Complex numbers Loci

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

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 1 2 + 𝑦 2 − 𝑦 1 2 SO 𝑑= 𝑧 2 − 𝑧 1 = 𝑥 2 − 𝑥 1 2 + 𝑦 2 − 𝑦 1 2

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 1 2 + 𝑦− 𝑦 1 2 =𝑟 Square both sides 𝑥− 𝑥 1 2 + 𝑦− 𝑦 1 2 = 𝑟 2 This is the Cartesian equation of a circle with centre (x1,y1) and radius r

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 ?

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−4 2 + 𝑦−0 2 = 5 2 Where 0−4 2 + 𝑦 2 = 5 2 Where 𝑦=±3 points (0, 3) and (0, -3) The values of z are z = 3i and z = -3i

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−5 2 + 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 = 0.540 𝑐 So max angle 𝐵𝑂𝐴= 2×arctan 3 5 = 1.08 𝑐

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 = 12 2 + 5 2 =13 So 𝑧 𝑚𝑖𝑛 = 13 – 3 = 10 𝑧 𝑚𝑎𝑥 =𝑂𝑌=13+3=16

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

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 3 2 , − 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

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 1 3 and goes through the origin So has equation y= 1 3 𝑥 The lines meet where 1 3 𝑥=−3𝑥+4 solves to give Point 6 5 , 2 5 so the minimum distance is 6 5 2 + 2 5 2 = 2 5 10

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+4 the same answer as our geometric solution Often squaring both sides is the quickest way to a solution

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

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

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 –

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