absolute value r is mathematically expressed as:

r = ${\left| z\right|}$ = ${\left| \sqrt{a^{2}+b^{2}}\right|}$ 

The absolute value is also known as the radius vector or the magnitude of the number.

Argument

In the above graph, point P makes an angle θ with the positive x-axis. It is called the argument of the complex number, denoted by arg(z).

Here, tanθ = ${\dfrac{VD} {HD}}$ = ${\dfrac{4}{5}}$, here VD = Vertical distance, HD = Horizontal distance

⇒ arg(z) = ${\theta =\tan ^{-1}\left( \dfrac{4}{5}\right)}$

Thus, for any complex number z, the argument is expressed as:

arg(z) = ${\theta =\tan ^{-1}\left( \dfrac{b}{a}\right)}$

Modulus and Argument of a Complex Number

There is another representation of a complex number where polar coordinates are used. They are represented as (r, θ) in the Argand plane, where r is the magnitude of the complex number, and θ is the argument angle.

Any complex number, z = a + ib, is represented in the polar form as z = r(Cosθ + isinθ).

For example,

${\cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2}}$, and ${\cos \dfrac{\pi }{4}+i\sin \dfrac{\pi }{4}}$

In addition to the polar form, complex numbers are represented in the exponential form using Euler’s formula. It shows the relation between the complex exponent and the trigonometric ratios, sin and cos.

In this form, a complex number is represented by re, where r is the distance and θ is the angle between the radius vector and the x-axis.

For example,

${e^{i\left( \dfrac{\pi }{2}\right) }}$, and ${5e^{i\left( \dfrac{\pi }{4}\right) }}$.

Also, all three forms of complex numbers (standard, polar, and exponential) can easily be converted from one form to another.

Properties

Conjugates

Let z = -2 + 7i be a complex number.

By changing the sign in the middle, the number becomes 

${\overline{z}}$ = -2 – 7i, which is the conjugate of z.

Now, what happens if we add or multiply a complex number with its conjugate?

On adding -2 + 7i and -2 – 7i, the sum gives

(-2 + 7i) + (-2 – 7i)

= -2 + 7i – 2 – 7i

= (-2 – 2) + (7i – 7i)

= -4

On multiplying -2 + 7i and -2 – 7i, the product gives

(-2 + 7i)(-2 – 7i)

= -2(-2 – 7i) + 7i(-2 – 7i)

= 4 + 14i – 14i – 49i2

= 4 + 49

= 53

Thus, for z = a + ib, the conjugate of z is represented by

${\overline{z}}$ = a – ib.

We also conclude that the sum and the product of a pair of conjugates always give a real number, where the sum is z + ${\overline{z}}$ = (a + ib) + (a – ib) = 2a, and the product is ${z\cdot \overline{z}=\left( a+ib\right) \times \left( a-ib\right)}$ = a2 + b2.

Reciprocal 

To find the reciprocal, we interchange the complex number’s numerator and denominator. For example, the reciprocal of z = 2 + 3i is  ${\dfrac{1}{2+3i}}$

To simplify further, we rationalize this fraction by multiplying the numerator and the denominator with its conjugate 2 – 3i.

${\dfrac{1}{2+3i}\times \dfrac{2-3i}{2-3i}}$

= ${\dfrac{2-3i}{\left( 2\right) ^{2}-\left( 3i\right) ^{2}}}$

= ${\dfrac{2-3i}{4-9i^{2}}}$

= ${\dfrac{2-3i}{4+9}}$

= ${\dfrac{2}{13}-\dfrac{3}{13}i}$, is the simplified form.

Thus, the reciprocal of any complex number z (= a + ib) is

${\dfrac{1}{a+ib}}$ = ${\dfrac{a}{a^{2}+b^{2}}+i\left( \dfrac{b}{a^{2}+b^{2}}\right)}$, and ${z\neq z^{-1}}$.

Find the conjugate and the reciprocal of the complex number 9 – 4i.

Solution:

As we know, the conjugate of a complex number z is ${\overline{z}}$ = a – ib. Thus, the conjugate of (9 – 4i) is 9 + 4i.
The reciprocal of a complex number z = ${z^{-1}=\dfrac{1}{a+ib}}$
Thus, the reciprocal of (9 – 4i) is ${\dfrac{1}{9-4i}}$
= ${\dfrac{1}{9-4i}\times \dfrac{9+4i}{9+4i}}$
= ${\dfrac{9+4i}{\left( 9\right) ^{2}-\left( 4i\right) ^{2}}}$
= ${\dfrac{9+4i}{81+16}}$
= ${\dfrac{9}{97}+\left( \dfrac{4}{97}\right) i}$

Equality

Like real numbers, complex numbers follow the rule of equality. 

If two complex numbers are z1 = a1 + ib1 and z2 = a2 + ib2, they are considered equal if the real and imaginary parts are also equal: a1 = a2, b1 = b2.

Also, two complex numbers in the polar form are equal if they have the same absolute value and their argument (θ) differs by an integral multiple of 2ℼ.

Check whether the complex numbers -4 – i and i – 4 are equal.

Solution:

As we know, if two complex numbers are equal, then the real and the imaginary parts must be equal.
Here, the real parts of the given complex numbers are -4 and -4, which are equal.
Again, the imaginary parts are -i and i, respectively, which are not equal.
Thus, the given complex numbers -4 – i and i – 4 are unequal.

Identities

All algebraic identities can be applied to complex numbers. They are:

(z1 + z2)2z12 + 2z1z2 + z22
(z1 – z2)2z12 – 2z1z2 + z22
(z1 + z2)(z1 – z2)z12 – z22
(z1 + z2 + z3)2z12 + z22 + z32 + 2z1z2 + 2z2z3 + 2z3z1
(z1 + z2)3z13 + 3z12z2 + 3z1z22 + z23
(z1 – z2)3z13 – 3z12z2 + 3z1z22 – z23

Solve: (-1 – i)(-1 + i)

Solution:

Applying the identity (z1 + z2)(z1 – z2) = z12 – z22, we get
(-1 – i)(-1 + i)
= (-1)2 – i2
Substituting the value of i2 = -1,
1 + 1
=2

Simplifying

Adding

Like the real numbers, the complex numbers follow all the properties of addition:

  • Closure Property: z1 + z2 = (a1 + a2) + i(b1 + b2)
  • Commutative Property: z1 + z2 = z2 + z1
  • Associative Law: (z1 + z2) + z3 = z1 + (z2 + z3)
  • Additive Identity: z + 0 = 0 + z = z
  • Additive Inverse: z + (-z) = (-z) + z = 0

Let us add two complex numbers z1 = 5 + 3i and z2 = -2 – i

 (5 + 3i) + (-2 – i)

= 5 + 3i – 2 – i

= (5 – 2) + (3i – i)

= 3 + 2i

Thus, adding z1 = a1 + ib1 and z2 = a2 + ib2, we get

z1 + z2 = (a1 + a2) + i(b1 + b2)

Add: 2 + 5i and 7 – 3i

Solution:

(2 + 5i) + (7 – 3i)
= 2 + 5i + 7 – 3i
= (2 + 7) + (5 – 3)i = 9 + 2i

Subtracting

Again, subtracting the previous two numbers 5 + 3i and -2 – i,

(5 + 3i) – (-2 – i)

= 5 + 3i + 2 + i

= (5 + 2) + (3i + i)

= 7 + 4i

The difference of the complex numbers z1 = a1 + ib1 and z2 = a2 + ib2 is of the form

z1 – z2 = (a1 + ib1) – (a2 + ib2) = (a1 – a2) + i(b1 – b2)

Thus, the subtraction of complex numbers follows the closer property.

Subtract (1 + 2i) from (8 + 5i).

Solution:

(8 + 5i) – (1 + 2i)
= (8 – 1) + (5i – 2i)
= 7 + 3i

Multiplying

To get the product of two complex numbers, we use the FOIL method, similar to the multiplication of two binomials. Here, we often need to substitute any power of i, formed by the repeated multiplication of i with itself.

Power of i (Iota)

  • i = ${\sqrt{i}}$
  • i2 = i ⋅ i = ${\sqrt{i}}$ ⋅ ${\sqrt{i}}$ = -1
  • i3  = i ⋅ i2 = i ⋅ (-1) = -i
  • i4 = (i2)2 = (-1)2 = 1

Now, on multiplying 5 + 3i and -2 – i, 

(5 + 3i)(-2 – i)

= 5(-2 – i) + 3i(-2 – i)

= -10 – 5i – 6i – 3i2

= -10 – 5i – 6i + 3

= (-10 + 3) + (-5i – 6i)

= -7 – 11i

Thus, for any two numbers z1 = a1 + ib1 and z2 = a2 + ib2, the product is

${z_{1}\cdot z_{2}}$ = (a1a2 – b1b2) + i(a1b2 + a2b1)

Like addition, multiplication of the complex numbers follows all the following properties:

  • Closure Property: ${z_{1}\cdot z_{2}}$ = (a1a2 – b1b2) + i(a1b2 + a2b1)
  • Commutative Property: ${z_{1}\cdot z_{2}=z_{2}\cdot z_{1}}$
  • Associative Law: ${\left( z_{1}\cdot z_{2}\right) \cdot z_{3}}$ = ${z_{1}\cdot \left( z_{2}\cdot z_{3}\right)}$
  • Multiplicative Identity: ${z\cdot 1=1\cdot z=z}$
  • Multiplicative Inverse: z ⋅ z-1 = z-1 ⋅ z = 1, where ${z^{-1}=\dfrac{1}{a+ib}}$

Multiply the complex numbers 3 + 4i and 1 + 2i.

Solution:

(3 + 4i)(1 + 2i)
= 3 + 4i + 6i + 8i2
= 3 + (4 + 6)i – 8
= -5 + 10i

Dividing

The division of complex numbers uses the reciprocal formula. 

If z1 = 5 + 3i is divided by z2 = -2 – i, the quotient is

${\dfrac{z_{1}}{z_{2}}}$

= ${\dfrac{5+3i}{-2-i}}$

= ${\left( 5+3i\right) \times \dfrac{1}{\left( -2-i\right) }}$

= ${\left( 5+3i\right) \times \dfrac{1}{\left( -2-i\right) }\times \dfrac{\left( -2+i\right) }{\left( -2+i\right) }}$

= ${\dfrac{\left( 5+3i\right) \times \left( -2+i\right) }{( -2) ^{2}-i^{2}}}$

= ${\dfrac{-10-6i+5i+3i^{2}}{4-i^{2}}}$

= ${\dfrac{-10-i-3}{4+1}}$

= ${\dfrac{-13-i}{5}}$

= ${\dfrac{-13}{5}-\dfrac{i}{5}}$

Thus, if z1 (= a1 + ib1) is divided by  z2 (= a2 + ib2), the quotient is

${\dfrac{z_{1}}{z_{2}}}$ = ${\dfrac{a_{1}+ib_{1}}{a_{2}+ib_{2}}}$

= ${\dfrac{\left( a_{1}+ib_{1}\right) \times \left( a_{2}-ib_{2}\right) }{a_{2}^{2}+b_{2}^{2}}}$

Divide (-3 + 3i) by (1 + i)

Solution:

The quotient is ${\dfrac{-3+3i}{1+i}}$
= ${\dfrac{-3+3i}{1+i}\times \dfrac{1-i}{1-i}}$
= ${\dfrac{\left( -3+3i\right) \left( 1-i\right) }{1^{2}-i^{2}}}$
= ${\dfrac{-3+3i+3i-3i^{2}}{1-i^{2}}}$
= ${\dfrac{-3+6i+3}{1+1}}$
= ${\dfrac{6i}{2}}$
= 3i

Last modified on February 22nd, 2024

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