AOS3 Topic 1: Complex Number Basics

Definition

A complex number is a number that can be expressed in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit, satisfying the equation \(i^{2} = -1\).

Real and Imaginary Parts

In a complex number \(a + bi\), \(a\) is called the real part, and \(b\) is called the imaginary part. The real part represents the horizontal component, and the imaginary part represents the vertical component.

Powers of Imaginary Unit(iota)

Powers of the imaginary unit, denoted by \(i\), follow a cyclic pattern. When \(i\) is raised to powers of integers, it repeats in a cycle of four:

  • \(i^{1} = i\)
  • \(i^{2} = -1\)
  • \(i^{3} = -i\)
  • \(i^{4} = 1\)

This cycle repeats for higher powers of i.

Complex Number Operations

Addition:

The sum of two complex numbers \(z_1 = a_1 + b_1i\) and \(z_2 = a_2 + b_2i\) is given by:

\(z_1 + z_2 = (a_1 + a_2) + (b_1 + b_2)i\)

Subtraction:

The difference of two complex numbers \(z_1 = a_1 + b_1i\) and \(z_2 = a_2 + b_2i\) is given by:

\(z_1 - z_2 = (a_1 - a_2) + (b_1 - b_2)i\)

Multiplication:

The product of two complex numbers \(z_1 = a_1 + b_1i\) and \(z_2 = a_2 + b_2i\) is given by:

\(z_1 \cdot z_2 = (a_1a_2 - b_1b_2) + (a_1b_2 + a_2b_1)i\)

Division:

The quotient of two complex numbers \(z_1 = a_1 + b_1i\) and \(z_2 = a_2 + b_2i\) is given by:

\(z_1 ÷ z_2 = \frac{{(a_1a_2 + b_1b_2) + (a_2^2 + b_2^2)}}{{(a_2^2 + b_2^2)}} + \frac{{(a_2a_1 - b_1b_2) + (a_2^2 + b_2^2)}}{{(a_2^2 + b_2^2)}}i\)

Examples

Complex Number Real Part Imaginary Part
\(2 + 3i\) \(2\) \(3i\)
\(-4 - 2i\) \(-4\) \(-2i\)
\(5i\) \(0\) \(5i(Purely Imaginary)\)
\(6\) \(6\) \(0(Purely Real)\)
\(3 + 4i\) \(3\) \(4i\)


Created with GeoGebra®, by Ben Sparks, Link



Complex Conjugate

The complex conjugate of a complex number \(a + bi\) is denoted by \(a - bi\). It involves changing the sign of the imaginary part.

Created with GeoGebra®, by 서문정, Pratima Nayak, Link


Modulus of a Complex Number

The modulus of a complex number \(a + bi\) is denoted by \(|a + bi|\) and represents the distance of the complex number from the origin in the complex plane. It is calculated as \(\sqrt{a^{2} + b^{2}}.


Identities

  1. \( (z_1 + z_2)^2 = (z_1)^2 + (z_2)^2 + 2z_1 \times z_2 \)
  2. \( (z_1 - z_2)^2 = (z_1)^2 + (z_2)^2 - 2z_1 \times z_2 \)
  3. \( (z_1)^2 - (z_2)^2 = (z_1 + z_2)(z_1 - z_2) \)
  4. \( (z_1 + z_2)^3 = (z_1)^3 + 3(z_1)^2z_2 + 3(z_2)^2z_1 + (z_2)^3 \)
  5. \( (z_1 - z_2)^3 = (z_1)^3 - 3(z_1)^2z_2 + 3(z_2)^2z_1 - (z_2)^3 \)


Properties

The properties of complex numbers are as follows:

  1. The addition of two conjugate complex numbers results in a real number.
  2. The multiplication of two conjugate complex numbers also results in a real number.
  3. If \( x \) and \( y \) are real numbers and \( x + yi = 0 \), then \( x = 0 \) and \( y = 0 \).
  4. If \( p, q, r, \) and \( s \) are real numbers and \( p + qi = r + si \), then \( p = r \) and \( q = s \).
  5. Complex numbers obey the commutative law of addition and multiplication: \( z_1 + z_2 = z_2 + z_1 \) and \( z_1 \cdot z_2 = z_2 \cdot z_1 \).
  6. Complex numbers obey the associative law of addition and multiplication: \( (z_1 + z_2) + z_3 = z_1 + (z_2 + z_3) \) and \( (z_1 \cdot z_2) \cdot z_3 = z_1 \cdot (z_2 \cdot z_3) \).
  7. Complex numbers obey the distributive law: \( z_1 \cdot (z_2 + z_3) = z_1 \cdot z_2 + z_1 \cdot z_3 \).
  8. If the sum of two complex numbers is real and the product of two complex numbers is also real, then these complex numbers are conjugates of each other.
  9. For any two complex numbers \( z_1 \) and \( z_2 \), \( |z_1 + z_2| \leq |z_1| + |z_2| \).
  10. The result of the multiplication of two complex numbers and its conjugate value should result in a complex number and it should be a positive value.


Argument of Complex Numbers

The argument of a complex number refers to the angle that the vector representing the complex number makes with the positive real axis in the complex plane. It is usually measured in radians or degrees counterclockwise from the positive real axis.

For a complex number \( z = a + bi \), where \( a \) and \( b \) are real numbers and \( i \) is the imaginary unit, the argument (denoted as \( \text{arg}(z) \)) can be calculated using the arctangent function as follows:

\( \text{arg}(z) = \arctan\left(\frac{b}{a}\right) \)

However, since the arctangent function has a range of \( (-\pi, \pi] \) or \( (-180^\circ, 180^\circ] \), depending on the convention, the argument may fall in that range. To express the argument more uniquely, we often use the principal value of the argument, which is the value within the range \( (-\pi, \pi] \) or \( (-180^\circ, 180^\circ] \).


Argand Diagram

An Argand diagram is a visual representation of complex numbers in the complex plane. In this diagram, the horizontal axis represents the real part of the complex number, while the vertical axis represents the imaginary part. By plotting complex numbers as points in this plane, we can visually understand their relationships, such as addition, subtraction, multiplication, and division.


Try moving the complex number by dragging the a and b bars to see how the radius and arg change.


Created with GeoGebra®, by Folens Publishers, Link

Example 1

Find the conjugate of \( z_1 - z_2 \) if \( z_1 = 2 + 3i \) and \( z_2 = 5 + 2i \).

Solution:

Given,

\( z_1 = 2 + 3i \)

\( z_2 = 5 + 2i \)

\( z_1 - z_2 = (2 + 3i) - (5 + 2i) \)

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

\( = -3 + i \)

As we know the conjugate of \( z = x + iy = x - iy \).

Conjugate of \( z_1 - z_2 = -3 - i \)

Example 2

If \( z = 2 - 3i \), then find \( z^2 \).

Solution:

Given,

\( z = 2 - 3i \)

\( z^2 = z \cdot z \)

\( = (2 - 3i)(2 - 3i) \)

\( = 2(2) - 2(3i) - (3i)(2) + (3i)(3i) \)

\( = 4 - 6i - 6i + 9i^2 \) {since \( i^2 = -1 \)}

\( = 4 - 12i + 9(-1) \)

\( = 4 - 12i - 9 \)

\( = -5 - 12i \)

Therefore, \( z^2 = -5 - 12i \).

Example 3

2. Suppose \( z = (2 - i)^2 + \frac{7 - 4i}{2 + i} - 8 \), express \( z \) in the form of \( x + iy \) such that \( x \) and \( y \) are real numbers.

Solution:

Given,

$$ z = (2 - i)^2 + \frac{{7 - 4i}}{{2 + i}} - 8 $$

$$ = (2)^2 - 2(2)(i) + (i^2) + \left[\frac{{(7 - 4i)(2 - i)}}{{(2 + i)(2 - i)}}\right] - 8 $$

$$ = (4 - 4i - 1) + \left[\frac{{14 - 7i - 8i + 4i^2}}{{4 - i^2}}\right] - 8 $$

$$ = (3 - 4i) + \left[\frac{{14 - 15i - 4}}{{4 + 1}}\right] - 8 $$ {since \(i^2 = -1\)}

$$ = (3 - 4i) + \left[\frac{{10 - 15i}}{5}\right] - 8 $$

$$ = (3 - 4i) + (2 - 3i) - 8 $$

$$ = -3 - 7i $$

This is of the form \( x + iy \) such that \( x = -3 \) and \( y = -7 \).

Example 4

Simplify \( \frac{2+3i}{1-2i} - \frac{4-i}{3+i} \).

To simplify the expression \( \frac{2 + 3i}{1 - 2i} - \frac{4 - i}{3 + i} \), we'll start by rationalizing the denominators.

For the first fraction, we multiply both the numerator and the denominator by the conjugate of \( 1 - 2i \), which is \( 1 + 2i \). Similarly, for the second fraction, we multiply both the numerator and the denominator by the conjugate of \( 3 + i \), which is \( 3 - i \).

Here's how we do it step by step:

For the first fraction: \[ \frac{2 + 3i}{1 - 2i} \times \frac{1 + 2i}{1 + 2i} = \frac{(1-2i)(1+2i)}{(2+3i)(1+2i)} \]

For the second fraction: \[ \frac{4 - i}{3 + i} \times \frac{3 - i}{3 - i} = \frac{(3+i)(3-i)}{(4-i)(3-i)} \]

Now, let's simplify each fraction:

For the first fraction: \[ = \frac{2 + 4i + 3i + 6i^2}{2} = \frac{1-2i+2i-4i}{2} = \frac{2 + 7i - 6}{1 - 4i} = \frac{1 + 4}{5} - \frac{4 + 7i}{5} = \frac{5}{5} - \frac{4 + 7i}{5} \]

For the second fraction: \[ = \frac{9-3i+3i-i^2}{12-4i-3i+i^2} = \frac{9-3i+3i-(-1)}{12-7i-1} = \frac{9-(-1)}{12-7i-1} = \frac{9-(-1)}{10} = \frac{10}{10} - \frac{11-7i}{10} \]

Now, we have: \[ \frac{-4 + 7i}{5} - \frac{10}{10} - \frac{11 - 7i}{10} \]

To add or subtract complex numbers, we need to have a common denominator. In this case, the common denominator is \( 10 \).

So, our expression becomes: \[ \frac{(-8 + 14i) - (11 - 7i)}{10} \]

Now, we can combine the terms: \[ = \frac{( -8 + 14i ) - ( 11 - 7i )}{10} = \frac{10(-8+14i)-(11-7i)}{10} = \frac{-8 + 14i - 11 + 7i}{10} = \frac{-19 + 21i}{10} \]

Therefore, the simplified expression is \( \frac{-19 + 21i}{10} \).

Example 5

If \( z_1 = 2 + 8i \) and \( z_2 = 1 - i \), then find \( |z_1/z_2| \).

Solution:

Given,

\( z_1 = 2 + 8i \) and \( z_2 = 1 - i \)

\( \frac{z_1}{z_2} = \frac{2 + 8i}{1 - i} \)

\( = \frac{(2 + 8i)(1 + i)}{(1 - i)(1 + i)} \)

\( = \frac{2 + 2i + 8i + 8i^2}{1 - i^2} \)

\( = \frac{(2 + 10i - 8)}{(1 + 1)} \) {since \( i^2 = -1 \}

\( = \frac{(-6 + 10i)}{2} \)

\( = -3 + 5i \)

Now, \( |z_1/z_2| = \sqrt{(-3)^2 + (5)^2} \)

\( = \sqrt{9 + 25} \)

\( = \sqrt{34} \)

Example 6

If \( |z^2 - 1| = |z^2| + 1 \), then show that \( z \) lies on an imaginary axis.

Solution:

Let \( z = x + iy \) be the complex number.

Now, \( z^2 = z \cdot z = (x + iy)(x + iy) = x^2 + ixy + ixy + (iy)^2 = x^2 + 2ixy - y^2 \) {since \( i^2 = -1 \)}

So, \( z^2 - 1 = x^2 + 2ixy - y^2 - 1 = (x^2 - y^2 - 1) + i(2xy) \)

Thus, \( |z^2 - 1| = \sqrt{(x^2 - y^2 - 1)^2 + (2xy)^2} = \sqrt{(x^2 - y^2 - 1)^2 + 4x^2y^2} \)

Also, \( |z|^2 + 1 = [\sqrt{x^2 + y^2}]^2 + 1 = x^2 + y^2 + 1 \)

Given that \( |z^2 - 1| = |z^2| + 1 \)

So, \( \sqrt{(x^2 - y^2 - 1)^2 + 4x^2y^2} = x^2 + y^2 + 1 \)

Squaring on both sides, we get:

\( (x^2 - y^2 - 1)^2 + 4x^2y^2 = (x^2 + y^2 + 1)^2 \)

\( [x^2 - (y^2 + 1)]^2 + 4x^2y^2 = [x^2 + (y^2 + 1)]^2 \)

\( [x^2 - (y^2 + 1)]^2 - [x^2 + (y^2 + 1)]^2 + 4x^2y^2 = 0 \)

As we know, \( (a - b)^2 - (a + b)^2 = -4ab \),

\( -4x^2(y^2 + 1) + 4x^2y^2 = 0 \)

\( -4x^2y^2 - 4x^2 + 4x^2y^2 = 0 \)

\( 4x^2 = 0 \)

\( x = 0 \)

Therefore, \( z \) lies on the \( y \)-axis.

Example 7

Simplify \( i^{80} \)

Solution:

We know that,

  • \( i^1 = i \)
  • \( i^2 = -1 \)
  • \( i^3 = -i \)
  • \( i^4 = 1 \)

The powers of \( i \) repeat in a cycle of four: \( i \), \( -1 \), \( -i \), \( 1 \).

Since \( 80 = 4 \times 20 \), we can express \( i^{80} \) as \( (i^4)^{20} \). Since \( i^4 = 1 \), we have:

\( i^{80} = (i^4)^{20} = 1^{20} = 1 \)

Therefore, \( i^{80} = 1 \).

Example 8

Find real x and y if \( (x - iy)(3 + 5i) \) is the conjugate of -6 - 24i.


Solution:

Given:

\( (x - iy)(3 + 5i) = 3x + 5ix - 3iy - 5yi^2 \)

\( = 3x + i(5x - 3y) + 5y \) {since \( i^2 = -1 \)}

\( = (3x + 5y) + i(5x - 3y) \)

Given that \( (x - iy)(3 + 5i) \) is the conjugate of -6 - 24i.

Here, the conjugate of -6 - 24i = -6 + 24i.

So, \( 3x + 5y = -6 \)

\( 5x - 3y = 24 \)

Solving these two equations, we get; \( x = 3 \) and \( y = -3 \).

Example 9

If \( |z + 1| = z + 2 (1 + i) \), then find \( z \).

Solution:

Let \( z = x + iy \) be the complex number.

Given,

\( |z + 1| = z + 2 (1 + i) \)

⇒ \( |x + iy + 1| = x + iy + 2 (1 + i) \)

We know,

\( |z| = \sqrt{x^2 + y^2} \)

\( \sqrt{(x + 1)^2 + y^2} = (x + 2) + i(y + 1) \)

Comparing real and imaginary parts,

⇒ \( \sqrt{(x + 1)^2 + y^2} = x + 2 \)

And \( 0 = y + 2 \)

⇒ \( y = -2 \)

Substituting the value of \( y \) in \( \sqrt{(x + 1)^2 + y^2} = x + 2 \), we get;

\( (x + 1)^2 + (-2)^2 = (x + 2)^2 \)

\( x^2 + 2x + 1 + 4 = x^2 + 4x + 4 \)

⇒ \( 2x = 1 \)

⇒ \( x = \frac{1}{2} \)

Therefore, \( z = x + iy = \frac{1}{2} - 2i \).

Exercise &&1&& (&&1&& Question)

Find the conjugate of \( z = -1 + 5i \).

1
Submit

Exercise &&2&& (&&1&& Question)

Find the product of the complex numbers \( (2 + 3i) \) and \( (4 - 5i) \).

2
Submit

Exercise &&3&& (&&1&& Question)

Find the modulus of the complex number \( z = 3 + 4i \).

3
Submit

Exercise &&4&& (&&1&& Question)

Given the complex number \( z = 3 - 2i \),Find the real and imaginary part.

4
Submit

Exercise &&5&& (&&1&& Question)

Find the value of \( i^{20} \).

5
Submit

Exercise &&6&& (&&1&& Question)

If \(z_1\) and \(z_2\) are \(1 - i\), \(-2 + 4i\), respectively. Find \(Im( \frac{z_1.z_2}{z_1}) \)

6
Submit

Exercise &&7&& (&&1&& Question)

Find the argument of the complex number \( z = 2 + 3i \).

7
Submit