Properties and Arithmetic Revision Notes for AQA A-Level Further Maths

Properties and Arithmetic

Introduction to imaginary numbers

Certain equations cannot be solved using only real numbers. For example, the equation $x^2 = -1$ has no real solutions because no real number squared gives a negative result. Historically, mathematicians encountered such equations when solving real-world problems and needed a way to express these solutions.

To overcome this limitation, mathematicians introduced the imaginary unit, denoted by $i$ , which represents the square root of negative one. This seemingly simple idea allows all polynomial equations to have solutions, making it an essential tool in advanced mathematics.

The imaginary unit i

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Definition: The imaginary unit $i$ is defined as:

$i = \sqrt{-1}$

From this definition, we can derive an important property that you will use frequently.

Key property:

$i^2 = -1$

This follows directly from the definition since $i^2 = (\sqrt{-1})^2 = -1$ .

Writing square roots of negative numbers

Once we have defined $i = \sqrt{-1}$ , we can express the square root of any negative number in terms of $i$ .

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Method: For any positive real number $a$ :

$\sqrt{-a} = \sqrt{a} \cdot \sqrt{-1} = i\sqrt{a}$

Remember: $\sqrt{-a} = i\sqrt{a}$ , not $-\sqrt{a}$ . The $i$ comes in front of the square root.

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Worked Example 1: Solving equations with imaginary solutions

Solve the equation $x^2 = -9$ .

Solution:

Taking the square root of both sides:

$x = \pm\sqrt{-9}$

We can rewrite $\sqrt{-9}$ as $\sqrt{9} \cdot \sqrt{-1}$ :

$x = \pm\sqrt{9} \cdot \sqrt{-1}$

Since $\sqrt{9} = 3$ and $\sqrt{-1} = i$ :

$x = \pm 3i$

Therefore, the solutions are x = 3i and x = -3i.

Solving more complex equations

For equations like $(x-3)^2 = -5$ , we follow the same approach.

Taking the square root of both sides: $x - 3 = \pm\sqrt{-5}$

Since $\sqrt{-5} = \sqrt{5} \cdot \sqrt{-1} = i\sqrt{5}$ : $x - 3 = \pm i\sqrt{5}$

Therefore: $x = 3 \pm i\sqrt{5}$

The solutions are $x = 3 + i\sqrt{5}$ and $x = 3 - i\sqrt{5}$ .

Complex numbers

When we combine real numbers with imaginary numbers, we create complex numbers.

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Definition: A complex number is a number that can be written in the form:

$a + bi$

where $a, b \in \mathbb{R}$ (both $a$ and $b$ are real numbers).

The value $a$ is called the real part of the complex number
The value $b$ is called the imaginary part of the complex number (note: it's just $b$ , not $bi$ )

Notation: The set of all complex numbers is denoted by $\mathbb{C}$ .

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Important: Every real number is also a complex number (with $b = 0$ ).

For example, $5 = 5 + 0i$ .

Similarly, every imaginary number is a complex number (with $a = 0$ ).

For example, $3i = 0 + 3i$ .

Arithmetic with complex numbers

Complex numbers can be added, subtracted, and multiplied by constants in the same way as algebraic expressions. The key is to keep real parts and imaginary parts separate.

Addition and subtraction

When adding or subtracting complex numbers, combine the real parts together and the imaginary parts together.

Multiplication by a constant

To multiply a complex number by a real constant, multiply both the real part and the imaginary part by that constant.

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Worked Example 2: Simplifying complex expressions

Simplify the expression $3(4-7i) - 2(3-2i)$ .

Solution:

First, multiply each complex number by its constant:

$3(4-7i) - 2(3-2i) = (12 - 21i) - (6 - 4i)$

Expand the brackets (be careful with the minus sign):

$= 12 - 21i - 6 + 4i$

Collect the real parts together: $12 - 6 = 6$

Collect the imaginary parts together: $-21i + 4i = -17i$

Therefore:

$3(4-7i) - 2(3-2i) = \boxed{6 - 17i}$

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Exam tip: Always write your final answer in the form $a + bi$ with the real part first. If there's no real part, write $0 + bi$ or just $bi$ .

Multiplying complex numbers

When multiplying complex numbers together, expand the brackets as you would with algebraic expressions, then use the fundamental property i² = -1 to simplify.

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Worked Example 3: Solving quadratic equations with complex solutions

Solve the equation $(x+7)^2 = -16$ .

Solution:

Taking the square root of both sides:

$x + 7 = \pm\sqrt{-16}$

Rewrite $\sqrt{-16}$ as $\sqrt{16} \cdot \sqrt{-1}$ :

$x + 7 = \pm\sqrt{16} \cdot \sqrt{-1}$

Since $\sqrt{16} = 4$ and $\sqrt{-1} = i$ :

$x + 7 = \pm 4i$

Therefore:

$x = -7 \pm 4i$

The solutions are x = -7 + 4i and x = -7 - 4i.

Complex conjugates

When working with fractions that have complex numbers in the denominator, we need a method to simplify them. This is similar to rationalising surds (removing square roots from denominators).

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Definition: The complex conjugate of the complex number $a + bi$ is $a - bi$ .

In other words, to find the complex conjugate, keep the real part the same and change the sign of the imaginary part.

Simplifying fractions with complex denominators

Key method: To simplify $\dfrac{1}{a+bi}$ , multiply both the numerator and denominator by the complex conjugate $a - bi$ .

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Why this works: When we multiply $(a+bi)(a-bi)$ , the imaginary parts cancel out, leaving only a real number in the denominator.

The general result is: $(a+bi)(a-bi) = a^2 + b^2$

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Worked Example 4: Rationalising complex denominators

Simplify $\dfrac{1+3i}{1-2i}$ , giving your answer in the form $a + bi$ .

Solution:

Multiply both numerator and denominator by the complex conjugate of the denominator. The conjugate of $1-2i$ is $1+2i$ :

$\dfrac{1+3i}{1-2i} = \dfrac{(1+3i)(1+2i)}{(1-2i)(1+2i)}$

Expand the numerator:

$(1+3i)(1+2i) = 1 + 2i + 3i + 6i^2 = 1 + 5i + 6i^2$

Using $i^2 = -1$ :

$= 1 + 5i + 6(-1) = 1 + 5i - 6 = -5 + 5i$

Expand the denominator:

$(1-2i)(1+2i) = 1 + 2i - 2i - 4i^2 = 1 - 4i^2$

Using $i^2 = -1$ :

$= 1 - 4(-1) = 1 + 4 = 5$

Therefore:

$\dfrac{1+3i}{1-2i} = \dfrac{-5 + 5i}{5} = \dfrac{-5}{5} + \dfrac{5i}{5} = \boxed{-1 + i}$

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Exam tip: Always check your final answer is in the form $a + bi$ . The denominator should be a positive real number.

Solving equations with complex numbers

When solving equations that involve complex numbers, we use a powerful strategy based on the fact that two complex numbers are equal if and only if their real parts are equal AND their imaginary parts are equal.

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Strategy for solving equations:

Write all numbers and expressions in the form $a + bi$
Equate the real parts on both sides of the equation
Equate the imaginary parts on both sides of the equation
Solve the resulting equations simultaneously

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Worked Example 5: Finding unknown real values

Find the real numbers $a$ and $b$ such that $(a+5i)(2-i) = 9+bi$ .

Solution:

Step 1: Expand the left-hand side:

$(a+5i)(2-i) = 2a - ai + 10i - 5i^2$

Step 2: Simplify using $i^2 = -1$ :

$= 2a - ai + 10i - 5(-1)$

$= 2a - ai + 10i + 5$

Step 3: Collect real and imaginary terms:

$= (2a + 5) + (10 - a)i$

Step 4: We now have $(2a+5) + (10-a)i = 9 + bi$

Equate the real parts:

$2a + 5 = 9$

$2a = 4$

$a = 2$

Equate the imaginary parts:

$10 - a = b$

Substitute $a = 2$ :

$b = 10 - 2 = 8$

Therefore, a = 2 and b = 8.

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Worked Example 6: Finding complex solutions

Find the complex numbers $z$ such that $z^2 = 5 + 12i$ .

Solution:

Step 1: Let $z = a + bi$ where $a, b \in \mathbb{R}$

Then $z^2 = (a+bi)^2 = a^2 + 2abi + b^2i^2$

Step 2: Simplify using $i^2 = -1$ :

$z^2 = a^2 + 2abi + b^2(-1) = a^2 + 2abi - b^2$

Rearrange:

$z^2 = (a^2 - b^2) + 2abi$

Step 3: We have $(a^2 - b^2) + 2abi = 5 + 12i$

Equate the real parts:

$a^2 - b^2 = 5 \quad \text{...(1)}$

Equate the imaginary parts:

$2ab = 12$

$ab = 6$

$a = \dfrac{6}{b} \quad \text{...(2)}$

Step 4: Substitute equation (2) into equation (1):

$\left(\dfrac{6}{b}\right)^2 - b^2 = 5$

$\dfrac{36}{b^2} - b^2 = 5$

Multiply through by $b^2$ :

$36 - b^4 = 5b^2$

$b^4 + 5b^2 - 36 = 0$

Step 5: This is a quadratic in $b^2$ . Let $u = b^2$ :

$u^2 + 5u - 36 = 0$

$(u + 9)(u - 4) = 0$

So $u = -9$ or $u = 4$

Step 6: Since $b$ is real, $b^2$ must be non-negative, so $b^2 = 4$ .

Therefore $b = 2$ or $b = -2$ .

Step 7: From equation (2), $a = \dfrac{6}{b}$ :

If $b = 2$ , then $a = 3$
If $b = -2$ , then $a = -3$

Therefore, z = 3 + 2i or z = -3 - 2i.

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Exam tip: When solving equations involving complex numbers, always remember to give both solutions if they exist. Check your answer by substituting back into the original equation.

Common exam mistakes to avoid

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Common Mistakes to Watch Out For:

Forgetting i² = -1: This is the most fundamental property. Always replace i² with -1 when simplifying.
Incorrect conjugate: The conjugate of $a + bi$ is $a - bi$ (change only the sign of the imaginary part).
Not writing in standard form: Always express your final answer as $a + bi$ with the real part first.
Confusion about the imaginary part: The imaginary part of $3 + 5i$ is 5, not $5i$ .
Sign errors when expanding: Be especially careful with negative signs when multiplying complex numbers.

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Key Points to Remember:

The imaginary unit is defined as $i = \sqrt{-1}$ , which gives the fundamental property i² = -1
Complex numbers have the form $a + bi$ where $a$ and $b$ are real numbers; $a$ is the real part and $b$ is the imaginary part
To simplify fractions with complex denominators, multiply both numerator and denominator by the complex conjugate (change the sign of the imaginary part)
When solving equations with complex numbers, equate the real parts and imaginary parts separately, then solve simultaneously
Always write your final answer in the standard form $a + bi$ and check that your denominator is a positive real number when simplifying fractions

Properties and Arithmetic (AQA A-Level Further Maths): Revision Notes