On a single Argand diagram, sketch the following loci:

  1. (i)

    ∣z−3∣=5|z - 3| = 5

  2. (ii)

    arg⁡(z−8)=π−tan⁡−1(12)\arg(z - 8) = \pi - \tan^{-1}\left(\frac{1}{2}\right)

  3. (iii)

    ∣z+2∣=∣z−3+5i∣|z + 2| = |z - 3 + 5i|

  4. (iv)

    State the complex number pp that represents the point of intersection of the loci in (i) and (ii), in the form a+iba + ib, where aa and bb are exact integers to be determined.

  5. (v)

    The complex number ww satisfies the inequalities
    ∣w−3∣≤5|w - 3| \le 5
    arg⁡(w−8)≤π−tan⁡−1(12)\arg(w - 8) \le \pi - \tan^{-1}\left(\frac{1}{2}\right)
    ∣w+2∣≥∣w−3+5i∣.|w + 2| \ge |w - 3 + 5i|.
    Illustrate the locus of ww on the same Argand diagram.

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Worked solution

11(i)

1. Identify the shape, centre, and radius of the locus

The equation is in the standard form ∣z−z0∣=r|z - z_0| = r, where z0=3z_0 = 3 and r=5r = 5.
This represents a circle in the Argand plane with centre at 33 (the point (3,0)(3, 0)) and radius 55.

2. Determine key intersections to sketch the circle accurately

Intersections with the real axis (where the imaginary part is 00):
∣x−3∣=5  ⟹  x−3=5 or x−3=−5|x - 3| = 5 \implies x - 3 = 5 \text{ or } x - 3 = -5
  ⟹  x=8 or x=−2\implies x = 8 \text{ or } x = -2

Intersections with the imaginary axis (where the real part is 00):
∣iy−3∣=5  ⟹  (−3)2+y2=5  ⟹  9+y2=25  ⟹  y2=16  ⟹  y=±4|iy - 3| = 5 \implies \sqrt{(-3)^2 + y^2} = 5 \implies 9 + y^2 = 25 \implies y^2 = 16 \implies y = \pm 4

Draw axes, mark the centre (3,0)(3,0) and intercepts −2,8,4i,−4i-2, 8, 4i, -4i, and sketch the circle passing through them.

Answer

A circle on the Argand diagram with centre at (3,0)(3, 0) and radius 55, intersecting the real axis at −2-2 and 88, and the imaginary axis at 4i4i and −4i-4i.

11(ii)

1. Identify the base point and the shape of the locus.

The equation arg⁡(z−8)=α\arg(z - 8) = \alpha represents a locus that is a half-line starting from the point corresponding to z=8z = 8. On the Argand diagram, this is the point (8,0)(8, 0).

2. Find the gradient and a reference point to sketch the half-line accurately.

The angle π−tan⁡−1(12)\pi - \tan^{-1}\left(\frac{1}{2}\right) indicates that the half-line points towards the second quadrant relative to (8,0)(8,0).

The gradient of the half-line is tan⁡(π−tan⁡−1(12))=−12\tan\left(\pi - \tan^{-1}\left(\frac{1}{2}\right)\right) = -\frac{1}{2}.

The line equation from (8,0)(8,0) with gradient −12-\frac{1}{2} is y−0=−12(x−8)y - 0 = -\frac{1}{2}(x - 8), which simplifies to y=−12x+4y = -\frac{1}{2}x + 4.

The half-line passes through the imaginary axis at y=4y = 4, which corresponds to the point (0,4)(0, 4).

3. State the sketching instructions.

Sketch a straight half-line originating from an empty (unshaded) circle at (8,0)(8, 0), passing through (0,4)(0, 4), and continuing infinitely.

Answer

A half-line starting from an open circle at (8,0)(8,0), passing through (0,4)(0,4), and extending indefinitely.

11(iii)

1. Identify the geometric meaning of the given locus equation.

The equation ∣z+2∣=∣z−3+5i∣|z + 2| = |z - 3 + 5i| can be rewritten as ∣z−(−2)∣=∣z−(3−5i)∣|z - (-2)| = |z - (3 - 5i)|.
This represents the perpendicular bisector of the line segment joining the points A(−2,0)A(-2, 0) and B(3,−5)B(3, -5).

2. Determine the Cartesian equation of the locus to identify key passing points for an accurate sketch.

The midpoint of AA and BB is (−2+32,0−52)=(12,−52)\left(\frac{-2+3}{2}, \frac{0-5}{2}\right) = \left(\frac{1}{2}, -\frac{5}{2}\right).
The gradient of the line segment ABAB is −5−03−(−2)=−1\frac{-5 - 0}{3 - (-2)} = -1.
Therefore, the gradient of the perpendicular bisector is 11.
The equation of the perpendicular bisector is:
y−(−52)=1(x−12)y - \left(-\frac{5}{2}\right) = 1\left(x - \frac{1}{2}\right)
y=x−3y = x - 3
The locus is a straight line passing through (3,0)(3, 0) and (0,−3)(0, -3).

3. Sketch the locus on the Argand diagram.

Draw a straight line passing through the real axis at 33 and the imaginary axis at −3-3. Note that this line passes exactly through the centre (3,0)(3, 0) of the circle from part (i).

Answer

A straight line representing the perpendicular bisector of the line segment joining (−2,0)(-2, 0) and (3,−5)(3, -5), passing through the points (3,0)(3, 0) and (0,−3)(0, -3).

11(iv)

1. Formulate Cartesian equations for the loci described in (i) and (ii).

Let z=x+iyz = x + iy.
Locus (i) is a circle: ∣(x−3)+iy∣=5  ⟹  (x−3)2+y2=25|(x-3) + iy| = 5 \implies (x-3)^2 + y^2 = 25.
Locus (ii) is a half-line: arg⁡((x−8)+iy)=π−tan⁡−1(12)\arg((x-8) + iy) = \pi - \tan^{-1}\left(\frac{1}{2}\right).
This gives yx−8=−12\frac{y}{x-8} = -\frac{1}{2} with x<8x < 8 (since the angle is in the second quadrant relative to (8,0)(8,0)).
So, y=−12(x−8)y = -\frac{1}{2}(x-8) for x<8x < 8.

2. Solve the equations simultaneously to find the intersection points.

Substitute the equation of the half-line into the circle's equation:
(x−3)2+(−12(x−8))2=25(x-3)^2 + \left(-\frac{1}{2}(x-8)\right)^2 = 25
(x−3)2+14(x−8)2=25(x-3)^2 + \frac{1}{4}(x-8)^2 = 25
4(x2−6x+9)+(x2−16x+64)=1004(x^2 - 6x + 9) + (x^2 - 16x + 64) = 100
4x2−24x+36+x2−16x+64=1004x^2 - 24x + 36 + x^2 - 16x + 64 = 100
5x2−40x=05x^2 - 40x = 0
5x(x−8)=05x(x-8) = 0
x=0x = 0 or x=8x = 8.

3. Apply the domain restriction to identify the valid intersection point and state $p$.

Since the half-line is restricted to x<8x < 8, we reject x=8x = 8.
For x=0x = 0, y=−12(0−8)=4y = -\frac{1}{2}(0-8) = 4.
Therefore, the point of intersection is (0,4)(0, 4), which corresponds to the complex number p=0+4ip = 0 + 4i.

Answer

0+4i0 + 4i

11(v)

1. Interpret the first inequality geometrically.

The inequality ∣w−3∣≤5|w - 3| \le 5 refers to the set of points with a distance of 55 or less from the point 33. This represents the region inside and on the boundary of the circle with centre 33 and radius 55.

2. Interpret the second inequality geometrically.

The inequality arg⁡(w−8)≤π−tan⁡−1(12)\arg(w - 8) \le \pi - \tan^{-1}\left(\frac{1}{2}\right) refers to angles smaller than or equal to the angle of the half-line in part (ii). This corresponds to the region swept clockwise from the half-line starting at z=8z = 8.

3. Interpret the third inequality geometrically.

The inequality ∣w+2∣≥∣w−3+5i∣|w + 2| \ge |w - 3 + 5i| requires the distance from ww to −2-2 to be greater than or equal to the distance from ww to 3−5i3-5i. The boundary is the perpendicular bisector from part (iii). The valid region is the half-plane containing the point 3−5i3-5i.

4. Identify and describe the final region to be shaded.

The locus of ww is the intersection of these three regions. On the Argand diagram, shade the area that is simultaneously inside the circle, below the half-line, and on the 3−5i3-5i side (below/right) of the perpendicular bisector.

Answer

The correct region is shaded on the Argand diagram (inside the circle, clockwise/below the half-line, and on the side of the perpendicular bisector containing 3−5i3-5i).

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