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Loci in the Complex Plane — Full Worked Solutions
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Sketch and describe the locus |z − 2| = 3.
Write z = x + iy. The locus |z − 2| = 3 says the distance from z to the fixed point 2 equals 3.
√[(x − 2)² + y²] = 3
Squaring: (x − 2)² + y² = 9
Circle, centre (2, 0), radius 3.
Key points on the circle: (−1, 0), (5, 0), (2, 3), (2, −3).
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Sketch and describe arg(z) = π/4.
arg(z) = π/4 means the argument (angle from positive real axis) of z is 45°.
This is a ray from the origin (origin excluded) at angle 45°.
Write z = x + iy with |z| = r > 0: arg(z) = arctan(y/x) = π/4 requires y/x = tan(π/4) = 1, so y = x with x > 0.
Cartesian equation: y = x, x > 0 (and y > 0). Open endpoint at the origin.
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Find the Cartesian equation for |z − 1 − i| = 2.
The fixed point is a = 1 + i, corresponding to (1, 1). Radius = 2.
Write z = x + iy:
|z − 1 − i| = |(x − 1) + i(y − 1)| = √[(x − 1)² + (y − 1)²] = 2
Squaring: (x − 1)² + (y − 1)² = 4
This is a circle, centre (1, 1), radius 2.
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Describe and find the equation of |z − 3| = |z + 1|.
Geometric description: Perpendicular bisector of the segment joining 3 and −1 on the real axis.
Write z = x + iy:
√[(x−3)² + y²] = √[(x+1)² + y²]
Squaring: x² − 6x + 9 + y² = x² + 2x + 1 + y²
−6x + 9 = 2x + 1
8 = 8x ⇒ x = 1
The locus is the vertical line x = 1, passing through the midpoint of −1 and 3.
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Find and sketch the region |z − 2i| ≤ 3.
The fixed point is 2i = (0, 2). The condition |z − 2i| ≤ 3 means distance from (x, y) to (0, 2) is at most 3.
Cartesian: √[x² + (y−2)²] ≤ 3, i.e. x² + (y − 2)² ≤ 9
Closed disk (filled circle including boundary): centre (0, 2), radius 3.
The region extends from x = −3 to x = 3, and y = −1 to y = 5. The boundary is included (solid line in sketch).
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Describe and sketch Im(z − 1 + 2i) = 0.
Write z = x + iy:
z − 1 + 2i = (x − 1) + i(y + 2)
Im(z − 1 + 2i) = y + 2 = 0 ⇒ y = −2
Geometric description: A horizontal line (parallel to the real axis) passing through the point −2i. This is the line Im(z) = −2, extending infinitely in both directions.
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Find all z satisfying |z| = |z − 4| and Im(z) = 2.
Write z = x + iy.
Locus 1: |z| = |z − 4|
√[x² + y²] = √[(x−4)² + y²]
Squaring: x² = x² − 8x + 16 ⇒ 8x = 16 ⇒ x = 2
Locus 2: Im(z) = 2 ⇒ y = 2
Intersection: x = 2 and y = 2, so z = 2 + 2i.
Verification: |2+2i| = √8, |2+2i−4| = |−2+2i| = √8. ✓ Im(2+2i) = 2. ✓
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Sketch the region {z : |z + i| ≤ 2 and Re(z) ≥ 0}.
Condition 1: |z + i| ≤ 2 ⇒ |z − (−i)| ≤ 2
Closed disk centred at (0, −1), radius 2: x² + (y + 1)² ≤ 4.
Extends horizontally from x = −2 to x = 2, vertically from y = −3 to y = 1.
Condition 2: Re(z) ≥ 0 ⇒ x ≥ 0 (right half-plane, including boundary).
Intersection: The right half of the closed disk. Boundary consists of:
- The arc of the circle x² + (y+1)² = 4 for x ≥ 0 (from (0, 1) to (0, −3) via (2, −1))
- The chord along x = 0 from (0, −3) to (0, 1)
All boundary included (solid lines). Key points: (0, 1), (2, −1), (0, −3).
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Write arg(z − 1) = π/3 parametrically and as a Cartesian equation.
Let w = z − 1, so arg(w) = π/3 means w = t e^(iπ/3) = t(cos(π/3) + i sin(π/3)) = t(1/2 + i√3/2) for t > 0.
Then z = 1 + w:
Parametric form: x(t) = 1 + t/2, y(t) = t√3/2, t > 0.
Cartesian equation: From y = t√3/2 ⇒ t = 2y/√3. Substitute into x = 1 + t/2:
x = 1 + y/√3 ⇒ x − 1 = y/√3 ⇒ y = √3(x − 1)
Restriction: x > 1 and y > 0 (since t > 0). The starting point (1, 0) is excluded.
The ray has gradient √3 (= tan 60°), starting at (1, 0) (open), extending into the first quadrant.
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Apollonius circle: find centre and radius of {z : |z − 1|/|z + 1| = 2}.
The condition is |z − 1| = 2|z + 1|. Write z = x + iy:
√[(x−1)² + y²] = 2√[(x+1)² + y²]
Squaring both sides:
(x−1)² + y² = 4[(x+1)² + y²]
x² − 2x + 1 + y² = 4x² + 8x + 4 + 4y²
0 = 3x² + 10x + 3 + 3y²
Divide by 3: x² + (10/3)x + y² = −1
Complete the square in x:
(x + 5/3)² − 25/9 + y² = −1
(x + 5/3)² + y² = 25/9 − 9/9 = 16/9
Centre: (−5/3, 0) Radius: 4/3
This is an Apollonius circle: the set of points whose ratio of distances to 1 and −1 is 2:1.