1. Complex numbers
Loci

2. Complex numbers: Loci in the Argand diagram
KUS objectives
BAT Use complex numbers to represent a locus of points on an Argand diagram
Starter:

3. WB1 distance between two points
𝑧− 𝑧 1 represents the distance between points z and 𝑧 1
𝑧 2 − 𝑧 1 = 𝑥 2 − 𝑥 2 +𝑖 𝑦 2 − 𝑦 2
Using Pythagoras
𝑑= 𝑥 2 − 𝑥 𝑦 2 − 𝑦 1 2 SO
𝑑= 𝑧 2 − 𝑧 1 = 𝑥 2 − 𝑥 𝑦 2 − 𝑦 1 2

4. describes the circle with centre z1 and radius r
WB2 Loci of a circle with complex numbers
locus of z
In GCSE, you learnt that the circle is the set of points equidistant from a point
the statement 𝑧− 𝑧 1 =𝑟
describes the circle with centre z1 and radius r
This can be easily proved
𝑧− 𝑧 1 = 𝑥− 𝑥 𝑦− 𝑦 =𝑟
Square both sides
𝑥− 𝑥 𝑦− 𝑦 = 𝑟 2
This is the Cartesian equation of a circle with centre (x1,y1) and radius r

5. WB2 (cont) Draw each circle
ii) 𝑧−3 =2
i) 𝑧−2𝑖 =4
i) 𝑧−3−2𝑖 =1
i) 𝑧−5−12𝑖 =4
Can you give the Cartesian equation of each circle ?

6. 𝑥−4 2 + 𝑦−0 2 = 5 2 WB3 Given that z satisfies 𝑧−4 =5
Sketch the locus of z on an Argand diagram
Find the values of z that satisfy both 𝑧−4 =5 and I𝑚 𝑧 =0
Find the values of z that satisfy both 𝑧−4 =5 and 𝑅𝑒 𝑧 =0
a) 𝑧−4 =5 is a circle radius 5
With centre (0, 4) Im 4 5 Re
b) 𝐼𝑚 (𝑧)=0 represents the real axis
The points where the circle cuts the x-axis are
(-1, 0) and (9, 0)
So the values of z at these points are z = -1 , and z= 9
c) Re(𝑧)=0 represents the Imaginary axis
The points where the circle cuts the y-axis are
𝑥− 𝑦−0 2 = 5 2
Where 0− 𝑦 2 = 5 2
Where 𝑦=± points (0, 3) and (0, -3)
The values of z are z = 3i and z = -3i

7. WB4 A complex number is represented by the point P in the Argand diagram.
Given that 𝑧−5−3𝑖 =3
Sketch the locus of P
Find the Cartesian equation of this locus
Find the maximum value of arg z in the interval −𝜋, 𝜋 Im 5 3 Re P
𝑧−5−3𝑖 = 𝑧−(5+3𝑖) =3
This distance is always 3
so a circle radius 3, centre (5, 3)
b) 𝑥− 𝑦−3 2 = 3 2
The maximum angle is the angle OA
Makes with the +ve Real axis
line OC bisects this angle Im 5 3 Re A B
𝜃 2 O C
Angle 𝐵𝑂𝐶= arctan 3 5 = 𝑐
So max angle 𝐵𝑂𝐴= 2×arctan 3 5 = 1.08 𝑐

8. WB5 Given that the complex number z=x+yi satisfies 𝑧−12−5𝑖 =3
Find the minimum value of 𝑧 and maximum value of 𝑧
𝑧−12−5𝑖 = 𝑧−(12+5𝑖) =3
This distance is always 3
so a circle radius 3, centre (12, 5) Im
C(12, 5) 3 Re X Y O
𝑧 𝑚𝑖𝑛 =𝑂𝑋=𝑂𝐶−𝐶𝑋
Distance OC = =13
So 𝑧 𝑚𝑖𝑛 = 13 – 3 = 10
𝑧 𝑚𝑎𝑥 =𝑂𝑌=13+3=16

9. 𝑧− 𝑧 1 = 𝑧− 𝑧 2 WB6 Perpendicular bisector locus of z
In GCSE, you learnt that a point is equidistant from 2 points if it lies on the perpendicular bisector of the line segment between them.
If you are on this line,
the distance to each point is the same
Hence the statement
𝑧− 𝑧 1 = 𝑧− 𝑧 2
describes the perpendicular bisector of the line segment between z1 and z2

10. WB7a Given that 𝑧−3 = 𝑧+𝑖
a) Sketch the locus of z and find the Cartesian equation of this locus
b) Find the least possible value of 𝑧
𝑧−3 = 𝑧+𝑖 is the perpendicular bisector of the line segment joining the points (3, 0) and (0, -1) Im
(3, 0) Re O
Gradient 1 3
(0, -1)
3 2 , − 1 2
Gradient −3
The midpoint is , − 1 2
The gradient is the negative reciprocal of the line joining (3, 0) and (0, -1)
The perpendicular line has equation
𝑦+ 1 2 =−3 𝑥− 3 2
Or 𝑦=−3𝑥+4

11. WB7b Given that 𝑧−3 = 𝑧+𝑖
a) Sketch the locus of z and find the Cartesian equation of this locus
b) Find the least possible value of 𝑧
The perpendicular line has equation
𝑦=−3𝑥+4
The least value of 𝑧 is the minimum distance from O to this line Im
(3, 0) Re O
Gradient 1 3
(0, -1)
3 2 , − 1 2
Gradient −3
This distance is a line parallel to the line from (0, -1) to (3, 0) so has gradient and goes through the origin
So has equation y= 1 3 𝑥
The lines meet where 𝑥=−3𝑥+4
solves to give Point , 2 5
so the minimum distance is =

12. WB7a (revisited) Given that 𝑧−3 = 𝑧+𝑖
a) Sketch the locus of z and find the Cartesian equation of this locus
b) Find the least possible value of 𝑧
𝑧−3 = 𝑧+𝑖 can also be dealt with by squaring both sides using 𝑧=𝑥+𝑦𝑖 Im
(3, 0) Re O
Gradient 1 3
(0, -1)
3 2 , − 1 2
Gradient −3
𝑥+𝑦𝑖−3 2 = 𝑥+𝑦𝑖+𝑖 2
(𝑥−3)+𝑦𝑖 2 = 𝑥+𝑖(𝑦+1) 2
Square the Real and imaginary parts
(𝑥−3) 2 + 𝑦 2 = 𝑥 2 + (𝑦+1) 2
𝑥 2 −6𝑥+9+ 𝑦 2 = 𝑥 2 + 𝑦 2 +2𝑦+1
Rearranges to 𝑦=−3𝑥 the same answer as our geometric solution
Often squaring both sides is the quickest way to a solution

13. WB8 𝑧−6 =2 𝑧+6−9𝑖 Find the Cartesian equation of the
locus of z and sketch it on a Argand diagram
Collecting real and imaginary parts
Squaring
locus of z
Circle centre (-10,12) radius 10

14. WB9 𝑧+2+3𝑖 = 𝑧+1−4𝑖 Find the Cartesian equation of the
locus of z and sketch it on a Argand diagram
Collecting real and imaginary parts
Squaring
locus of z

15. KUS objectives
BAT Use complex numbers to represent a locus of points on an Argand diagram
self-assess
One thing learned is –
One thing to improve is –

16. END