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 $\arg (z - (1 + 5 i)) = \frac{2 π}{3}$

Examiner Tip

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

Worked example

On separate axes, sketch the locus of points representing complex numbers, $z$ , that satisfy the following equations:

Im z = 2

al-fm-1-1-5-loci-in-argand-diagrams-we-solution-a

| z + 5 - 12 i | = 8

al-fm-1-1-5-loci-in-argand-diagrams-we-solution-b

| z - 4 i | = | z + i - 1 |

edexcel-fm-core-pure-loci-in-argand-diagrams-fix1

\arg (z + i) = - \frac{π}{3}

edexcel-fm-core-pure-loci-in-argand-diagrams-fix2

Loci in Argand Diagrams (Edexcel A Level Further Maths: Core Pure)

Revision Note