Loci in Argand Diagrams

How do I sketch the locus of $Re z = k$ or $Im z = k$ on an Argand diagram?

All complex numbers, $z = x + i y$ , that satisfy the equation $Re z = k$ lie on a vertical line with Cartesian equation $x = k$
- Any complex number along this vertical line will have a real part of $k$
All complex numbers, $z = x + i y$ , that satisfy the equation $Im z = k$ lie on a horizontal line with Cartesian equation $y = k$
- Any complex number along this horizontal line will have an imaginary part of $k$
E.g. The loci $Re z = 4$ and $Im z = - 3$ are represented by the vertical line $x = 4$ and the horizontal line $y = - 3$

8-2-4_notes_fig1

Sketching the loci of $Re z = 4$ and $Im z = - 3$

How do I sketch the locus of $|z - a| = k$ on an Argand diagram?

All complex numbers, $z$ , that satisfy the equation $|z| = k$ lie on a circle of radius $k$ about the origin
- E.g. the locus of $|z| = 10$ is a circle of radius 10, centred at the origin, as every complex number on that circle has a modulus of 10
For a given complex number, $a$ , all complex numbers, $z$ , that satisfy the equation $|z - a| = k$ lie on a circle of radius $k$ about the centre $a$
- This is because $|z - a|$ represents the distance between complex numbers $z$ and $a$
- E.g. the locus of $|z - (3 + 4 i)| = 10$ is a circle of radius 10 about $(3 + 4 i)$
Many equations need to be adjusted algebraically into the correct $|z - a|$ form
- E.g. to find the centre of the circle $|z - 8 i + 3| = 12$ , first rewrite it as $|z - (- 3 + 8 i)| = 12$ , giving the centre as $- 3 + 8 i$
- E.g. to find the centre of the circle $|z + i| = 2$ , first rewrite it as $|z - (- i)| = 2$ , giving the centre as $(- i)$
- Note that the centre of the circle $|z| = 5$ is the origin (it can be thought of as $|z - 0| = 5$ )
In order to sketch correctly, check whether the origin lies outside, on or inside the circle
- E.g. for the locus of $|z - (3 + 4 i)| = 10,$ the distance from the centre of the circle, $3 + 4 i$ , to the origin is 5 (by Pythagoras), which is less than the radius of 10; a sketch must therefore show the origin inside the circle
By knowing the radius and centre of a circle, the Cartesian equation of the circle can be found
- The circle $|z - (3 + 4 i)| = 10$ has a radius of 10 and centre of $(3, 4)$ in coordinates, so the equation of the circle is ${(x - 3)}^{2} + {(y - 4)}^{2} = 100$

8-2-4_notes_fig2

Sketching the loci of $| z | = 10$ and $| z - (3 + 4 i) | = 10$

How do I sketch the locus of $|z - a| = | z - b |$ on an Argand diagram?

For two given complex numbers, $a$ and $b$ , all complex numbers, $z$ , that satisfy the equation $|z - a| = | z - b |$ lie on the perpendicular bisector of $a$ and $b$
- This is because the distance from $z$ to $a$ must equal the distance from $z$ to $b$
  - a condition that is satisfied by all the complex numbers, $z$ , on the perpendicular bisector of $a$ and $b$
- E.g. the locus of $|z - 3 + 2 i| = | z + 8 |$ can be rewritten as $|z - (3 - 2 i)| = |z - (- 8)|$ which is the perpendicular bisector of the points $3 - 2 i$ and $- 8$
A sketch of the perpendicular bisector is sufficient, without finding its exact equation (though this could be found using coordinate geometry methods)

8-2-4_notes_fig3

Sketching the loci of $| z - i | = |z - 3 i|$ and $| z - (3 - 2 i) | = |z - (- 8)|$

How do I sketch the locus of $\arg (z - a) = α$ on an Argand diagram?

All complex numbers, $z$ , that satisfy the equation $\arg z = α$ lie on a half-line from the origin at an angle of $α$ to the positive real axis
- Although the half-line starts at the origin, the origin itself ( $z = 0$ ) does not satisfy the equation $\arg z = α$ as $\arg 0$ is undefined (there is no angle at the origin)
- To show the exclusion of $z = 0$ from the locus of $\arg z = α$ , a small open circle at the origin is used
- E.g. the locus of $\arg z = \frac{π}{4}$ is a half-line of angle $\frac{π}{4}$ to the positive real axis, starting from the origin, with an open circle at the origin
For a given complex number, $a$ , all complex numbers, $z$ , that satisfy the equation $\arg (z - a) = α$ lie on a half-line from the point $a$ at an angle of $α$ to the positive real axis, with an open circle to show the exclusion of $z = a$
- E.g. the locus of $\arg (z - 1 - 5 i) = \frac{2 π}{3}$ can be rewritten as $\arg (z - (1 + 5 i)) = \frac{2 π}{3}$ , which is a half-line of angle $\frac{2 π}{3}$ measured from the point $1 + 5 i$ , with an open circle at $1 + 5 i$ to show its exclusion
In some cases, the equation of the half-line can be found using a sketch to help
- E.g. the locus of $\arg z = \frac{π}{4}$ is the half-line $y = x$ for $x > 0$
- E.g. the locus of $\arg (z - (8 + 5 i) = - \frac{π}{4}$ can be thought of, in coordinate geometry, as the half-line through $(8, 5)$ with gradient -1, giving $y = - x + 13$ for $x > 8$
- Whilst not examinable, the half-line equation for a more general angle, $\arg z = α$ , is $y = (\tan α) x$ for $x > 0$ , as the $gradient = \frac{opposite}{adjacent} = \tan α$

8-2-4_notes_fig4

Sketching the loci of $\arg z = \frac{π}{4}$ and $\arg (z - (1 + 5 i)) = \frac{2 π}{3}$

Worked example

8-2-4_example_fig1-part-1

8-2-4_example_fig1-part-2

Examiner Tip

In the exam, do not worry about making your diagrams perfect.
A quick sketch with all the key features is sufficient.

Loci in Argand Diagrams (CIE A Level Maths: Pure 3)

Revision Note

How do I sketch the locus of or on an Argand diagram?

How do I sketch the locus of on an Argand diagram?

How do I sketch the locus of on an Argand diagram?

How do I sketch the locus of on an Argand diagram?

You've read 0 of your 5 free revision notes this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

1. Algebra & Functions

1.1 Modulus Functions

1.1.1 Modulus Functions - Sketching Graphs

1.1.2 Modulus Functions - Solving Equations

1.2 Polynomials

1.2.1 Polynomial Division

1.2.2 Factor & Remainder Theorem

1.2.3 Factorisation

1.2.4 Rational Expressions

1.2.5 Top Heavy Rational Expressions

1.3 Partial Fractions

1.3.1 Linear Denominators

1.3.2 Squared Linear Denominators

1.3.3 Quadratic Denominators

1.4 Further Modelling with Functions

1.4.1 Further Modelling with Functions

1.5 General Binomial Expansion

1.5.1 General Binomial Expansion

1.5.2 General Binomial Expansion - Subtleties

1.5.3 General Binomial Expansion - Multiple

1.5.4 Approximating values

2. Logs & Exponentials

2.1 Logarithmic & Exponential Function

2.1.1 Exponential Functions

2.1.2 Logarithmic Functions

2.1.3 "e"

2.1.4 Derivatives of Exponential Functions

2.2 Laws of Logarithms

2.2.1 Laws of Logarithms

2.2.2 Exponential Equations

2.3 Modelling with Logs & Exponentials

2.3.1 Exponential Growth & Decay

2.3.2 Using Exps & Logs in Modelling

2.3.3 Using Log Graphs in Modelling

3. Trigonometry

3.1 Reciprocal Trigonometric Functions

3.1.1 Reciprocal Trig Functions - Definitions

3.1.2 Reciprocal Trig Functions - Graphs

3.1.3 Trigonometry - Further Identities

3.2 Compound & Double Angle Formulae

3.2.1 Compound Angle Formulae

3.2.2 Double Angle Formulae

3.2.3 R addition formulae Rcos Rsin etc

3.3 Further Trigonometric Equations

3.3.1 Strategy for Further Trigonometric Equations

3.4 Trigonometric Proof

3.4.1 Trigonometric Proof

4. Differentiation

4.1 Further Differentiation

4.1.1 Differentiating Other Functions (Trig, ln & e etc)

4.1.2 Product Rule

4.1.3 Quotient Rule

4.2 Implicit Differentiation

4.2.1 Implicit Differentiation

4.3 Differentiation of Parametric Equations

4.3.1 Parametric Equations - Basics

4.3.2 Parametric Equations - Eliminating the Parameter

4.3.3 Parametric Equations - Sketching Graphs

4.3.4 Parametric Differentiation

5. Integration

5.1 Further Integration

5.1.1 Integrating Other Functions (Trig, ln & e etc)

5.1.2 Integrating with Trigonometric Identities

5.1.3 f'(x)/f(x)

5.1.4 Substitution (Reverse Chain Rule)

5.1.5 Harder Substitution

5.1.6 Integration by Parts

5.1.7 Integration using Partial Fractions

5.1.8 Integration Decision Making

5.2 Differential Equations

5.2.1 General Solutions

How do I sketch the locus of $Re z = k$ or $Im z = k$ on an Argand diagram?

How do I sketch the locus of $|z - a| = k$ on an Argand diagram?

How do I sketch the locus of $|z - a| = | z - b |$ on an Argand diagram?

How do I sketch the locus of $\arg (z - a) = α$ on an Argand diagram?