Complex loci and proof: 7 exam questions solved step by step

Q1: |z − 3 + 2i| = 4

• Modulus squared: no i.
• Watch the centre’s signs.

1. z − 3 + 2i = (x − 3) + (y + 2)i
2. (x − 3)² + (y + 2)² = 16modulus squared
3. x² − 6x + 9 + y² + 4y + 4 = 16
4. x² + y² − 6x + 4y = 3expanded form
5. z − (3 − 2i): centre 3 − 2inot −3 + 2i
6. circle, centre (3, −2), radius 4
7. |(7 − 2i) − (3 − 2i)| = |4| = 4

Q2: |z + 1 − 2i| = |z − 3|

• Square both sides.
• x² and y² cancel.

1. (x + 1)² + (y − 2)²
2. = (x − 3)² + y²
3. x² + 2x + 1 + y² − 4y + 4expand left
4. = x² − 6x + 9 + y²expand right
5. 2x − 4y + 5 = −6x + 9x², y² cancel
6. 8x − 4y − 4 = 0
7. y = 2x − 1
8. midpoint (1, 1): 2(1) − 1 = 1
9. 2 × (−1/2) = −1perpendicular

Q2: the sketch

• Mark −1 + 2i and 3.
• Join them, find the midpoint.
• The perpendicular bisector.

Q3: w = (3 + ki)/(2 − i)

• Conjugate first.
• Purely imaginary: Re = 0.

1. × (2 + i)/(2 + i)conjugate
2. (2 − i)(2 + i) = 4 − i² = 5
3. (3 + ki)(2 + i)
4. = 6 + 3i + 2ki + ki²expand
5. = (6 − k) + (3 + 2k)ii² = −1, group
6. w = ((6 − k) + (3 + 2k)i)/5
7. 6 − k = 0, so k = 6purely imaginary
8. Im: 3 + 12 = 15 ≠ 0so w is not 0
9. w = 15i/5 = 3i

Q4: the imaginary part

• The new bottom is real.
• Only the top decides.

1. z = x + iy, z ≠ 2bottom not 0
2. w = (x + (y − 4)i)/((x − 2) + yi)
3. × ((x − 2) − yi) top and bottom
4. bottom: (x − 2)² + y² > 0real
5. Im(top) = (y − 4)(x − 2) − xy
6. = xy − 2y − 4x + 8 − xyexpand
7. = 8 − 4x − 2yxy cancels

Q4: purely real, so Im = 0

• Set Im = 0.
• Leave out z = 2.

1. Im(top) = 8 − 4x − 2y
2. 8 − 4x − 2y = 0purely real
3. y = 4 − 2x
4. x = 0: y = 4, the point 4i
5. x = 2: y = 0, the point 2
6. z = 2w undefined there
7. z = 1 + 2i: w = (1 − 2i)/(−1 + 2i)
8. = −1, real
9. line through 4i and 2, z ≠ 2

Q5: a proof for every k

• Examples do not prove it.
• Complete the square in k.

1. trying k = 1, 2, 3not a proof
2. a = 1, b = k − 3, c = −k
3. b² − 4ac = (k − 3)² − 4(1)(−k)
4. = k² − 6k + 9 + 4kexpand
5. = k² − 2k + 9collect
6. = (k − 1)² + 8complete the square
7. (k − 1)² ≥ 0, so b² − 4ac ≥ 8 > 0
8. two distinct real roots, all k

Q6: |z − 2| = 2|z + 1|

• Square the 2 as well.
• The squares stay: a circle.

1. (x − 2)² + y² = 4((x + 1)² + y²)
2. x² − 4x + 4 + y²expand left
3. = 4x² + 8x + 4 + 4y²expand right
4. 0 = 3x² + 12x + 3y²all to the right
5. x² + 4x + y² = 0÷ 3
6. (x + 2)² + y² = 4complete the square
7. circle: centre −2, radius 2
8. z = 0: |−2| = 2 = 2|1|

Q7: set up the proof

• Use |z| = 1 as x² + y² = 1.
• Need Re = 0.

1. z = x + iy, x² + y² = 1|z| = 1
2. w = ((x − 1) + iy)/((x + 1) + iy)
3. × ((x + 1) − iy) top and bottom
4. bottom: (x + 1)² + y² > 0z ≠ −1
5. need Re(top) = 0purely imaginary
6. Re(top) = (x − 1)(x + 1) + y²

Q7: finish the proof

• Use x² + y² = 1.
• End with the sentence.

1. Re(top) = (x − 1)(x + 1) + y²
2. = x² − 1 + y²
3. = (x² + y²) − 1regroup
4. = 1 − 1 = 0x² + y² = 1
5. Im(top) = y(x + 1) − y(x − 1) = 2y
6. w = 2yi/((x + 1)² + y²)
7. Re(w) = 0, w ≠ 0y ≠ 0 as z ≠ ±1
8. so w is purely imaginary