Pure Mathematics | Further complex numbers

🚀 Welcome to Further Complex Numbers! (FP2)

Hello mathematicians! You've already mastered the basics of complex numbers, plotting them on the Argand diagram, and understanding their polar form. Now, we are taking things up a notch!

This chapter, Further Complex Numbers, introduces powerful tools that allow us to calculate high powers and roots of complex numbers efficiently, and truly understand the geometry behind these fascinating numbers. Don't worry if the concepts look intimidating at first—we'll break down the barriers together using simple steps and helpful analogies. Let's dive in!

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1. Review: Essential Prerequisites

Before tackling the new material, we absolutely must be comfortable with the Polar Form of a complex number \(z\).

1.1 Polar and Exponential Form

If \(z\) has modulus \(r = |z|\) and argument \(\theta = \arg(z)\), we write:

• Standard Polar Form (Modulus-Argument Form):
  \(z = r(\cos \theta + i \sin \theta)\)
• Euler's Form (The Super Shortcut):
  \(z = r e^{i\theta}\)

Quick Review Tip: Remember that \(\theta\) must be in the range \(-\pi < \theta \le \pi\) unless specified otherwise.

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2. De Moivre’s Theorem: The Power Shortcut

Imagine trying to calculate \((1 + i)^{10}\) using binomial expansion—it would take forever! De Moivre's Theorem gives us a simple, elegant shortcut when the complex number is in polar form.

2.1 The Theorem Statement

If \(n\) is any rational number, then:

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

In words: To raise a complex number in polar form to the power \(n\), you raise the modulus \(r\) to the power \(n\), and you multiply the argument \(\theta\) by \(n\).

Analogy: The Angle Amplifier

Think of the power \(n\) as an amplifier for the angle. It only interacts with the angle, leaving the \(r^n\) modulus outside.

2.2 Using Euler’s Form (The Logic Behind It)

If you use Euler's form, the theorem is crystal clear:

\((r e^{i\theta})^n = r^n (e^{i\theta})^n = r^n e^{i(n\theta)}\)

This demonstrates why the angle simply gets multiplied by the power \(n\). This is why mastering Euler's form is so useful in FP2!

2.3 Application: Proving Trig Identities

De Moivre’s Theorem is powerful for deriving identities for \(\cos(n\theta)\) or \(\sin(n\theta)\) in terms of powers of \(\sin \theta\) and \(\cos \theta\).

Step-by-Step for \(\cos(3\theta)\) and \(\sin(3\theta)\):

1. Start with De Moivre's Theorem for \(n=3\):
   \(\cos(3\theta) + i \sin(3\theta) = (\cos \theta + i \sin \theta)^3\)
2. Expand the Right Hand Side (RHS) using the Binomial Theorem \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\):
   \(RHS = (\cos \theta)^3 + 3(\cos \theta)^2 (i \sin \theta) + 3(\cos \theta)(i \sin \theta)^2 + (i \sin \theta)^3\)
3. Simplify using \(i^2 = -1\) and \(i^3 = -i\):
   \(RHS = \cos^3 \theta + i(3 \cos^2 \theta \sin \theta) - 3 \cos \theta \sin^2 \theta - i \sin^3 \theta\)
4. Group the Real and Imaginary parts:
   \(RHS = (\cos^3 \theta - 3 \cos \theta \sin^2 \theta) + i (3 \cos^2 \theta \sin \theta - \sin^3 \theta)\)
5. Equate the Real parts to find \(\cos(3\theta)\):
   \(\cos(3\theta) = \cos^3 \theta - 3 \cos \theta \sin^2 \theta\)
6. Equate the Imaginary parts to find \(\sin(3\theta)\):
   \(\sin(3\theta) = 3 \cos^2 \theta \sin \theta - \sin^3 \theta\)

Key Takeaway for De Moivre's: It connects powers (on the outside) to multiples of the angle (on the inside). Always ensure your complex number is in the \((\cos \theta + i \sin \theta)\) format before applying the theorem!

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3. Finding Roots of Complex Numbers (Solving \(z^n = w\))

This is the inverse operation of De Moivre's theorem. When solving an equation like \(z^4 = -16\), we are looking for the four complex roots. Because complex numbers "cycle" every \(2\pi\) radians, there will always be \(n\) distinct roots for an equation \(z^n = w\).

3.1 The Crucial Step: The General Argument

When finding roots, you must remember that an angle \(\theta\) is the same as \(\theta + 2k\pi\) for any integer \(k\). This periodicity gives us all the different roots.

If \(w = r(\cos \alpha + i \sin \alpha)\), we must rewrite \(w\) in its General Polar Form:

\(w = r(\cos (\alpha + 2k\pi) + i \sin (\alpha + 2k\pi))\)

where \(k\) is an integer.

3.2 The Root Calculation Process

To solve \(z^n = w\):

1. Express \(w\) in General Polar Form: Use \(r\) and the general argument \(\alpha + 2k\pi\).
2. Apply the Root: Take the \(n\)-th root of both sides.
   \(z = w^{1/n} = r^{1/n} \left(\cos \left(\frac{\alpha + 2k\pi}{n}\right) + i \sin \left(\frac{\alpha + 2k\pi}{n}\right)\right)\)
3. Find the \(n\) Distinct Roots: Substitute integer values for \(k\):
   \(k = 0, 1, 2, \ldots, n-1\)
   You only need \(n\) values of \(k\), as the roots repeat after \(k = n-1\).
4. Simplify and Present: Convert the arguments back into the principal range \((-\pi, \pi]\) if required.

💡 Memory Aid: Slicing the Pizza

If you are finding \(n\) roots (e.g., \(n=5\)), visualize the roots on the Argand diagram. They will always lie on a circle of radius \(r^{1/n}\) and be evenly spaced by an angle of \(\frac{2\pi}{n}\). You are essentially slicing the complex plane pizza into \(n\) equal slices!

3.3 Special Case: Roots of Unity (\(z^n = 1\))

The roots of the equation \(z^n = 1\) are called the roots of unity. Since \(1 = 1 e^{i(0 + 2k\pi)}\), the roots are:

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

• One root is always \(z_0 = 1\).
• The roots are vertices of a regular \(n\)-sided polygon inscribed in the unit circle.
• If \(\omega\) is the first non-trivial root (\(k=1\)), the other roots are powers of \(\omega\): \(1, \omega, \omega^2, \ldots, \omega^{n-1}\).
• The sum of the roots of unity is always zero.

Key Takeaway for Roots: Finding roots means dividing the angle by \(n\), but you must first account for the infinite possible representations of the original angle by adding \(2k\pi\).

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4. Loci in the Argand Diagram

In the Argand diagram, a locus (plural: loci) is a set of points \(z\) that satisfy a given condition. These conditions are usually expressed using modulus or argument notation. Understanding these geometrical conditions is vital for FP2.

4.1 Locus 1: The Circle (Constant Distance)

The expression \(|z - z_1|\) represents the distance between the point \(z\) and the fixed point \(z_1\).

\(|z - z_1| = r\)

This describes the locus of points \(z\) whose distance from the fixed point \(z_1\) is always equal to the constant radius \(r\).

• Locus Description: A circle.
• Centre: \(z_1\)
• Radius: \(r\)

Analogy: Imagine a dog on a leash of length \(r\), tied to a fixed post \(z_1\). The path the dog can trace is a circle.

If the condition is an inequality, like \(|z - z_1| \le r\), this describes the closed disk (the circle and everything inside it).

4.2 Locus 2: The Perpendicular Bisector (Equidistant)

The expression \(|z - z_1| = |z - z_2|\) describes points \(z\) that are the same distance from two fixed points, \(z_1\) and \(z_2\).

\(|z - z_1| = |z - z_2|\)

• Locus Description: A straight line.
• Geometrical Property: It is the perpendicular bisector of the line segment connecting \(z_1\) and \(z_2\).

Tip for Struggling Students: Cartesian Method

If the geometrical interpretation is difficult, always substitute \(z = x + iy\), \(z_1 = x_1 + iy_1\), etc., and square both sides to find the Cartesian equation:

Example: \(|x + iy - 2| = |x + iy - 4i|\)
\((x-2)^2 + y^2 = x^2 + (y-4)^2\)
Expanding this will give you a simple straight-line equation (e.g., \(y = mx + c\)).

4.3 Locus 3: The Half-Line (Constant Angle)

The argument \(\arg(z - z_1)\) represents the angle the vector \(\vec{z_1 z}\) makes with the positive real axis.

\(\arg(z - z_1) = \alpha\)

This describes points \(z\) such that the line starting at \(z_1\) and passing through \(z\) always makes a fixed angle \(\alpha\) with the real axis.

• Locus Description: A half-line (or ray).
• Starting Point: \(z_1\) (This point is usually excluded from the locus because \(\arg(0)\) is undefined).
• Direction: Determined by the angle \(\alpha\).

Important Note: The locus starts *at* \(z_1\) and extends outwards in one direction. It is not a full line.

4.4 Locus 4: Combined Conditions (Regions)

You may be asked to sketch the region defined by two conditions simultaneously, for example:

\(|z - 3i| \le 2\) AND \(\frac{\pi}{6} < \arg(z - 3) \le \frac{\pi}{3}\)

Break it down:

1. The first part is the closed disk centred at \(3i\) with radius 2.
2. The second part is the region between two half-lines starting at \(z=3\), restricted by the angles \(\pi/6\) and \(\pi/3\).
3. The solution region is the area where the disk and the angular region overlap.

Key Takeaway for Loci: Modulus means distance (related to circles/lines). Argument means angle (related to half-lines/sectors). Always identify the fixed point \(z_1\) first!

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🎉 Summary and Final Thoughts

You have successfully tackled the further theory of complex numbers! These tools—De Moivre's Theorem and Loci—are fundamental to solving geometric problems in mathematics and physics.

Keep practising converting between Cartesian and Polar forms, and remember that when finding roots, the Argand diagram is your best friend for visualization! Good luck!