IB Mathematics AI SL Complex numbers Study Notes - New Syllabus
IB Mathematics AI SL Complex numbers Study Notes
LEARNING OBJECTIVE
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Key Concepts:
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- IBDP Maths AI SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IBDP Maths AI SL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Maths AI HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Maths AI HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Maths AI HL- IB Style Practice Questions with Answer-Topic Wise-Paper 3
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/ExplanationSolution: \( z_1 = z_2 \quad \Leftrightarrow \quad \begin{cases} 3 = a \\ 4 = 3b – 2 \end{cases} \quad \Leftrightarrow \quad \begin{cases} a = 3 \\ b = 2 \end{cases} \) |
♦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/ExplanationSolution: Addition: Subtraction: Multiplication: Division: Verification: |
♦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 \( (3 + 4i)(3 – 4i) = ? \) ▶️Answer/ExplanationSolution: \( (3 + 4i)(3 – 4i) = 9 + 16 = 25 \)
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The powers of \( i \) cycle every four exponents: |
Example Calculate: ▶️Answer/ExplanationSolution: (a) Expand \( (2 + i)^3 \): (b) Divide \( z \) by \( (1 – i) \): |
Example Find \( z \) if: ▶️Answer/ExplanationSolution: Method A (Analytical): Method B (Quicker): |
♦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/ExplanationSolution: \( 3 + 4i \) has modulus \( |3 + 4i| = \sqrt{25} = 5 \). |
♦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 \( z, \overline{z}, -z, -\overline{z} \).
▶️Answer/ExplanationSolution: All these have the same modulus \( 5 \) (distance from 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).