Multiplying and Dividing Complex Numbers Revision Notes for Leaving Cert Mathematics

Multiplying and Dividing Complex Numbers

Multiplying complex numbers

Multiplication of complex numbers follows the same process as multiplying algebraic expressions. We use the distributive property and apply the fundamental rule that $i^2 = -1$ .

The key principle is to multiply complex numbers exactly like you would multiply two algebraic expressions such as $(2x + 4)(x - 3)$ , but remember to substitute $i^2 = -1$ wherever it appears.

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Method for multiplication:

Use the distributive property (also known as FOIL for binomials)
Multiply each term in the first complex number by each term in the second
Combine like terms
Replace any $i^2$ with -1
Write the final answer in the form $a + bi$

Worked examples

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Worked Example 1: Multiply $i(3 - 2i)$

Step 1: Apply distributive property $i(3 - 2i) = 3i - 2i^2$

Step 2: Substitute $i^2 = -1$ $= 3i - 2(-1) = 3i + 2$

Step 3: Write in standard form $= 2 + 3i$

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Worked Example 2: Multiply $(2 + 3i)(-4 + 5i)$

Step 1: Apply distributive property $(2 + 3i)(-4 + 5i) = 2(-4 + 5i) + 3i(-4 + 5i)$

Step 2: Expand each term $= -8 + 10i - 12i + 15i^2$

Step 3: Combine like terms and substitute $i^2 = -1$ $= -8 - 2i + 15(-1) = -8 - 2i - 15$

Step 4: Simplify to standard form $= -23 - 2i$

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Worked Example 3: Multiply $(3 - 2i)(5 + 3i)$

Step 1: Apply distributive property $(3 - 2i)(5 + 3i) = 3(5 + 3i) - 2i(5 + 3i)$

Step 2: Expand $= 15 + 9i - 10i - 6i^2$

Step 3: Combine like terms and substitute $i^2 = -1$ $= 15 - i - 6(-1) = 15 - i + 6$

Step 4: Simplify to standard form $= 21 - i$

The conjugate of a complex number

Definition

If $z = a + bi$ is a complex number, then $a - bi$ is called the complex conjugate of $z$ , written as $\overline{z}$ .

To find the conjugate of a complex number, change the sign of the imaginary part only.

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Examples of conjugates:

The conjugate of $3 - 4i$ is $3 + 4i$
The conjugate of $-2 + 6i$ is $-2 - 6i$
The conjugate of $5 + 7i$ is $5 - 7i$

Key properties of conjugates

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When you add or multiply a complex number with its conjugate, the result is always a real number.

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Worked Example: If $z = 3 - 4i$ , find $z + \overline{z}$ and $z \times \overline{z}$

For $z + \overline{z}$ : $z + \overline{z} = (3 - 4i) + (3 + 4i)$ $= 3 - 4i + 3 + 4i = 6 + 0i = 6$ (a real number)

For $z \times \overline{z}$ : $z \times \overline{z} = (3 - 4i)(3 + 4i)$ $= 9 + 12i - 12i - 16i^2$ $= 9 + 0i + 16 = 25$ (a real number)

Powers of i

The powers of $i$ follow a cyclic pattern that repeats every four terms:

Power	Value	Calculation
$i^1$	$i$	$i$
$i^2$	$-1$	$i \times i = -1$
$i^3$	$-i$	$i^2 \times i = (-1) \times i = -i$
$i^4$	$1$	$i^3 \times i = (-i) \times i = -i^2 = -(-1) = 1$
$i^5$	$i$	$i^4 \times i = 1 \times i = i$

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The pattern $i, -1, -i, 1$ then repeats. To find any power of $i$ , divide the exponent by 4 and use the remainder to determine which value in the cycle applies.

Dividing complex numbers

Division by a real number

When dividing a complex number by a real number, divide each term separately.

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Worked Example: Express $\frac{4 + 10i}{5}$ in the form $a + bi$

$\frac{4 + 10i}{5} = \frac{4}{5} + \frac{10i}{5} = \frac{4}{5} + 2i$

Division by a complex number

When dividing by a complex number, multiply both numerator and denominator by the conjugate of the denominator. This creates a real number in the denominator.

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Method for dividing by complex numbers:

Identify the conjugate of the denominator
Multiply both numerator and denominator by this conjugate
Expand the numerator using multiplication rules
Simplify the denominator (which becomes real)
Write in the form $a + bi$

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Worked Example: Express $\frac{6 + 2i}{2 - 3i}$ in the form $a + bi$

Step 1: The conjugate of $(2 - 3i)$ is $(2 + 3i)$

Step 2: Multiply numerator and denominator by $(2 + 3i)$ : $\frac{6 + 2i}{2 - 3i} \times \frac{2 + 3i}{2 + 3i}$

Step 3: Expand the numerator: $(6 + 2i)(2 + 3i) = 12 + 18i + 4i + 6i^2$ $= 12 + 22i + 6(-1) = 12 + 22i - 6 = 6 + 22i$

Step 4: Expand the denominator: $(2 - 3i)(2 + 3i) = 4 + 6i - 6i - 9i^2$ $= 4 - 9(-1) = 4 + 9 = 13$

Step 5: Final answer: $\frac{6 + 22i}{13} = \frac{6}{13} + \frac{22i}{13}$

Exam tips

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Critical points to remember:

Always remember that $i^2 = -1$ - this is the most important rule
When multiplying complex numbers, use the same method as algebraic multiplication
For conjugates, only change the sign of the imaginary part
When dividing by complex numbers, always multiply by the conjugate of the denominator
Check your final answers are in the form $a + bi$
Remember that $z + \overline{z}$ and $z \times \overline{z}$ always give real numbers

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Key Points to Remember:

Multiplication: Use distributive property and remember $i^2 = -1$
Conjugate: Change the sign of the imaginary part only (if $z = a + bi$ , then $\overline{z} = a - bi$ )
Addition/multiplication with conjugates: Always produces a real number result
Division by real numbers: Divide each term separately
Division by complex numbers: Multiply by conjugate of denominator to make it real

Multiplying and Dividing Complex Numbers (Leaving Cert Mathematics): Revision Notes