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19 Jan 2026

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4 pages

Understanding Argand Diagrams and Complex Number Loci

Los

@los

Ever wondered how angles work in pure maths beyond degrees?... Show more

1 / 4

$# RADIANS $ \frac{\theta}{360} = \frac{r}{2\pi r} $ 1 radian $\approx$ 57.3° 1° $\approx$ 57.3° curcumfererence = 2$\pi$r $ \frac{57}{3$

Understanding Radians

Radians are just another way to measure angles, and they're actually more natural than degrees in many maths problems. Think of it this way: one radian is the angle you get when the arc length equals the radius of a circle.

The key relationship is θ/360° = arc length/(2πr). This means 1 radian ≈ 57.3°, whilst a full circle is 2π radians instead of 360°. You'll need to memorise some common conversions: 180° = π radians, 90° = π/2 radians, and 60° = π/3 radians.

The formula arc length = θr (where θ is in radians) makes calculations much simpler than using degrees. This is why radians pop up everywhere in A-level maths - they make the algebra cleaner and more elegant.

Quick tip: When you see π in an angle, you're definitely working in radians!

Introduction to Argand Diagrams

Argand diagrams let you plot complex numbers like z = 3 + 2i on a coordinate system. The horizontal axis represents the real part, whilst the vertical axis shows the imaginary part.

Every complex number z = a + bi becomes a point (a, b) on this diagram. The distance from the origin to this point is called the modulus |z| = √ $a² + b²$ . Meanwhile, the angle from the positive real axis is the argument arg(z).

This visual approach makes complex number operations much easier to understand. Instead of just manipulating algebra, you can actually see what's happening geometrically when you add, multiply, or transform complex numbers.

$# RADIANS $ \frac{\theta}{360} = \frac{r}{2\pi r} $ 1 radian $\approx$ 57.3° 1° $\approx$ 57.3° curcumfererence = 2$\pi$r $ \frac{57}{3$

Polar Form of Complex Numbers

You can write any complex number in polar form: z = r $cos θ + i sin θ$ , where r is the modulus and θ is the argument. This form is brilliant for multiplication and division because the moduli multiply/divide whilst the arguments add/subtract.

For example, if z₁ = 2 $cos 30° + i sin 30°$ and z₂ = 3 $cos 60° + i sin 60°$ , then z₁z₂ = 6 $cos 90° + i sin 90°$ . The argument rules are: arg(z₁z₂) = arg(z₁) + arg(z₂) and arg $z₁/z₂$ = arg(z₁) - arg(z₂).

Converting between forms is straightforward. From rectangular form a + bi: r = √ $a² + b²$ and θ = tan⁻¹ $b/a$ . From polar back to rectangular: a = r cos θ and b = r sin θ.

Remember: Always check which quadrant your complex number is in when finding the argument!

Loci and Geometric Properties

The locus |z - z₁| = r represents a circle with centre z₁ and radius r on an Argand diagram. This is incredibly useful for solving geometric problems involving complex numbers.

If |z - z₁| = |z - z₂|, you get the perpendicular bisector of the line segment joining z₁ and z₂. The Cartesian equation becomes $x - x₁$ ² + $y - y₁$ ² = r² for a circle.

These geometric interpretations help you visualise complex equations. Instead of just solving algebraically, you can sketch the locus and find intersections, maximum distances, or specific values that satisfy multiple conditions.

$# RADIANS $ \frac{\theta}{360} = \frac{r}{2\pi r} $ 1 radian $\approx$ 57.3° 1° $\approx$ 57.3° curcumfererence = 2$\pi$r $ \frac{57}{3$

Advanced Loci Problems

Complex loci can form various shapes depending on the conditions. The equation |z - a| = |z - b| creates a straight line (the perpendicular bisector), whilst arg $z - z₁$ = α gives you a half-line from point z₁.

When finding maximum and minimum values of |z|, draw your locus first. The maximum distance from the origin occurs at the point furthest from O, whilst the minimum is at the closest point. For circles, this means |z|max = |centre| + radius and |z|min = ||centre| - radius|.

Combined conditions often appear in exam questions. You might need to find points satisfying both |z - a| = r and arg $z - b$ = θ simultaneously. Sketch both loci and look for their intersection points.

The gradient of a line in the complex plane relates to arguments. If arg $z - z₁$ = α, then the Cartesian equation is y - y₁ = tan(α) $x - x₁$ , but remember to specify the correct domain.

Exam tip: Always sketch the locus first - it makes finding intersections and extreme values much easier!

Solving Complex Geometric Problems

Real exam questions combine multiple conditions like |z - 4 - 2i| ≤ 2 (inside a circle) with arg $z - 2 - 2i$ = π/4 $on a half-line$ . Intersection problems require you to solve these simultaneously.

To find specific complex numbers satisfying multiple conditions, use the geometric constraints to set up equations. If you know the modulus and argument, you can write z in polar form and convert back to rectangular form.

Distance relationships like |z - 4| < |z - 6| represent regions where points are closer to one complex number than another. These create half-planes bounded by the perpendicular bisector.

$# RADIANS $ \frac{\theta}{360} = \frac{r}{2\pi r} $ 1 radian $\approx$ 57.3° 1° $\approx$ 57.3° curcumfererence = 2$\pi$r $ \frac{57}{3$

Regions and Inequalities

Complex inequalities define regions on Argand diagrams. The condition |z - 4 - 2i| ≤ 2 represents the inside (including boundary) of a circle centred at 4 + 2i with radius 2.

Combined regions use intersection notation like {z ∈ ℂ: |z - 4 - 2i| ≤ 2} ∩ {z ∈ ℂ: |z - 4| < |z - 6|}. You need to find where all conditions are satisfied simultaneously by shading the overlapping areas.

Argument inequalities such as 0 ≤ arg $z - 2 - 2i$ ≤ π/4 create angular sectors. These are wedge-shaped regions between two half-lines emanating from a fixed point.

When sketching these regions, work systematically: draw each individual locus or boundary first, then identify the region satisfying each inequality, and finally find their intersection.

Visual strategy: Use different colours or shading patterns for each condition, then identify where they all overlap!

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19 Jan 2026

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4 pages

Understanding Argand Diagrams and Complex Number Loci

Los

@los

Ever wondered how angles work in pure maths beyond degrees? Radians and Argand diagrams are two powerful tools that'll help you tackle complex numbers and advanced trigonometry with confidence. These concepts might seem tricky at first, but they're actually quite... Show more

$# RADIANS $ \frac{\theta}{360} = \frac{r}{2\pi r} $ 1 radian $\approx$ 57.3° 1° $\approx$ 57.3° curcumfererence = 2$\pi$r $ \frac{57}{3$

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Understanding Radians

Quick tip: When you see π in an angle, you're definitely working in radians!

Introduction to Argand Diagrams

Argand diagrams let you plot complex numbers like z = 3 + 2i on a coordinate system. The horizontal axis represents the real part, whilst the vertical axis shows the imaginary part.

$# RADIANS $ \frac{\theta}{360} = \frac{r}{2\pi r} $ 1 radian $\approx$ 57.3° 1° $\approx$ 57.3° curcumfererence = 2$\pi$r $ \frac{57}{3$

Sign up to see the contentIt's free!

Access to all documents

Improve your grades

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Polar Form of Complex Numbers

Converting between forms is straightforward. From rectangular form a + bi: r = √ $a² + b²$ and θ = tan⁻¹ $b/a$ . From polar back to rectangular: a = r cos θ and b = r sin θ.

Remember: Always check which quadrant your complex number is in when finding the argument!

Loci and Geometric Properties

The locus |z - z₁| = r represents a circle with centre z₁ and radius r on an Argand diagram. This is incredibly useful for solving geometric problems involving complex numbers.

If |z - z₁| = |z - z₂|, you get the perpendicular bisector of the line segment joining z₁ and z₂. The Cartesian equation becomes $x - x₁$ ² + $y - y₁$ ² = r² for a circle.

$# RADIANS $ \frac{\theta}{360} = \frac{r}{2\pi r} $ 1 radian $\approx$ 57.3° 1° $\approx$ 57.3° curcumfererence = 2$\pi$r $ \frac{57}{3$

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Advanced Loci Problems

Exam tip: Always sketch the locus first - it makes finding intersections and extreme values much easier!

Solving Complex Geometric Problems

Distance relationships like |z - 4| < |z - 6| represent regions where points are closer to one complex number than another. These create half-planes bounded by the perpendicular bisector.

$# RADIANS $ \frac{\theta}{360} = \frac{r}{2\pi r} $ 1 radian $\approx$ 57.3° 1° $\approx$ 57.3° curcumfererence = 2$\pi$r $ \frac{57}{3$

Sign up to see the contentIt's free!

Access to all documents

Improve your grades

Join milions of students

By signing up you accept Terms of Service and Privacy Policy

Regions and Inequalities

Complex inequalities define regions on Argand diagrams. The condition |z - 4 - 2i| ≤ 2 represents the inside (including boundary) of a circle centred at 4 + 2i with radius 2.

Argument inequalities such as 0 ≤ arg $z - 2 - 2i$ ≤ π/4 create angular sectors. These are wedge-shaped regions between two half-lines emanating from a fixed point.

When sketching these regions, work systematically: draw each individual locus or boundary first, then identify the region satisfying each inequality, and finally find their intersection.

Visual strategy: Use different colours or shading patterns for each condition, then identify where they all overlap!