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IB Mathematics AI SL Complex numbers Study Notes - New Syllabus

IB Mathematics AI SL Complex numbers Study Notes

LEARNING OBJECTIVE

Key Concepts: 

MAI HL and SL Notes – All topics

COMPLEX NUMBERS – BASIC OPERATIONS

♦As we know, there are no real numbers of the form:
\( \sqrt{-1}, \sqrt{-4}, \sqrt{-9}, \sqrt{-5} \)

However, we agree to accept an imaginary number \( i \) such that:
\( i^2 = -1 \)
(so, in some way, the definition of \( i \) is: \( i = \sqrt{-1} \)).

♦The imaginary numbers mentioned above can be written as follows:
 Instead of \( \sqrt{-4} \), we write \( 2i \).
 Instead of \( \sqrt{-9} \), we write \( 3i \).
 Instead of \( \sqrt{-5} \), we write \( \sqrt{5}i \).

Consider now the equation:
\( x^2 – 4x + 13 = 0 \)

Since \( \Delta = -36 \), there are no real solutions. However, if we accept that \( \sqrt{\Delta} = i\sqrt{36} = 6i \), we obtain solutions of the form:
\( x = \frac{4 \pm \sqrt{\Delta}}{2} = \frac{4 \pm 6i}{2} = 2 \pm 3i \)

These “new” numbers are known as complex numbers.

In general, if \( \Delta < 0 \), the complex roots of the quadratic are given by:
\( x = \frac{-b \pm i \sqrt{|\Delta|}}{2a} \)

NOTICE for the GDC – For Casio:
 Use `shift-O` for \( i \).
 For a quadratic with \( \Delta < 0 \), you may obtain the complex roots if you use `SET UP – Complex mode: a+bi`.

♦THE DEFINITION

A number \( z \) of the form:
\( z = x + yi \)
where \( x, y \in \mathbb{R} \), is called a complex number. We also say:
 The real part of \( z \) is \( x \): \( \text{Re}(z) = x \).
 The imaginary part of \( z \) is \( y \): \( \text{Im}(z) = y \).

The set of all complex numbers is denoted by \( \mathbb{C} \). A real number \( x \) is also complex of the form \( x + 0i \) (it has no imaginary part).

♦THE CONJUGATE \( \overline{z} \)

The conjugate complex number of \( z = x + yi \) is given by:
\( \overline{z} = x – yi \)
(Sometimes, the conjugate number of \( z \) is denoted by \( z^* \).)

For \( z = 3 + 4i \), we write:
 \( \text{Re}(z) = 3 \),
 \( \text{Im}(z) = 4 \),
 \( \overline{z} = 3 – 4i \).

♦THE MODULUS \( |z| \)

The modulus of \( z = x + yi \) is defined by:
\( |z| = \sqrt{x^2 + y^2} \)

For example, if \( z = 2 + 3i \), then:
\( |z| = |2 + 3i| = \sqrt{2^2 + 3^2} = \sqrt{13} \)

Notice:
The following numbers all have the same modulus \( \sqrt{x^2 + y^2} \):
\( z = x + yi \)
 \( \overline{z} = x – yi \)
 \(-z = -x – yi \)
 \(-\overline{z} = -x + yi \)

Thus, \( 3 + 4i, 3 – 4i, -3 – 4i, -3 + 4i \) all have the same modulus:
\( \sqrt{3^2 + 4^2} = \sqrt{25} = 5 \)

Finally, observe that \( |3| = 3 \) and \( |-3| = 3 \). That is, the modulus generalizes the notion of the absolute value for real numbers.

♦EQUALITY: \( z_1 = z_2 \)

Two complex numbers are equal if they have equal real parts and equal imaginary parts. Let \( z_1 = x_1 + y_1i \) and \( z_2 = x_2 + y_2i \). Then:
\( z_1 = z_2 \quad \Leftrightarrow \quad \begin{cases} x_1 = x_2 \\ y_1 = y_2 \end{cases} \)

Thus, an equation of complex numbers must be thought of as a system of two simultaneous equations.

Example

Let \( z_1 = 3 + 4i \) and \( z_2 = a + (3b – 2)i \). Find \( a, b \) if \( z_1 = z_2 \).

▶️Answer/Explanation

Solution:

\( z_1 = z_2 \quad \Leftrightarrow \quad \begin{cases} 3 = a \\ 4 = 3b – 2 \end{cases} \quad \Leftrightarrow \quad \begin{cases} a = 3 \\ b = 2 \end{cases} \)

♦ADDITION, SUBTRACTION, MULTIPLICATION, DIVISION

The four operations for complex numbers follow the usual laws of algebra, keeping in mind that \( i^2 = -1 \).

Example

Consider the two complex numbers \( z = 7 + 4i \) and \( w = 2 + 3i \).

▶️Answer/Explanation

Solution:

Addition:
\( z + w = (7 + 4i) + (2 + 3i) = 9 + 7i \)
(Add real parts; add imaginary parts.)

Subtraction:
\( z – w = (7 + 4i) – (2 + 3i) = 5 + i \)

Multiplication:
\( zw = (7 + 4i)(2 + 3i) = 14 + 21i + 8i + 12i^2 = 14 + 21i + 8i – 12 = 2 + 29i \)

Division:
To express \( \frac{z}{w} = \frac{7 + 4i}{2 + 3i} \) in the form \( x + yi \), multiply numerator and denominator by the conjugate of the denominator \( \overline{w} = 2 – 3i \):
\( \frac{z}{w} = \frac{7 + 4i}{2 + 3i} \cdot \frac{2 – 3i}{2 – 3i} = \frac{14 – 21i + 8i – 12i^2}{4 – (3i)^2} = \frac{26 – 13i}{13} = 2 – i \)
Thus:
\( \frac{7 + 4i}{2 + 3i} = 2 – i \)

Verification:
Multiply \( (2 + 3i)(2 – i) = 4 + 6i – 2i – 3i^2 = 4 + 4i + 3 = 7 + 4i \).

♦NOTICE:
\( |z|^2 = z \overline{z} \)
Indeed, both sides are equal to \( x^2 + y^2 \). For \( z = x + yi \):
\( |z|^2 = x^2 + y^2 \)
\( z \overline{z} = (x + yi)(x – yi) = x^2 – y^2i^2 = x^2 + y^2 \)

Example

\( (3 + 4i)(3 – 4i) = ? \)
\( (1 + i)(1 – i) =? \)
\( (2 – i)(2 + i) = ? \)

▶️Answer/Explanation

Solution:

\( (3 + 4i)(3 – 4i) = 9 + 16 = 25 \)
\( (1 + i)(1 – i) = 1 + 1 = 2 \)
\( (2 – i)(2 + i) = 4 + 1 = 5 \)
The result is always a real number (the square of the modulus).

 

 

The powers of \( i \) cycle every four exponents:
\( i^0 = 1, \quad i^1 = i, \quad i^2 = -1, \quad i^3 = -i \)
\( i^4 = 1, \quad i^5 = i, \quad i^6 = -1, \quad i^7 = -i \)
\( i^8 = 1, \quad i^9 = i, \quad i^{10} = -1, \quad i^{11} = -i \)
For example:
\( i^{35} = i^{32 + 3} = i^3 = -i \)
(since 32 is a multiple of 4).

 

Example

Calculate:
(a) \( z = (2 + i)^3 \)
(b) \( w = \frac{(2 + i)^3}{1 – i} \)

▶️Answer/Explanation

Solution:

(a) Expand \( (2 + i)^3 \):
\( z = (2 + i)^2(2 + i) = (4 + 4i + i^2)(2 + i) = (3 + 4i)(2 + i) = 6 + 3i + 8i + 4i^2 = 2 + 11i \)

(b) Divide \( z \) by \( (1 – i) \):
\( w = \frac{2 + 11i}{1 – i} \cdot \frac{1 + i}{1 + i} = \frac{(2 + 11i)(1 + i)}{2} = \frac{-9 + 13i}{2} = -\frac{9}{2} + \frac{13}{2}i \)

 

Example

Find \( z \) if:
\( z(1 – i) = 2 + 11i \)

▶️Answer/Explanation

Solution:

Method A (Analytical):
Let \( z = x + yi \). Then:
\( (x + yi)(1 – i) = 2 + 11i \)
\( (x + y) + (y – x)i = 2 + 11i \)
This gives the system:
\( \begin{cases} x + y = 2 \\ y – x = 11 \end{cases} \)
Solving, \( x = -\frac{9}{2} \) and \( y = \frac{13}{2} \). Thus:
\( z = -\frac{9}{2} + \frac{13}{2}i \)

Method B (Quicker):
\( z = \frac{2 + 11i}{1 – i} \)
This is the same as part (b) of Example 6.

♦THE COMPLEX PLANE

The complex number \( z = x + yi \) can be represented on the Cartesian plane as follows:

 \( z = x + yi \) is the point \((x, y)\).
 Real part = \( x \)-coordinate, Imaginary part = \( y \)-coordinate.
 The modulus \( |z| = \sqrt{x^2 + y^2} \) is the distance from the origin.

Example

Points on the complex plane:

▶️Answer/Explanation

Solution:

\( 3 + 4i \) has modulus \( |3 + 4i| = \sqrt{25} = 5 \).
\( 3 \) (real number) has modulus \( |3| = 3 \).
 \( 4i \) has modulus \( |4i| = 4 \).
\(-5\) has modulus \( |-5| = 5 \).
\(-5 – 2i\) has modulus \( |-5 – 2i| = \sqrt{29} \).

 ♦The modulus is always the distance from the origin.

We may also think of \( z \) as a vector from the origin to the point \((x, y)\). Compare with vectors \(\begin{pmatrix} x \\ y \end{pmatrix}\) in paragraph 3.10.

NOTICE
We already know that the sets:
\( \mathbb{N} = \) natural numbers
\( \mathbb{Z} = \) integers
\( \mathbb{Q} = \) rational numbers
 \( \mathbb{R} = \) real numbers

can be represented on the real axis. We extend this representation here to the complex plane (considering an imaginary \( y \)-axis).

It also holds:
\( \mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R} \subset \mathbb{C} \)

Example

It is interesting to see the representations of \( z, \overline{z}, -z, -\overline{z} \).

 
\( z = 3 + 4i \)
\( \overline{z} = 3 – 4i \)
\(-z = -3 – 4i \)
 \(-\overline{z} = -3 + 4i \)

▶️Answer/Explanation

Solution:

All these have the same modulus \( 5 \) (distance from the origin).
\( \overline{z} \) is symmetric to \( z \) about the \( x \)-axis.
 \(-z \) is symmetric to \( z \) about the origin.

♦For real numbers:
The absolute value of \( 5 \) and \(-5\) is \( 5 \).
The conjugate of \( 5 \) is \( 5 \) itself (symmetric about \( x \)-axis).
The opposite of \( 5 \) is \(-5\) (symmetric about the origin).

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