1. Complex Loci – Modulus Question
|Z – (3 + 2i)| = |Z – (-2 + 1i)|
Means magnitude of vector joining variable vector Z to a fixed vector Z1 is equal in magnitude to vector joining variable vector Z to a fixed vector Z2 2 4 -2 -4 x y
Z1 = 3+2i
Z2 = i
Z – Z1
Z – Z2
Free vector Z
Z2 = i
Z1 = 3+2i
Z – Z1
Z – Z2

2. Complex Loci – Modulus Question
|Z – (3 + 2i)| = |Z – (-2 + 1i)|
Means magnitude of vector joining variable vector Z to a fixed vector Z1 is equal in magnitude to vector joining variable vector Z to a fixed vector Z2
Z1 = 3+2i 2 4 -2 -4 x y
Z2 = i
Free vector Z
Z – Z1
Z – Z2

3. Complex Loci – Modulus Question
|Z – (3 + 2i)| = |Z – (-2 + 1i)|
Means magnitude of vector joining variable vector Z to a fixed vector Z1 is equal in magnitude to vector joining variable vector Z to a fixed vector Z2
Z1 = 3+2i 2 4 -2 -4 x y
Z2 = i
Free vector Z
Z – Z1
Z – Z2

4. Complex Loci – Modulus Question
|Z – (3 + 2i)| = |Z – (-2 + 1i)|
Means magnitude of vector joining variable vector Z to a fixed vector Z1 is equal in magnitude to vector joining variable vector Z to a fixed vector Z2
Z1 = 3+2i 2 4 -2 -4 x y
Z2 = i
Free vector Z
Z – Z1
Z – Z2

5. Complex Loci – Modulus Question
|Z – (3 + 2i)| = |Z – (-2 + 1i)|
Means magnitude of vector joining variable vector Z to a fixed vector Z1 is equal in magnitude to vector joining variable vector Z to a fixed vector Z2
Z1 = 3+2i 2 4 -2 -4 x y
Z2 = i
Free vector Z
Z – Z1
Z – Z2

6. Complex Loci – Modulus Question
|Z – (3 + 2i)| = |Z – (-2 + 1i)|
Means magnitude of vector joining variable vector Z to a fixed vector Z1 is equal in magnitude to vector joining variable vector Z to a fixed vector Z2
Z1 = 3+2i 2 4 -2 -4 x y
Z2 = i
Free vector Z
Z – Z1
Z – Z2

7. Complex Loci – Modulus Question
|Z – (3 + 2i)| = |Z – (-2 + 1i)|
So Z lies on a the perpendicular bisector of the line joining Z1 and Z2
Z1 = 3+2i 2 4 -2 -4 x y
Z2 = i
Free vector Z
Z – Z1
Z – Z2

8. Using Algebra to deduce the equation of the locus
|Z – (3 + 2i)| = |Z – (– 2 + 1i)|
As Z is a variable vector let Z = x + iy
|x + iy – (3 + 2i)| = |x + iy – (– 2 + 1i)|
|(x – 3) + i(y – 2)| = |(x + 2) + i(y – 1)|
(x – 3)2 + (y – 2)2 = (x + 2)2 + (y – 1)2
x2 – 6x y2 – 4y + 4 = x2 + 4x y2 – 2y + 1
-2y = 10x – 8
y = -5x + 4

9. |Z – (3 + 2i)| = |Z – (-1 + 2i)|
So Z lies on a the perpendicular bisector of the line joining Z1 and Z2
Z1 = 3+2i 2 4 -2 -4 x y
y = -5x + 4
Z2 = i
Free vector Z
Z – Z1
Z – Z2