Simplifying Surds to Complex Numbers

From basic surd manipulation to solving quadratic equations with complex roots

📚 Further Mathematics 🎯 Difficulty: ⭐⭐ ⏱️ Reading time: 25 minutes 📋 Edexcel A-Level

📚 Introduction

This lesson bridges the gap between surd manipulation and complex number theory.
We'll explore how the fundamental rules of surds extend naturally into the complex plane, especially when working with negative square roots.
Understanding this progression is key to solving quadratic equations that have no real solutions.
It also lays the groundwork for more advanced topics in Further Mathematics.

🎯 Learning Objectives

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

Apply surd laws intuitively and manipulate complex surds
Express numbers in the form $a + bi$ where $a, b \in \mathbb{R}$
Solve quadratic equations with complex roots in the form $\pm hi$
Understand why negative discriminants produce $\pm$ complex solutions

💡 Clarification: What's a Surd Again?

A surd is an irrational root — for example, $\sqrt{2}$ or $\sqrt{5}$. You can't simplify them into whole numbers, but they follow specific algebraic rules (surd laws).
In this lesson, we’ll extend those rules to handle square roots of negative numbers, by introducing the concept of imaginary numbers.

💡 Why Complex Numbers?

Not all equations have real solutions. For example, the equation $x^2 + 4 = 0$ has no real answer because no real number squared gives a negative result.
But when we allow the idea of $\sqrt{-1} = i$, we can solve these equations — and that's where complex numbers become essential.

🔑 Key Concepts

Revisiting Surd Laws

Before we venture into complex territory, let's refresh the fundamental surd laws that you already know. These rules will extend naturally to complex numbers:

📖 Surd Laws - Remember, they go both ways.

Law 1: $\quad \text{If } a \text{ and } b \text{ are over } 0, \text{ then }\rightarrow \sqrt{a} \times \sqrt{b} \quad \text{can be written as: } \sqrt{a \times b}$
Law 2: $\quad \text{If } a \text{ and } b \text{ are over } 0, \text{ then }\rightarrow \sqrt{\frac{a}{b}}\quad \text{can be written as: } \frac{\sqrt{a}}{\sqrt{b}}$
Law 3: $\quad \text{If } a \text{ is over } 0, \text { then }\rightarrow (\sqrt{a})^2 \quad \text{ is just } a \text{ (exponent cancels square root)}$
Law 4: $\sqrt{a^2} \rightleftharpoons |a| \text { for any real value of } a$
Law 4 extended (modulus): $\text{For complex numbers: } |z| = |a + bi| = \sqrt{a^2 + b^2}$

The Bridge to Complex Numbers

Remember the golden rule: ($\rightleftharpoons$ means it goes both ways.)

i \rightleftharpoons \sqrt{-1} \quad \text{and} \quad i^2 \rightleftharpoons -1

📝 Surd Laws in Action

📋 1st Law:

Solution:

1st law:

Simplify $\sqrt{9} \times \sqrt{16}$

\text{Using Law 1: } \sqrt{a} \times \sqrt{b} = \sqrt{a \times b}$$ $$\text{So: }$$ $$\sqrt{9} \times \sqrt{16} = \sqrt{9 \times 16} = \sqrt{144} = \boxed{12}

📋 2nd Law:

Solution:

2nd law:

Simplify $\frac{\sqrt{25}}{\sqrt{16}}$

\text{Using Law 2: } \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$$ $$\text{So: }$$ $$\frac{\sqrt{25}}{\sqrt{16}} = \boxed{\frac{5}{4}}

📋 3rd Law:

Solution:

3rd law:

Simplify: $i^2$
Technically, Law 3 applies to positive real numbers. In the case of $i = \sqrt{-1}$, this extends the idea into complex numbers.

\text{Using Law 3: } (\sqrt{a})^2 = a$$ $$\text{And remember: } i = \sqrt{-1}$$ $$\text{So: } $$ $$i^2 = \sqrt{-1}^2 = \boxed{-1}

📋 4th Law:

Solution:

4th law:

$z = 4i.$ solve: $|z|$

\text{Using Law 4 extended: } |z| = |a + bi| = \sqrt{a^2 + b^2}$$ $$\text{For } z = 4i = 0 + 4i \text{, we have } a = 0 \text{ and } b = 4$$ $$|z| = |4i| = \sqrt{0^2 + 4^2} = \sqrt{0 + 16} = \sqrt{16} = \boxed{4}

Complex Number Form

A complex number is written in the form $z = a + bi$ where:

$a$ is the real part (Re$(z)$)
$b$ is the imaginary part (Im$(z)$)
$a, b \in \mathbb{R}$ (both are real numbers)

📖 Note

Pure imaginary numbers: When $a = 0$, we have $z = bi$ (purely imaginary because $bi$ is the imaginary part of the complex number)
Real numbers: When $b = 0$, we have $z = a$ (purely real because $a$ is the real part of the complex number)
$a, b \in \mathbb{R}$: simply means that $a$ and $b$ are members of real numbers ($\mathbb{R}$)
Symbol meanings:

$\mathbb{R} \quad \rightarrow $ Set of all real numbers
$\mathbb{C} \quad \rightarrow $ Set of all complex numbers
$\mathbb{I} \quad \rightarrow $ Set of all imaginary numbers
$\mathbb{Z} \quad \rightarrow $ Set of all integers
$\mathbb{Q} \quad \rightarrow $ Set of all rational numbers
$\mathbb{N} \quad \rightarrow $ Set of all natural numbers

📝 Worked Examples

📋 Example 1: Converting to Standard Form

Question: Write $\sqrt{-49}$ in the form $a + bi$

Solution:

Step 1:

Simplify what's under the square root:

\text{Remember the first law: } \sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$$ $$-49 \text{ can be written as: } 49 \times -1$$ $$\text{So this becomes: } \sqrt{49 \times -1}

Step 2:

Apply the surd law:

\text{Now we can apply the first law: }$$ $$=\sqrt{49 \times -1} \rightarrow \sqrt{49} \times \sqrt{-1}$$ $$\text{And: } \sqrt{-1} = i$$ $$\text{So it becomes: } \sqrt{49} \times i

Final Answer:

Answer:

\text{We have: } \sqrt{49} \times i$$ $$ \sqrt{49} = 7$$ $$\text{So it becomes: } 7 \times i = 7i$$ $$\text{In the form: } a + bi \rightarrow 0 + 7i \text{ (there is no a)}

📋 Example 2: Finding roots of complex numbers using surd simplifications

Question: Solve $g^2 + 4g + 13 = 0$

Solution:

Step 1:

Use the quadratic formula (simplest method)

\text{The Quadratic formula is:}$$ $$ \frac{-(b) \pm \sqrt{(b)^2 - 4(a)(c)}}{2(a)}$$ $$\quad$$ $$\text{Here, } a = 1, b = 4, c = 13. \text{ Let's plug them in:}$$ $$g = \frac{-4 \pm \sqrt{4^2 - 4(1)(13)}}{2(1)}

Step 2:

Simplify the square root:

\text{The square root is: } \sqrt{4^2 - 4(1)(13)}$$ $$\text{Simplified: } \sqrt{16 - 52} = \sqrt{-36}$$ $$\text{Use 1st law: } \sqrt{-36} = \sqrt{36} \times \sqrt{-1}$$ $$\sqrt{-1} = i, \text{ so } \sqrt{36} \times \sqrt{-1} = \sqrt{36} \times i$$ $$\text{Final simplified surd: } 6 \times i

Final Answer:

Throw it back into the quadratic formula:

\text{The surd was: }6 \times i \equiv 6i$$ $$\text{Putting it in the quadratic formula:}$$ $$g = \frac{-4 \pm 6i}{2} = \frac{-4}{2} \pm \frac{6i}{2}$$ $$\text{Remember, the } \pm \text{ means you must do it twice:}$$ $$\text{Leave your answer in exact form:}$$ $$g = -2 + 3i \text{ and } g = -2 - 3i$$ $$\text{So: } \boxed{g = -2 + 3i, -2 - 3i}

💡Look closely

The solutions were: $\boxed{-2 + 3i}$ and $\boxed{-2 - 3i}$.
Notice how the solutions are complex conjugates of each other? This feature is important.
If a quadratic has 1 complex root, the other root will also be complex and will be its conjugate.

📋 Example 3: Understanding the $\pm$ in Complex Solutions

Question: Solve $z^2 + 36 = 0$ and explain why we get $\pm$ solutions

Solution:

Step 1:

Rearrange: $z^2 = -36$

Step 2:

Take square roots of both sides:

z = \pm\sqrt{-36}$$ $$\text{Use the identity } \sqrt{-1} = i \text{ to simplify:}$$ $$\sqrt{-36} = \sqrt{36} \cdot \sqrt{-1} = 6i$$ $$\therefore z = \pm 6i

$\therefore \quad \longrightarrow \quad $ Means 'therefore'

Step 3:

Verification - let's check both solutions work:

\text{For } z = 6i: \quad (6i)^2 = 36i^2 = 36(-1) = \boxed{-36} \quad $$ $$\text{For } z = -6i: \quad (-6i)^2 = 36i^2 = 36(-1) = \boxed{-36} \quad

Explanation:

We get $\pm$ because when we square either $+6i$ or $-6i$, we get the same result: $-36$. This is why quadratic equations often have two solutions — both the positive and negative square roots satisfy the original equation.

⚠️ Common Mistakes

🚫 Mistake 1: Forgetting the $\pm$ in square roots

What students often do wrong: Writing $z = \sqrt{-16} = 4i$ and stopping there

Why it's wrong: Every quadratic equation has two solutions (unless it's a repeated root)

How to avoid it: Always write $z = \pm\sqrt{-16} = \pm 4i$ when solving $z^2 = -16$

🚫 Mistake 2: Incorrect surd manipulation with negatives

What students often do wrong: $\sqrt{-9} \times \sqrt{-4} = \sqrt{(-9) \times (-4)} = \sqrt{36} = 6$

Why it's wrong: The surd law $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$ only works when both $a$ and $b$ are non-negative

How to avoid it: Convert to $i$ form first: $\sqrt{-9} \times \sqrt{-4} = 3i \times 2i = 6i^2 = -6$

🚫 Mistake 3: Confusing the imaginary part

What students often do wrong: For $z = 3 + 7i$, saying the imaginary part is $7i$

Why it's wrong: The imaginary part is the coefficient of $i$, not including $i$ itself

How to avoid it: $Im(z) = 7$, not $7i$

💪 Practice Problems

Try these problems to test your understanding:

🎯 Practice Questions A

Write each of the following in the form $a + bi$ where $a, b \in \mathbb{R}$, and in simplest form:
Do not use any forms of a calculator.

(A) Simplify $\sqrt{-121}$
(B) Simplify $\sqrt{-16}$ (C) Simplify $\sqrt{-55}$
(D) Solve $z^2 + 16 = 0$

🔍 Show Solution

(A)

$$\sqrt{-121} = \sqrt{121} \times \sqrt{-1} = \boxed{11i}$$

(B)

$$\sqrt{-16} = \sqrt{16} \times \sqrt{-1} = \boxed{4i}$$

(C)

$$\sqrt{-55} = \sqrt{55} \times \sqrt{-1} = \boxed{\sqrt{55}i}$$

(D)

$$z^2 + 16 = 0 \Rightarrow z^2 = -16 \Rightarrow z = \pm\sqrt{-16} = \boxed{\pm 4i}$$

🎯 Practice Questions B

Solve the quadratic equations, giving your answers in the form $a + bi$ where $b$ is simplified:
Do not use any forms of a calculator.

(A) $z^2 + 64z + 16 = 0$ Find $z$ in the form $a + bi , a, b \in \mathbb{R}$
(B) $s^2 +6s + 25 = 0$ Find $s$ in the form $a + bi , a, b \in \mathbb{R}$
(C) $t^2 + 8t + 20 = 0$ Find $t$ in the form $a + bi , a, b \in \mathbb{R}$
(D) $f(z) = z^2 - 2z + 17$. Show $(1 - 4i)$ is a solution to $f(z) = 0$.

🔍 Show Solution

(A)

$$z = \frac{-(-64) \pm \sqrt{(-64)^2 - 4(1)(16)}}{2(1)} = \frac{64 \pm \sqrt{4096 - 64}}{2} = \frac{64 \pm \sqrt{4032}}{2} $$ $$= \frac{64 \pm 8\sqrt{63}}{2} = \boxed{32 \pm 4\sqrt{63} \rightarrow \text{purely real}}$$

(B)

$$s = \frac{-6 \pm \sqrt{6^2 - 4(1)(25)}}{2(1)} = \frac{-6 \pm \sqrt{36 - 100}}{2} = \frac{-6 \pm \sqrt{-64}}{2} = \frac{-6 \pm 8i}{2}$$ $$ = \boxed{-3 + 4i}, \boxed{-3 - 4i}$$

(C)

$$t = \frac{-8 \pm \sqrt{8^2 - 4(1)(20)}}{2(1)} = \frac{-8 \pm \sqrt{64 - 80}}{2} = \frac{-8 \pm \sqrt{-16}}{2} = \frac{-8 \pm 4i}{2} $$ $$= \boxed{-4 + 2i}, \boxed{-4 - 2i}$$

(D)

$$f(1 - 4i) = (1 - 4i)^2 - 2(1 - 4i) + 17 = (1 - 8i + 16i^2) - (2 - 8i) + 17$$ $$ = (1 - 16) + (-8i + 8i) + (17 - 2) = 0$$ $$\text{So, } 1 - 4i \text{ is indeed a solution to } f(z) = 0$$

📋 Summary

🎯 Key Takeaways

Complex numbers expand the real number system by introducing $\sqrt{-1} = i$
Any complex number can be written in the form $z = a + bi$ where $a, b \in \mathbb{R}$
Quadratic equations with negative discriminants have imaginary solutions of the form $\pm hi$
The $\pm$ appears because both $+hi$ and $-hi$ square to the same value
Always convert square roots of negative numbers to $i$ form before applying surd laws

📚 What's Next?

Now that you understand the basics of complex numbers, you're ready to move on to arithmetic with complex numbers (addition, multiplication, division), visualising them on the complex plane, and applying the quadratic formula to equations with complex roots. These tools are crucial for Further Maths topics such as complex analysis and advanced algebra.

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