IB Mathematics AI AHL Complex numbers Study Notes - New Syllabus

IB Mathematics AI AHL Complex numbers Study Notes

LEARNING OBJECTIVE

Complex numbers

Key Concepts:

Complex numbers
Cartesian form
The complex plane

MAI HL and SL Notes – All topics

♦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 $.

♦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).

$\text{GDC – STEUP}$

We go to SET UP
First press d followed by (L-p) and then scroll down to Complex Mode

Then we change it to a + bi by pressing w

Now, we can solve the equation again and get all solutions

Example

$4x^3 – 4x^2 + x – 1 = 0$

Solve the cubic equation using GDC – COMPLEX Mode.

▶️Answer/Explanation

Solution:

First, press e to CLEAR all entries, then type in the new coefficients,

GDC Input

Press q for SOLVE

GDC Solution

♦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 $.

Find Addition, Subtraction , multiplication of both z and w.

▶️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

Simplify the following

$ (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).

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 $

♦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} $

The Argand diagram (also known as the complex plane) is a geometric representation of complex numbers where: 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

Find Modulus of all the points.

▶️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

Let the complex number be defined as:

$z = 3 + 4i$

Plot the complex numbers $z$, $\overline{z}$, $-z$, and $-\overline{z}$ on an Argand diagram. Label each point clearly. Find the modulus of each of the complex numbers and state any observations.

▶️Answer/Explanation

Solution:

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

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).

♦What are Complex Roots?

A quadratic equation can have:

Two distinct real roots.
One repeated real root.
No real roots (when the graph does not touch or cross the x-axis).

When a quadratic has no real roots, the discriminant (Δ) is negative:

$\Delta = b^2 – 4ac < 0$

In this case, taking the square root of a negative number introduces complex numbers:

$i = \sqrt{-1}$

Complex numbers are used to express these non-real solutions.

♦How to Solve a Quadratic with Complex Roots

A quadratic equation has the form:

$ax^2 + bx + c = 0$

Methods:

Quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$

If $\sqrt{b^2 – 4ac}$ is negative, use:

$\sqrt{-k} = \sqrt{k} \cdot i$

Completing the square: Also works for complex roots.

Special case: If the equation is of the form $ax^2 + b = 0$, solve by rearranging:

$ax^2 = -b \quad \Rightarrow \quad x = \pm \sqrt{\frac{-b}{a}} = \pm \sqrt{\frac{b}{a}} \cdot i$

♦ Complex Conjugate Roots

If coefficients are real, complex roots come in conjugate pairs:

If $z = p + qi$, then $z^ = p – qi$ is also a root.
These roots are symmetric about the real axis.

The real part of the complex roots equals the x-coordinate of the turning point of the graph:

$x = -\frac{b}{2a}$

If the coefficients are not all real, the roots may not be conjugates.

♦ Factorising Quadratics with Complex Roots

Given a quadratic:

$az^2 + bz + c = 0$

With real coefficients and complex roots $z = p + qi$ and $z^ = p – qi$, we can factorise as:

$az^2 + bz + c = a(z – (p + qi))(z – (p – qi))$

Using the identity:

$(z – (p + qi))(z – (p – qi)) = (z – p)^2 + q^2$

This gives:

$az^2 + bz + c = a[(z – p)^2 + q^2]$

Example

Solve the quadratic equation

$x^2 + 4x + 5 = 0$

Also, write it in factorised form using its complex roots.

▶️Answer/Explanation

Solution:

Since the discriminant is negative, the equation has no real roots and two complex roots.

$
x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}
$

$
x = \frac{-4 \pm \sqrt{(4)^2 – 4(1)(5)}}{2(1)}
$

$
x = \frac{-4 \pm \sqrt{16 – 20}}{2}
$

$
x = \frac{-4 \pm \sqrt{-4}}{2}
$

$
x = \frac{-4 \pm 2i}{2}
$

$
x = -2 \pm i
$

So the roots are $x = -2 + i$ and $x = -2 – i$, which are complex conjugates.

$
x^2 + 4x + 5 = (x + 2 – i)(x + 2 + i)
$

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IB Mathematics AI AHL Complex numbers Study Notes - New Syllabus

IB Mathematics AI AHL Complex numbers Study Notes

COMPLEX NUMBERS

COMPLEX NUMBERS – BASIC OPERATIONS

COMPLEX NUMBERS – CONJUGATE ,MODULUS , EQUALITY

THE COMPLEX PLANE

Complex Roots of Quadratics

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