Complex loci, purely real conditions and proofs

|z − 2 − i| = 3

• Every z is 3 from 2 + i.
• Centre 2 + i, the point (2, 1).
• A circle of radius 3.

Into Cartesian form

• Group the two parts
• |a + bi|² = a² + b², no i

1. z − (2 + i) = (x + iy) − (2 + i)
2. = (x − 2) + (y − 1)igroup the parts
3. (x − 2)² + (y − 1)² = 9modulus squared = 3²
4. x² − 4x + 4 + y² − 2y + 1 = 9
5. x² + y² − 4x − 2y = 4expanded form

|z − 4| = |z − 2i|

• Equally far from 4 and from 2i.
• Join them and find the midpoint.
• The perpendicular bisector.

Square both sides

• x² and y² cancel
• Check with the midpoint

1. |z − 4|² = |z − 2i|²square both sides
2. (x − 4)² + y² = x² + (y − 2)²
3. x² − 8x + 16 + y²expand
4. = x² + y² − 4y + 4
5. −8x + 16 = −4y + 4x² and y² cancel
6. 4y = 8x − 12rearrange
7. y = 2x − 3
8. at (2, 1): 2(2) − 3 = 1midpoint lies on it
9. gradients 2 × (−1/2) = −1perpendicular

w = (k + 2i)/(1 + i)

• Conjugate first
• Real: Im(w) = 0
• Imaginary: Re(w) = 0

1. w = (k + 2i)/(1 + i)k real
2. × (1 − i)/(1 − i)conjugate
3. bottom: (1 + i)(1 − i) = 21 − i² = 2
4. top: (k + 2i)(1 − i)
5. = k − ki + 2i − 2i²expand
6. = (k + 2) + (2 − k)ii² = −1, group
7. w = ((k + 2) + (2 − k)i)/2
8. Im = 0: 2 − k = 0, so k = 2real: w = 2
9. Re = 0: k + 2 = 0, so k = −2imaginary: w = 2i

z + 4/z is purely real

• Write 4/z in x + iy form
• Take the imaginary part

1. w = z + 4/z, z = x + iy
2. 4/(x + iy) × (x − iy)/(x − iy)
3. = 4(x − iy)/(x² + y²)bottom x² + y²
4. w = x + iy + 4(x − iy)/(x² + y²)
5. Im(w) = y − 4y/(x² + y²)imaginary parts
6. y − 4y/(x² + y²) = 0purely real

Clear the fraction, factorise

• x² + y² is never 0 here
• Never divide by y

1. y − 4y/(x² + y²) = 0
2. y(x² + y²) − 4y = 0× (x² + y²)
3. y(x² + y² − 4) = 0factorise
4. y = 0 or x² + y² = 4
5. y = 0, z ≠ 0the real axis
6. x² + y² = 4, so |z| = 2circle, radius 2

What a proof needs: the 2025 error

• Prove: real roots for every real k.
• Trying some k is not a proof.
• It must work for any k.

x² + kx + (k − 1) = 0

• Write the discriminant
• A square is never negative
• End with the conclusion

1. a = 1, b = k, c = k − 1
2. b² − 4ac = k² − 4(1)(k − 1)
3. = k² − 4k + 4expand
4. = (k − 2)²factorise
5. (k − 2)² ≥ 0 for all real ka square is never < 0
6. so b² − 4ac ≥ 0
7. real roots for every real k

Prove: z is purely imaginary

• Substitute x + iy, square
• End with the sentence

1. |z − 1| = |z + 1|, z = x + iy
2. (x − 1)² + y² = (x + 1)² + y²
3. x² − 2x + 1 + y²expand
4. = x² + 2x + 1 + y²
5. −2x = 2xx², y² and 1 cancel
6. −4x = 0, so x = 0
7. Re(z) = 0the real part
8. so z is purely imaginary

The answers: questions 1 and 2

1. 1. z − 3i = x + (y − 3)isubstitute x + iy
2. x² + (y − 3)² = 2² = 4centre 3i, radius 2
3. 2. (x + 2)² + y² = (x − 4)² + y²square both sides
4. x² + 4x + 4 = x² − 8x + 16y² cancels, expand
5. 12x = 12x² cancels
6. x = 1a vertical line

The answers: questions 3 and 4

1. 3. (3 + i)(k + 2i)
2. = 3k + 6i + ki + 2i²expand
3. = (3k − 2) + (k + 6)ii² = −1, group
4. k + 6 = 0, so k = −6purely real: Im = 0
5. 3(−6) − 2 = −20the real number
6. 4. b² − 4ac = (k + 1)² − 4(k)(1)
7. = k² + 2k + 1 − 4kexpand
8. = k² − 2k + 1 = (k − 1)²factorise
9. (k − 1)² ≥ 0: real rootsa square is never < 0