Multiplying Complex Numbers

Master the art of multiplying complex numbers using simple bracket techniques

📚 Further Mathematics 🎯 Difficulty: ⭐⭐ ⏱️ Reading time: 20 minutes 📋 Edexcel & AQA

📚 Introduction

Multiplying complex numbers is just like expanding brackets in algebra - but with a twist! We use the same techniques you already know, but we need to remember that $i^2 = -1$. This lesson will show you how to multiply complex numbers step by step, making it feel as natural as regular algebra.

🎯 Learning Objectives

By the end of this lesson, you will be able to:

Multiply complex numbers using bracket expansion
Use the key fact that $i^2 = -1$ to simplify your answers
Express your final answers in the form $a + bi$
Handle powers of $i$ confidently

🔑 Key Concepts

The Golden Rule

Everything you need to know about multiplying complex numbers comes down to one simple fact:

⭐ The Most Important Rule

$$i^2 = -1$$

This means whenever you see $i \times i$, or $ii$, you can replace it with $-1$. This is because:

i = \sqrt{-1}$$ $$\text{So:}$$ $$i^2 = (\sqrt{-1})^2 = -1

The power of 2 cancels the square root, leaving just $-1$.

It's Just Like Expanding Brackets

When you multiply complex numbers, you expand the brackets exactly like you would with normal algebra. The only difference is that when you get $i^2$, you replace it with $-1$.

💡 Key Tip:

MAKE SURE YOU EQUATE IMAGINARY PARTS AND REAL PARTS SEPARATELY.
Ensure you are proficient with expanding brackets. If not, I'll guide you through it, but ensure you understand the basics first.
Its just like expanding brackets.
Remember if you see '$zw$' and $z$ and $w$ are given, it just means $z \times w$.
You can use FOIL or Grid method. I prefer FOIL which I will use for the examples as its easier for complex numbers.

Powers of $i$ - The Pattern

Understanding powers of $i$ makes everything easier. There's a simple pattern that repeats every 4 steps:

🔄 The $i$ Pattern

$i^1 = i$
$i^2 = -1$
$i^3 = i^2 \times i = -1 \times i = -i$
$i^4 = i^2 \times i^2 = (-1) \times (-1) = 1$
$i^5 = i^4 \times i = 1 \times i = i$ (back to the start!)

📝 Worked Examples

📋 Example 1: Basic Multiplication

Question: $z = 3 + 4i, w = 5 + 2i.$ Solve: $zw$

Solution:

Step 1:

Write clearly what the question asks

z = 3 + 4i$$ $$w = 5 + 2i$$ $$\text{So,}$$ $$zw = (3 + 4i)(5 + 2i)

Step 2:

Expand the brackets:

zw = (3 + 4i)(5 + 2i)$$ $$\text{Using FOIL:}$$ $$\text{First: } 3 \times 5 = \boxed{15}$$ $$\text{Outer: } 3 \times 2i = \boxed{6i}$$ $$\text{Inner: } 4i \times 5 = \boxed{20i}$$ $$\text{Last: } 4i \times 2i = \boxed{8i^2}

FOIL is an abbreviation of First, Outer, Inner, Last.

Step 3:

Use the fact that $i^2 = -1$ and simplify:

\text{So now we have:}$$ $$zw = 15 + 6i + 20i + 8i^{2}$$ $$\text{Use the fact: } i^{2} = -1\text{ so now we have:}$$ $$= 15 + 6i + 20i + 8(-1)$$ $$\text{which equals:}$$ $$= 15 + 6i + 20i - 8

Step 4:

Collect like terms

\text{Now add the real and imaginary parts separately:}$$ $$(15 - 8) + (6i + 20i) = \boxed{7 + 26i}

📋 Example 2: Squaring a Complex Number

Question: $g = (7 - 4i)$. find $g^2$ in the form $a + bi$

....

Solution:

Step 1:

Write the question out clearly:

g = (7 - 4i) \text{, so:} $$ $$g^2 = (7 - 4i)^2 = (7 - 4i)(7 - 4i)

Step 2:

Expand the brackets

(7 - 4i)(7 - 4i) =$$ $$\text{first: } 7 \times 7 = 49$$ $$\text{outer: } 7 \times (-4i) = -28i$$ $$\text{inner: } (-4i) \times 7 = -28i$$ $$\text{last: } (-4i) \times (-4i) = 16i^2

Step 3:

Add like-terms, simplify each term, and remember that $i^2 = -1$:

\text{We have: } 49 - 28i - 28i + 16i^2$$ $$i^2 = -1\text{, so: }16i^2 = 16(-1) = -16$$ $$\text{So now we have: } 49 - 28i - 28i - 16$$ $$\text{real and imaginary parts combined: }$$ $$49 - 28i - 28i - 16 = (49 - 16) + (-28i - 28i)

Final Answer:

49 - 16 = \boxed{33}$$ $$-28i - 28i = \boxed{-56i}$$ $$\text{So the final answer is:}$$ $$\boxed{33 - 56i}

📋 Example 3: Powers of $i$

Question: Simplify:
a) $i^3$
b) $i^4$
c) $(2i)^5$

Solution:

Part a:

$$ i = \sqrt{-1} $$ $$\text{So: } i^3 = iii = i^2 \times i = (-1)(\sqrt{-1}) = \boxed{-1i \equiv -i}$$

💡 Tip:

You can use your calculator to easily compute this.
Ensure you are in 'complex mode' and just type in $i^{\text{(any number)}}$
Casio Fx-991CW and Casio Fx-991EX highly recommended.
This symbol $\equiv$ means 'is equivalent to' or 'is the same as'.

Part b:

$$i^4 = i^2 \times i^2 = (-1) \times (-1) = \boxed{1}$$

Part c:

$$(2i)^5 \equiv 2^5 \times i^5$$ $$\text{First: } 2^5 = 32$$ $$\text{Next: } i^5 = i^4 \times i$$ $$i^4 \times i = i^2 \times i^2 \times i \quad \equiv \quad (-1)(-1)(i) = i$$ $$\text{So: } (2i)^5 = 32 \times i = 32i$$

Final Answers:

\text{a) } \boxed{-i} \quad \text{b) } \boxed{1} \quad \text{c) } \boxed{32i}

⚠️ Common Mistakes

🚫 Mistake 1: Forgetting to use $i^2 = -1$

What students often do wrong: Leave $i^2$ in their final answer

Why it's wrong: The answer should be in the form $a + bi$ with real numbers $a$ and $b$

How to avoid it: Always replace $i^2$ with $-1$ before collecting like terms

🚫 Mistake 2: Not collecting like terms properly

What students often do wrong: Leave the answer as $5 + 3i - 2 + 7i$ instead of simplifying

Why it's wrong: The final answer must be in the standard form $a + bi$

How to avoid it: Always group real parts together and imaginary parts together: $(5 - 2) + (3i + 7i) = 3 + 10i$

💪 Practice Problems

Try these problems to test your understanding:

🎯 Practice Questions A

Simplify each of the following, giving your answers in the form $a + bi$:

(A) Solve: $i^2$

(B) Solve: $8i^2$

(C) Solve: $5i \times -5i$

(D) Solve: $(5 + i)(3 + 4i)$

🔍 Show Solution

(A) $i^2 = \boxed{-1}$

(B) $8i^2 = 8(-1) = \boxed{-8}$

(C) $5i \times -5i = -25i^2 = -25(-1) = \boxed{25}$

(D) $(5 + i)(3 + 4i) = 15 + 20i + 3i + 4i^2 = 15 + 23i + 4(-1) = \boxed{11 + 23i}$

🎯 Practice Questions B

Find the following powers:

(A) Solve: $(i^6)^2$

(B) $z = 4 - 2i, w = 4i$. Find $zw$.

(C) Solve: $(4 + 5i)(2 - 3i)(1 + 9i)$

🔍 Show Solution

(A) $i^6 = i^4 \times i^2 = 1 \times (-1) = \boxed{-1}$

(B) $z = 4 - 2i, w = 4i$. Find $zw$: $zw = (4 - 2i)(4i) = 16i - 8i^2 = 16i + 8 = \boxed{8 + 16i}$

(C) $(4 + 5i)(2 - 3i)(1 + 9i) = (8 - 12i + 10i - 15i^2)(1 + 9i)$
$(8 - 12i + 10i - 15i^2)(1 + 9i) = (23 - 2i)(1 + 9i) = 23 + 207i - 2i - 18 = \boxed{5 + 205i}$

🏆 Challenge Exam Question

📋 Exam-Style Question

$z = 4 - 3i, f = 2 + 5i$, $j = z^*$.

(A) Solve: $zf$ (2 marks)

(B) Solve: $zj$ (2 marks)

(C) Hence, or further, solve: $jfz$ (4 marks)

💡 Hint: $z^* \equiv$ conjugate of z

🔍 Show Solution

(A) $$zf = (4 - 3i)(2 + 5i) = 8 + 20i - 6i - 15i^2 = 8 + 14i + 15 =$$ $$ \boxed{23 + 14i}$$

(B) $$j = z^* = 4 + 3i \text{. So, } zj = (4 - 3i)(4 + 3i) = 16 + 12i - 12i - 9i^2 = $$ $$ 16 + 9 = \boxed{25}$$

(C) $$\text{ We can use our previous answers:}$$ $$jfz = j \times f \times z = (4 + 3i)(23 + 14i)$$ $$= 4 \times 23 + 4 \times 14i + 3i \times 23 + 3i \times 14i$$ $$= 92 + 56i + 69i + 42i^2$$ $$= 92 + 125i + 42(-1)$$ $$= 92 + 125i - 42 = \boxed{50 + 125i}$$

📋 Summary

🎯 Key Takeaways

Multiplying complex numbers is just like expanding brackets in regular algebra
The golden rule: $i^2 = -1$ - use this to simplify your answers
Always express your final answer in the form $a + bi$ where $a$ and $b$ are real numbers
Powers of $i$ follow a pattern that repeats every 4: $i, -1, -i, 1, i, -1, -i, 1, ...$
When you multiply a complex number by its conjugate, you get a real number

📚 What's Next?

It was simple, wasnt it? You should now be confident with multiplying complex numbers. I suggest learning how to divide complex numbers next, its abit more tricky but my flawless guide will help you through it.
Click on me to go to my guide on dividing complex numbers.

⬆️ Back to Top