How do I sketch the locus of?oron 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 ofon 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 (-i)
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 (3,4)?in coordinates, so the equation of the circle is
Sketching the loci ofand
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 -8
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
E.g. the locus of can be thought of, in coordinate geometry, as the half-line through (8,5)with gradient -1, giving
Whilst not examinable, the half-line equation for a more general angle,, is ? , as the
Sketching the loci ofand
Exam 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,, that satisfy the following equations:
a)
b)
c)
d)
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