Loci in Argand Diagrams (Edexcel A Level Further Maths): Revision Note

Loci in Argand Diagrams

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

• All complex numbers, , that satisfy the equation  lie on a vertical line with Cartesian equation 

  • Any complex number along this vertical line will have a real part of 

• All complex numbers, , that satisfy the equation  lie on a horizontal line with Cartesian equation 

  • Any complex number along this horizontal line will have an imaginary part of 

• E.g. The loci  and  are represented by the vertical line  and the horizontal line  

Sketching the loci of  and 

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

• All complex numbers, , that satisfy the equation  lie on a circle of radius about the origin

  • E.g. the locus of  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, , all complex numbers, , that satisfy the equation  lie on a circle of radius  about the centre 

  • This is because  represents the distance between complex numbers  and  

  • E.g. the locus of  is a circle of radius 10 about

• Many equations need to be adjusted algebraically into the correct form

  • E.g. to find the centre of the circle , first rewrite it as , giving the centre as 

  • E.g. to find the centre of the circle , first rewrite it as , giving the centre as 

  • Note that the centre of the circle  is the origin (it can be thought of as )

• In order to sketch correctly, check whether the origin lies outside, on or inside the circle

  • E.g. for the locus of  the distance from the centre of the circle, , 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  has a radius of 10 and centre of  in coordinates, so the equation of the circle is 

Sketching the loci of  and 

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

• For two given complex numbers, and , all complex numbers, , that satisfy the equation  lie on the perpendicular bisector of and 

  • This is because the distance from  to  must equal the distance from to 

    • a condition that is satisfied by all the complex numbers, , on the perpendicular bisector of  and 

  • E.g. the locus of  can be rewritten as  which is the perpendicular bisector of the points and 

• A sketch of the perpendicular bisector is sufficient, without finding its exact equation (though this could be found using coordinate geometry methods)

Sketching the loci of  and 

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

• All complex numbers, , that satisfy the equation  lie on a half-line from the origin at an angle of  to the positive real axis

  • Although the half-line starts at the origin, the origin itself () does not satisfy the equation   as  is undefined (there is no angle at the origin)

  • To show the exclusion of  from the locus of , a small open circle at the origin is used

  • E.g. the locus of  is a half-line of angle  to the positive real axis, starting from the origin, with an open circle at the origin

• For a given complex number, , all complex numbers, , that satisfy the equation  lie on a half-line from the point  at an angle of  to the positive real axis, with an open circle to show the exclusion of 

  • E.g. the locus of  can be rewritten as , which is a half-line of angle measured from the point , with an open circle at  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 is the half-line  for  

  • E.g. the locus of  can be thought of, in coordinate geometry, as the half-line through with gradient -1, giving  for  

  • Whilst not examinable, the half-line equation for a more general angle, , is  for , as the 

Sketching the loci of  and