Mathematics - Further (9231) | Complex numbers

Complex Numbers (9231 Further Pure Mathematics 2, Topic 2.5)

Hello! Welcome to the world of Complex Numbers. If real numbers let you measure distances along a line, complex numbers allow you to move and measure across a whole plane. This chapter takes your prior knowledge (from A Level Mathematics Pure 3) and turbocharges it, focusing on how these numbers behave when multiplied, divided, and raised to powers. Mastering this topic is essential for handling advanced trigonometry and series summation in Further Maths.

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1. Quick Review: Cartesian and Polar Forms

A complex number \(z\) can be written in two main forms. Since the core of this chapter (De Moivre's Theorem) relies on the polar form, let's quickly review the basics.

1.1 The Forms of a Complex Number

• Cartesian Form: \(z = x + iy\)
  (Where \(x\) is the real part, \(\text{Re}(z)\), and \(y\) is the imaginary part, \(\text{Im}(z)\)).
• Polar Form: \(z = r(\cos \theta + i \sin \theta)\) or \(z = r \text{cis} \theta\)
  (Where \(r\) is the modulus or magnitude, \(r = \sqrt{x^2 + y^2}\), and \(\theta\) is the argument or angle, \(\tan \theta = y/x\)).

Analogy: Think of the complex plane like a map. Cartesian coordinates (\(x, y\)) tell you how many steps east and north to take. Polar coordinates (\(r, \theta\)) tell you to face a certain direction (\(\theta\)) and walk a certain distance (\(r\)).

1.2 Argument Convention

In Further Mathematics, the principal argument, Arg(\(z\)), is usually restricted to the interval \(-\pi < \theta \le \pi\) (or sometimes \(0 \le \theta < 2\pi\), always check the question context, but the standard is \(-\pi\) to \(\pi\)).

Key Takeaway: Multiplication and powers are much easier in polar form because we simply manipulate the modulus (\(r\)) and argument (\(\theta\)).

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2. De Moivre's Theorem (DMT)

De Moivre's Theorem is the powerhouse of complex numbers, allowing us to find powers and roots very quickly.

2.1 Statement of De Moivre's Theorem

For any complex number \(z = r(\cos \theta + i \sin \theta)\) and any rational number \(n\):

$$z^n = r^n (\cos(n\theta) + i \sin(n\theta))$$

For powers of complex numbers on the unit circle (\(r=1\)), the formula simplifies to:

$$(\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta)$$

The syllabus requires you to understand and apply DMT for positive integer, negative integer, and rational exponents (\(n \in \mathbb{Q}\)).

2.2 Geometrical Interpretation (The "Why")

The key geometric effect of De Moivre's Theorem is multiplication becomes rotation and scaling.

• Scaling: The modulus \(r\) is raised to the power \(n\).
• Rotation: The argument \(\theta\) is multiplied by the power \(n\).

Example: If you square a complex number \(z\), you square its distance from the origin (\(r^2\)) and double its angle (\(2\theta\)).

When you multiply two complex numbers, \(z_1 z_2\):

$$r_1 \text{cis} \theta_1 \times r_2 \text{cis} \theta_2 = (r_1 r_2) \text{cis}(\theta_1 + \theta_2)$$

When you divide two complex numbers, \(z_1 / z_2\):

$$\frac{r_1 \text{cis} \theta_1}{r_2 \text{cis} \theta_2} = \frac{r_1}{r_2} \text{cis}(\theta_1 - \theta_2)$$

This geometric understanding is crucial for visualizing the roots of unity later!

2.3 Proof of De Moivre’s Theorem (Positive Integer \(n\))

You must be able to prove DMT for a positive integer exponent \(n\) using the method of Mathematical Induction.

Step 1: Base Case (n = 1)

If \(n=1\), LHS \( = (\cos \theta + i \sin \theta)^1\). RHS \( = \cos(1\theta) + i \sin(1\theta)\). LHS = RHS. The statement holds for \(n=1\).

Step 2: Assumption (n = k)

Assume the result is true for some positive integer \(k\):
$$(\cos \theta + i \sin \theta)^k = \cos(k\theta) + i \sin(k\theta)$$

Step 3: Inductive Step (n = k + 1)

We need to show it holds for \(n = k+1\):

$$(\cos \theta + i \sin \theta)^{k+1} = (\cos \theta + i \sin \theta)^k \cdot (\cos \theta + i \sin \theta)^1$$

Substitute the assumption from Step 2:

$$ = (\cos(k\theta) + i \sin(k\theta)) \cdot (\cos \theta + i \sin \theta)$$

Expand the product (using basic multiplication, noting \(i^2 = -1\)):

$$ = \cos(k\theta)\cos \theta - \sin(k\theta)\sin \theta + i(\sin(k\theta)\cos \theta + \cos(k\theta)\sin \theta)$$

Now, use the standard trigonometric addition formulas (which you should recall):

$$ \cos(A+B) = \cos A \cos B - \sin A \sin B $$ $$ \sin(A+B) = \sin A \cos B + \cos A \sin B $$

Applying these gives:

$$ = \cos(k\theta + \theta) + i \sin(k\theta + \theta) $$ $$ = \cos((k+1)\theta) + i \sin((k+1)\theta) $$

Since the result holds for \(n=1\), and if it holds for \(n=k\) it also holds for \(n=k+1\), by induction, De Moivre's Theorem is proven for all positive integers \(n\).

Key Takeaway: DMT allows us to raise a complex number to a power by multiplying the angle by the power. Remember the steps for the proof by induction, particularly the use of the trigonometric addition formulas in the final step.

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3. Advanced Trigonometric Applications

The real power of DMT lies in deriving complex trigonometric identities quickly, a core skill required in FP2.

3.1 The Magic Identifiers

We work exclusively with complex numbers on the unit circle, \(z = \cos \theta + i \sin \theta\). Using DMT for \(n=-1\):

$$z^{-1} = \cos(-\theta) + i \sin(-\theta) = \cos \theta - i \sin \theta$$

By adding and subtracting \(z\) and \(z^{-1}\), we get two vital relationships:

1. $$z + z^{-1} = 2 \cos \theta$$
2. $$z - z^{-1} = 2i \sin \theta$$

More generally, using DMT:

1. $$z^n + z^{-n} = 2 \cos(n\theta)$$
2. $$z^n - z^{-n} = 2i \sin(n\theta)$$

3.2 Expressing Multiple Angles in Terms of Powers of Trig Ratios (e.g., \(\cos 5\theta\))

To express \(\cos(n\theta)\) or \(\sin(n\theta)\) in terms of \(\cos \theta\) and \(\sin \theta\), we use the direct application of DMT combined with the Binomial Theorem.

Step-by-Step Example (General Approach):

1. Start with DMT: $$\cos(n\theta) + i \sin(n\theta) = (\cos \theta + i \sin \theta)^n$$
2. Expand the Right-Hand Side (RHS) using the Binomial Theorem: $$(a+b)^n = \sum_{r=0}^n \binom{n}{r} a^{n-r} b^r$$
3. Group the terms on the RHS into Real and Imaginary parts (remembering \(i^2=-1, i^3=-i, i^4=1\)).
4. Equate Real Parts to find \(\cos(n\theta)\).
5. Equate Imaginary Parts to find \(\sin(n\theta)\).

Common Mistake to Avoid: When equating imaginary parts, DO NOT include the \(i\). If the imaginary part of RHS is \(i V\), then \(\sin(n\theta) = V\).

3.3 Expressing Powers of Trig Ratios in Terms of Multiple Angles (Linearization)

This application is required for integration and summing series. We aim to convert expressions like \(\cos^5 \theta\) into a sum of terms involving \(\cos(k\theta)\).

Step-by-Step Example (General Approach):

1. Use the magic identifiers: To evaluate \(\cos^n \theta\), use $$(2 \cos \theta)^n = (z + z^{-1})^n$$
   To evaluate \(\sin^n \theta\), use $$(2i \sin \theta)^n = (z - z^{-1})^n$$
2. Expand the RHS using the Binomial Theorem.
3. Collect the terms into conjugate pairs: \((z^k + z^{-k})\) or \((z^k - z^{-k})\).
4. Substitute back using the magic identifiers (e.g., \(z^k + z^{-k} = 2 \cos(k\theta)\)).
5. Isolate \(\cos^n \theta\) or \(\sin^n \theta\) by dividing by \(2^n\) or \((2i)^n\).

Memory Aid: If you are dealing with Cos, you Collect terms that are conjugates (\(z^n\) and \(z^{-n}\)).

Key Takeaway: Binomial expansion is the link between DMT and advanced trig identities. Know the formulas \(z^n \pm z^{-n}\) and practice equating real/imaginary parts carefully.

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4. Finding the \(n\)th Roots of Complex Numbers

Finding the \(n\)th roots of a complex number \(w\) means solving the equation \(z^n = w\). This requires the application of DMT for rational exponents (\(1/n\)).

4.1 The General Root Formula

If we want to find the \(n\) roots of \(w = R(\cos \phi + i \sin \phi)\), we must first write the angle using the general form, adding multiples of \(2\pi\):

$$w = R(\cos (\phi + 2k\pi) + i \sin (\phi + 2k\pi))$$

Applying DMT with exponent \(1/n\), the roots \(z_k\) are given by:

$$z_k = R^{1/n} \left[ \cos \left(\frac{\phi + 2k\pi}{n}\right) + i \sin \left(\frac{\phi + 2k\pi}{n}\right) \right]$$

We use \(k = 0, 1, 2, ..., n-1\) to generate the \(n\) distinct roots. (Once \(k=n\) is reached, the roots start repeating.)

4.2 Roots of Unity (\(z^n = 1\))

This is a special, very common case where \(w=1\). Here, the modulus \(R=1\) and the argument \(\phi=0\). The roots of unity all lie on the unit circle and are the vertices of a regular \(n\)-sided polygon.

For \(z^n = 1\), the roots are:

$$z_k = \cos \left(\frac{2k\pi}{n}\right) + i \sin \left(\frac{2k\pi}{n}\right), \quad k = 0, 1, ..., n-1$$

Did you know? The sum of the \(n\)th roots of unity is always zero (except for the trivial case \(n=1\)). This relates to the symmetry of the polygon vertices.

Key Property of Roots of Unity: If \(\omega\) is a non-real \(n\)th root of unity (i.e., the root generated by \(k=1\)), then all \(n\) roots can be written as \(1, \omega, \omega^2, \omega^3, ..., \omega^{n-1}\).

Key Takeaway: To find roots, use the general argument \(\theta + 2k\pi\). This ensures you capture all \(n\) geometrically distinct solutions, which always form a regular polygon in the Argand diagram.

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5. Summation of Series using Complex Numbers (C + iS Method)

The C + iS method is an elegant application of complex numbers used to sum series involving sines and cosines, often where the angles are in arithmetic progression (AP).

Let C be the sum of the cosine series and S be the sum of the sine series you wish to find.

5.1 The Method

We combine C and S into a single complex series, C + iS. If the individual terms of C and S are related by DMT, the combined series often becomes a standard Geometric Progression (GP) in the complex plane.

Step-by-Step Process:

1. Define C and S: Write down the target series C and the corresponding series S (where cosine terms are replaced by sine terms, and vice versa).
2. Form C + iS: Combine them. $$C + iS = \sum_{r=1}^n (\cos(r\theta) + i \sin(r\theta))$$
3. Convert to Complex Notation: Let \(z = \cos \theta + i \sin \theta\). Using DMT, \(\cos(r\theta) + i \sin(r\theta) = z^r\). $$C + iS = \sum_{r=1}^n z^r = z + z^2 + z^3 + ... + z^n$$
4. Sum the GP: This is a GP with first term \(a=z\), common ratio \(R=z\), and \(n\) terms. The sum is: $$S_n = \frac{a(1 - R^n)}{1 - R} = \frac{z(1 - z^n)}{1 - z}$$
5. Convert Back: Substitute \(z = \cos \theta + i \sin \theta\) and \(z^n = \cos(n\theta) + i \sin(n\theta)\).
6. Simplify and Rationalize: Manipulate the expression to isolate the real and imaginary parts. Often, this involves multiplying the numerator and denominator by the conjugate of the denominator, or using trigonometric half-angle identities to factorize the numerator and denominator.
7. Equate Parts:
   • If you want C, equate the Real Part of the final simplified expression.
   • If you want S, equate the Imaginary Part of the final simplified expression.

5.2 Example Setup (Avoiding Common Errors)

Suppose you are asked to sum \(C = \sum_{r=1}^n \cos(r\theta)\).

You define \(S = \sum_{r=1}^n \sin(r\theta)\).

$$C + iS = z + z^2 + ... + z^n$$

If the series starts at \(r=0\), the first term is \(z^0 = 1\), so \(a=1\). If the series starts at \(r=1\), the first term is \(z\), so \(a=z\).

The trickiest part is usually Step 6, simplification. Remember that expressions involving \(1 - \cos \theta - i \sin \theta\) can often be simplified using half-angle identities and factorization.

Key Takeaway: The C + iS method is just summing a complex GP. Define C and S correctly, identify the first term and ratio, and then carefully separate the real and imaginary parts at the end.