Exponential Form Revision Notes for Edexcel A-Level Further Mathematics

1.2.1 Exponential Form

Overview

The exponential form of complex numbers is a powerful tool in mathematics, particularly in handling complex number operations like multiplication and division. This form is closely tied to Euler's Formula, which connects the exponential function with trigonometric functions.

Let's start by exploring the background and deriving the exponential form step-by-step.

The Exponential Function

The exponential function $e^x$ can be expanded using its Maclaurin series as:

e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots + \frac{x^r}{r!} + \ldots \quad \text{for all real or complex } x.

For real numbers, this function grows exponentially, but it can also be applied when $x$ is a complex number.

Maclaurin Series for $\sin x$ and $\cos x$

Similarly, we have the Maclaurin expansions for $\sin x$ and $\cos x$ :

\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots + (-1)^r \frac{x^{2r+1}}{(2r+1)!} + \ldots

\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots + (-1)^r \frac{x^{2r}}{(2r)!} + \ldots

These expansions define the trigonometric functions in terms of powers of $x$ .

The Key Insight: Let $x = i\theta$

To understand the relationship between exponentials and trigonometric functions, we substitute $x = i\theta$ (where $i$ is the imaginary unit) into the expansion for $e^x$ :

e^{i\theta} = 1 + i\theta + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \frac{(i\theta)^4}{4!} + \ldots

Now, simplify each term:

$(i\theta)^2 = i^2 \theta^2 = -\theta^2$
$(i\theta)^3 = i^3 \theta^3 = -i \theta^3$
$(i\theta)^4 = i^4 \theta^4 = \theta^4$
$(i\theta)^5 = i^5 \theta^5 = i \theta^5$ So,

e^{i\theta} = 1 + i\theta - \frac{\theta^2}{2!} - \frac{i\theta^3}{3!} + \frac{\theta^4}{4!} + \frac{i\theta^5}{5!} - \ldots

Group real and imaginary parts:

e^{i\theta} = \left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \ldots \right) + i\left( \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \ldots \right)

But from the Maclaurin expansions, we recognise these as:

e^{i\theta} = \cos \theta + i \sin \theta

This is Euler's Formula:

e^{i\theta} = \cos \theta + i \sin \theta

The Exponential Form of a Complex Number

For a complex number $z$ , written in polar form as:

z = r (\cos \theta + i \sin \theta),

we can equivalently write:

z = r e^{i\theta}.

Here:

$r$ is the modulus (or magnitude) of the complex number: $r = |z|$
$\theta$ is the argument (or angle) of the complex number: $\theta = \arg(z)$

Why Use Exponential Form?

The exponential form simplifies many operations:

Multiplication of complex numbers:

z_1 = r_1 e^{i\theta_1}, \quad z_2 = r_2 e^{i\theta_2}

Then:

z_1 z_2 = (r_1 r_2) e^{i(\theta_1 + \theta_2)}

The moduli multiply, and the arguments add.

Division of complex numbers:

\frac{z_1}{z_2} = \frac{r_1}{r_2} e^{i(\theta_1 - \theta_2)}

Finding powers ( $z^n$ ) and roots ( $z^{1/n}$ ) becomes straightforward using De Moivre's Theorem.

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Example 1: Converting to Exponential Form Let's convert the complex number $z = 1 + i$ to exponential form.

Step 1: Find the modulus $( r )$ :

r = |z| = \sqrt{1^2 + 1^2} = \sqrt{2}

Step 2: Find the argument $( \theta )$ :

\theta = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4} \text{ radians}

Step 3: Write in exponential form:

z = \sqrt{2} e^{i \frac{\pi}{4}}

So the exponential form of $1 + i$ is $\sqrt{2} e^{i \frac{\pi}{4}}$

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Example 2: Multiplying Complex Numbers in Exponential Form Multiply $z_1 = 2e^{i \frac{\pi}{6}}$ and $z_2 = 3e^{i \frac{\pi}{3}}$

Step 1: Multiply the moduli:

r_1 \times r_2 = 2 \times 3 = 6

Step 2: Add the arguments:

\theta_1 + \theta_2 = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}

Step 3: Write the result:

z_1 \times z_2 = 6 e^{i \frac{\pi}{2}}

So the product is $6 e^{i \frac{\pi}{2}}$

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Key Takeaways:

The exponential form of a complex number is written as $z = r e^{i \theta}$ , where $r$ is the modulus and $\theta$ is the argument.
The exponential form simplifies operations like multiplication, division, and raising complex numbers to powers.
Euler's formula, $e^{i \theta} = \cos \theta + i \sin \theta$ , is the key to converting between trigonometric and exponential forms of complex numbers. This form is especially helpful when dealing with advanced operations in complex numbers, as it makes calculations much easier!

Applications of Complex Numbers to Trig Identities

Using the exponential form of complex numbers, we can derive important trigonometric identities involving $\sin\theta$ and $\cos\theta$ . These identities help simplify trigonometric expressions and connect them with exponential functions.

Key Identity: Euler's Formula

From Euler's formula, we know:

e^{i\theta} = \cos \theta + i \sin \theta \quad \text{and} \quad e^{-i\theta} = \cos(-\theta) + i \sin(-\theta).

Using the even-odd properties of $\cos$ and $\sin$ :

$\cos(-\theta) = \cos(\theta)$ (even function).
$\sin(-\theta) = -\sin(\theta)$ (odd function). Thus:

e^{-i\theta} = \cos \theta - i \sin \theta.

Deriving Trigonometric Identities

Identity 1: $\cos \theta$ in terms of exponentials

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Let's calculate:

\frac{1}{2} \left( e^{i\theta} + e^{-i\theta} \right).

Step 1: Substitute the expressions for $e^{i\theta}$ and $e^{-i\theta}$ :

\frac{1}{2} \left( e^{i\theta} + e^{-i\theta} \right) = \frac{1}{2} \left( (\cos \theta + i \sin \theta) + (\cos \theta - i \sin \theta) \right).]

Step 2: Simplify the terms:

= \frac{1}{2} \left( \cos \theta + i \sin \theta + \cos \theta - i \sin \theta \right).

The imaginary parts $i \sin \theta$ and $-i \sin \theta$ cancel out:

= \frac{1}{2} \left( 2 \cos \theta \right) = \cos \theta.

Thus:

\frac{1}{2} \left( e^{i\theta} + e^{-i\theta} \right) = \cos \theta.

Identity 2: $\sin \theta$ in terms of exponentials

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Now consider:

\frac{1}{2} \left( e^{i\theta} - e^{-i\theta} \right).

Step 1: Substitute the expressions for $(e^{i\theta})$ and $(e^{-i\theta})$ :

\frac{1}{2} \left( e^{i\theta} - e^{-i\theta} \right) = \frac{1}{2} \left( (\cos \theta + i \sin \theta) - (\cos \theta - i \sin \theta) \right).

Step 2: Simplify the terms:

= \frac{1}{2} \left( \cos \theta + i \sin \theta - \cos \theta + i \sin \theta \right).

The real parts $\cos \theta$ and $-\cos \theta$ cancel out:

= \frac{1}{2} \left( 2i \sin \theta \right) = i \sin \theta.

Thus:

\frac{1}{2} \left( e^{i\theta} - e^{-i\theta} \right) = i \sin \theta.

Final Results

We've derived two important identities:

\cos \theta = \frac{1}{2} \left( e^{i\theta} + e^{-i\theta} \right)

\sin \theta = \frac{1}{2i} \left( e^{i\theta} - e^{-i\theta} \right)

These identities are fundamental for simplifying trigonometric expressions and solving equations involving complex exponentials.

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Example: Express $\cos 2\theta$ in exponential form.

Using the identity:

\cos \theta = \frac{1}{2} \left( e^{i\theta} + e^{-i\theta} \right),

we replace $\theta$ with $2\theta$ :

\cos 2\theta = \frac{1}{2} \left( e^{i(2\theta)} + e^{-i(2\theta)} \right).

This is the exponential form of $\cos 2\theta$

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Example: Derive the Double Angle Formula for $\cos 2\theta$

From Euler's formula:

e^{i\theta} = \cos \theta + i \sin \theta.

Squaring $e^{i\theta}$

(e^{i\theta})^2 = (\cos \theta + i \sin \theta)^2.

Expand:

e^{i2\theta} = \cos^2 \theta + 2i \cos \theta \sin \theta - \sin^2 \theta.

Using $\cos 2\theta = \text{Re}(e^{i2\theta})$

\cos 2\theta = \cos^2 \theta - \sin^2 \theta.

This confirms the familiar double-angle identity of cosine.

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Example: Write $\cos3\theta$ solely in terms of $\cos\theta$ .

\cos3\theta + i\sin3\theta

is our starting point.

Notice that $\cos3\theta$ is the real part of the above expression.

Using De Moivre's Theorem, we get:

\cos(3\theta) + i\sin(3\theta) \equiv \left [ \cos(\theta) + i\sin(\theta)\right ]^3

Using the binomial expansion on this, we get:

\cos^3(\theta) + 3\cos^2(\theta)(i\sin(\theta)) + 3\cos(\theta)(i\sin(\theta))^2 + (i\sin(\theta))^3

= \cos^3(\theta) + 3i\cos^2(\theta)\sin(\theta) - 3\cos(\theta)\sin^2(\theta) - i\sin^3(\theta)

Gathering together the real and imaginary parts:

= \cos^3(\theta) - 3\cos(\theta)\sin^2(\theta) + i(3\cos^2(\theta)\sin(\theta) - \sin^3(\theta))

= \cos(3\theta) + i\sin(3\theta)

Thus

\cos(3\theta) \equiv \cos^3(\theta) - 3\cos(\theta)\sin^2(\theta)

\equiv \cos^3(\theta) - 3\cos(\theta)(1 - \cos^2(\theta))

\equiv \cos^3\theta - 3\cos\theta+3\cos^3\theta

\equiv 4\cos^3(\theta) - 3\cos(\theta)

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Example: Write $\cos(\theta)$ in terms of $\cos(n\theta)$ , where $n \in \mathbb{N}$ :

Using the identity:

\cos(\theta) \equiv \frac{1}{2} \left(e^{i\theta} + e^{-i\theta}\right)

We get:

\bigg [\frac{1}{2} \left(e^{i\theta} + e^{-i\theta}\right) \bigg]^6

\cos^6(\theta) \equiv \frac{1}{64} \left(e^{i\theta} + e^{-i\theta}\right)^6

Expanding:

= \frac{1}{64} \bigg(e^{i6\theta} + 6e^{i5\theta}e^{-i\theta} + 15e^{i4\theta}e^{-i2\theta} + 20e^{i3\theta}e^{-i3\theta} + 15e^{-i2\theta}e^{-i4\theta} + 6e^{-i\theta}e^{-i5\theta} + e^{-i6\theta}\bigg)

= \frac{1}{64} \bigg(e^{i6\theta} + 6e^{i4\theta} + 15e^{i2\theta} + 20 + 15e^{-i2\theta} + 6e^{-i4\theta} + e^{-i6\theta}\bigg)

Now gathering together terms that contain powers of the same magnitude:

\Rightarrow \cos^6(\theta) \equiv \frac{1}{64} \bigg( \left( e^{i6\theta} + e^{-i6\theta} \right) + 6 \left( e^{i4\theta} + e^{-i4\theta} \right) + 15 \left( e^{i2\theta} + e^{-i2\theta} \right) + 20 \bigg)

\text {Assume} \quad 2\cos 6\theta = \left( e^{i6\theta} + e^{-i6\theta} \right) \quad \quad 2\cos 4\theta = \left( e^{i4\theta} + e^{-i4\theta} \right)\quad \\\quad 2\cos 2\theta = \left( e^{i2\theta} + e^{-i2\theta} \right)

\equiv \frac{1}{64} \bigg( 2\cos(6\theta) + 12\cos(4\theta) + 30\cos(2\theta) + 20 \bigg)

\equiv \frac{1}{32} \bigg( \cos(6\theta) + 6\cos(4\theta) + 15\cos(2\theta) + 10 \bigg)

Key Takeaways

Key Exponential Identities:

\cos \theta = \frac{1}{2} (e^{i\theta} + e^{-i\theta}), \quad \sin \theta = \frac{1}{2i} (e^{i\theta} - e^{-i\theta}).

Even and Odd Functions:

$\cos(-\theta) = \cos(\theta)$
$\sin(-\theta) = -\sin(\theta)$ Common Mistake: Forgetting that $i^2 = -1$ when simplifying exponential terms.

Proof

Exponential forms of complex numbers simplify operations like multiplication and division, making it easy to prove key results and solve problems. Here, we'll go through proofs and examples with clear explanations.

Proof: Argument of a Product

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Let:

z_1 = r e^{i\alpha}, \quad z_2 = s e^{i\beta}

We'll find the argument of $z_1 z_2$

Step 1: Multiply the complex numbers

z_1 z_2 = (r e^{i\alpha})(s e^{i\beta}) = rs e^{i(\alpha + \beta)}

Using Euler's formula:

z_1 z_2 = rs \left( \cos(\alpha + \beta) + i \sin(\alpha + \beta) \right)

This shows that the argument of $z_1 z_2 is (\alpha + \beta)$

Argument of $z_1$ : $\arg(z_1) = \alpha$
Argument of $z_2$ : $\arg(z_2) = \beta$
Argument of $z_1 z_2$ : $\arg(z_1 z_2) = \alpha + \beta$

Thus:

\arg(z_1 z_2) = \arg(z_1) + \arg(z_2) \quad

Proof: Modulus of a Product

The modulus of a complex number $z = re^{i\theta}$ is given by $|z| = r$

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Example: For $z_1 = r e^{i\alpha}$ and $z_2 = s e^{i\beta}$

Step 1: Multiply $z_1$ and $z_2$ :

z_1 z_2 = rs e^{i(\alpha + \beta)}

Step 2: Take the modulus:

|z_1 z_2| = |rs e^{i(\alpha + \beta)}| = rs

Since $|z_1| = r$ and $|z_2| = s$ , it follows that:

|z_1 z_2| = |z_1| \times |z_2| \quad

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Example: Illustrating $e^{i\pi}$ on an Argand Diagram

Using Euler's formula:

e^{i\pi} = \cos(\pi) + i\sin(\pi) = -1 + 0i

On the Argand diagram:

$e^{i\pi}$ corresponds to the point $(-1, 0)$ , which lies on the negative real axis.

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Example: Calculating $i^i$

To evaluate $i^i$ , we use the exponential form of $i$ :

i = e^{i\frac{\pi}{2}}

Raise both sides to the power of $i$ :

i^i = \left( e^{i\frac{\pi}{2}} \right)^i

Simplify using $(a^b)^c = a^{bc}$

i^i = e^{i^2 \frac{\pi}{2}} = e^{-\frac{\pi}{2}}

Thus:

i^i = e^{-\frac{\pi}{2}} \approx 0.2079

This surprising result shows that $i^i$ is a real number!

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Example: Finding $\ln(i)$

To evaluate $\ln(i)$ , recall that:

i = e^{i\frac{\pi}{2}}

Taking the natural logarithm of both sides:

\ln(i) = \ln\left( e^{i\frac{\pi}{2}} \right)

Using the property $\ln(e^x) = x$

\ln(i) = i\frac{\pi}{2}

Thus:

\ln(i) = \frac{\pi}{2}i

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Key Takeaways

Argument of a Product:

\arg(z_1 z_2) = \arg(z_1) + \arg(z_2)

Modulus of a Product:

|z_1 z_2| = |z_1| \times |z_2|

Exponential Form (Edexcel A-Level Further Mathematics): Revision Notes

1.2.1 Exponential Form

Overview

The Exponential Function

Maclaurin Series for $\sin x$ and $\cos x$

The Key Insight: Let $x = i\theta$

The Exponential Form of a Complex Number

Why Use Exponential Form?

Key Takeaways:

Applications of Complex Numbers to Trig Identities

Key Identity: Euler's Formula

Deriving Trigonometric Identities

Identity 1: $\cos \theta$ in terms of exponentials

Identity 2: $\sin \theta$ in terms of exponentials

Final Results

Key Takeaways

Proof

Proof: Argument of a Product

Proof: Modulus of a Product

Example: Illustrating $e^{i\pi}$ on an Argand Diagram

Example: Calculating $i^i$

Example: Finding $\ln(i)$

Key Takeaways

Explore Edexcel A-Level Further Mathematics Model Answers by Topics

Complex Numbers & Argand Diagrams

Exponential Form & de Moivre's Theorem

Explore Edexcel A-Level Further Mathematics Quizzes by Topics

Complex Numbers & Argand Diagrams

Exponential Form & de Moivre's Theorem

Explore Edexcel A-Level Further Mathematics Flashcards by Topics

Complex Numbers & Argand Diagrams

Exponential Form & de Moivre's Theorem

Explore Edexcel A-Level Further Mathematics Exam Questions by Topics

Complex Numbers & Argand Diagrams

Exponential Form & de Moivre's Theorem

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Exponential Form (Edexcel A-Level Further Mathematics): Revision Notes

1.2.1 Exponential Form

Overview

The Exponential Function

Maclaurin Series for sin⁡x\sin xsinx and cos⁡x\cos xcosx

The Key Insight: Let x=iθx = i\thetax=iθ

The Exponential Form of a Complex Number

Why Use Exponential Form?

Key Takeaways:

Applications of Complex Numbers to Trig Identities

Key Identity: Euler's Formula

Deriving Trigonometric Identities

Identity 1: cos⁡θ\cos \thetacosθ in terms of exponentials

Identity 2: sin⁡θ\sin \thetasinθ in terms of exponentials

Final Results

Key Takeaways

Proof

Proof: Argument of a Product

Proof: Modulus of a Product

Example: Illustrating eiπe^{i\pi}eiπ on an Argand Diagram

Example: Calculating iii^iii

Example: Finding ln⁡(i)\ln(i)ln(i)

Key Takeaways

Explore Edexcel A-Level Further Mathematics Model Answers by Topics

Complex Numbers & Argand Diagrams

Exponential Form & de Moivre's Theorem

Explore Edexcel A-Level Further Mathematics Quizzes by Topics

Complex Numbers & Argand Diagrams

Exponential Form & de Moivre's Theorem

Explore Edexcel A-Level Further Mathematics Flashcards by Topics

Complex Numbers & Argand Diagrams

Exponential Form & de Moivre's Theorem

Explore Edexcel A-Level Further Mathematics Exam Questions by Topics

Complex Numbers & Argand Diagrams

Exponential Form & de Moivre's Theorem

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Maclaurin Series for $\sin x$ and $\cos x$

The Key Insight: Let $x = i\theta$

Identity 1: $\cos \theta$ in terms of exponentials

Identity 2: $\sin \theta$ in terms of exponentials

Example: Illustrating $e^{i\pi}$ on an Argand Diagram

Example: Calculating $i^i$

Example: Finding $\ln(i)$