In a single Argand diagram, sketch the following loci, labelling each locus clearly.

  1. ((i))

    ∣z∣=5,|z| = 5,

  2. ((ii))

    ∣z+8∣=∣z−8i∣.|z + 8| = |z - 8i|.

  3. ((iii))

    The two complex numbers that satisfy the above equations are represented by pp and qq, where arg⁡(pq)>0\arg\left(\frac{p}{q}\right) > 0. Find pp and qq exactly.

  4. ((iv))

    State the exact value of arg⁡(5−p5−q)\arg\left(\frac{5-p}{5-q}\right).

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Worked solution

3((i))

1. Identify the shape and key features of the locus.

The equation ∣z∣=5|z| = 5 represents the locus of points zz in the complex plane whose distance from the origin is 55.

This forms a circle with centre 00 and radius 55.

2. Determine the intersections with the axes for the sketch.

Since the radius is 55 and the centre is at the origin, the circle intersects the axes at:

  • Real axis: 55 and −5-5
  • Imaginary axis: 5i5i and −5i-5i

The sketch will be a circle passing through these points.

Answer

A circle on the Argand diagram centered at the origin (0,0)(0,0) with radius 55, intersecting the real axis at 55 and −5-5, and the imaginary axis at 5i5i and −5i-5i.

3((ii))

1. Express the given equation in Cartesian coordinates by substituting $z = x + iy$.

Let z=x+iyz = x + iy.
∣x+iy+8∣=∣x+iy−8i∣|x + iy + 8| = |x + iy - 8i|
∣(x+8)+iy∣=∣x+i(y−8)∣|(x + 8) + iy| = |x + i(y - 8)|

2. Evaluate the moduli and simplify to find the Cartesian equation of the locus.

(x+8)2+y2=x2+(y−8)2\sqrt{(x + 8)^2 + y^2} = \sqrt{x^2 + (y - 8)^2}
(x+8)2+y2=x2+(y−8)2(x + 8)^2 + y^2 = x^2 + (y - 8)^2
x2+16x+64+y2=x2+y2−16y+64x^2 + 16x + 64 + y^2 = x^2 + y^2 - 16y + 64
16x=−16y16x = -16y
y=−xy = -x

3. Identify and describe the locus for sketching.

The locus is the perpendicular bisector of the line segment joining (−8,0)(-8, 0) and (0,8)(0, 8), which simplifies to the straight line y=−xy = -x passing through the origin. This line should be drawn on the Argand diagram and labelled.

Answer

A straight line with equation y=−xy = -x, sketched on the Argand diagram.

3((iii))

1. Express locus (ii) as a straight line

Let z=x+iyz = x + iy. Then ∣z+8∣=∣z−8i∣|z+8| = |z-8i| gives
(x+8)2+y2=x2+(y−8)2(x+8)^2 + y^2 = x^2 + (y-8)^2
x2+16x+64+y2=x2+y2−16y+64x^2 + 16x + 64 + y^2 = x^2 + y^2 - 16y + 64
16x=−16y  ⟹  y=−x.16x = -16y \implies y = -x.

2. Find the intersection with the circle |z| = 5

Substitute y=−xy = -x into x2+y2=25x^2 + y^2 = 25:
x2+x2=25  ⟹  2x2=25  ⟹  x=±52=±522.x^2 + x^2 = 25 \implies 2x^2 = 25 \implies x = \pm\frac{5}{\sqrt{2}} = \pm\frac{5\sqrt{2}}{2}.
So the intersection points are
(522, −522)and(−522, 522).\left(\tfrac{5\sqrt{2}}{2},\,-\tfrac{5\sqrt{2}}{2}\right) \quad\text{and}\quad \left(-\tfrac{5\sqrt{2}}{2},\,\tfrac{5\sqrt{2}}{2}\right).

3. Write the two points as complex numbers and find their arguments

The two complex numbers are
z1=522−522i(arg⁡z1=−π4),z_1 = \frac{5\sqrt{2}}{2} - \frac{5\sqrt{2}}{2}i \quad (\arg z_1 = -\tfrac{\pi}{4}),
z2=−522+522i(arg⁡z2=3π4).z_2 = -\frac{5\sqrt{2}}{2} + \frac{5\sqrt{2}}{2}i \quad (\arg z_2 = \tfrac{3\pi}{4}).

4. Assign p and q using the argument condition

arg⁡ ⁣(pq)=arg⁡p−arg⁡q>0  ⟹  arg⁡p>arg⁡q.\arg\!\left(\dfrac{p}{q}\right) = \arg p - \arg q > 0 \implies \arg p > \arg q.
Hence pp is the point with the larger argument and qq the smaller:
p=−522+522i,q=522−522i.p = -\frac{5\sqrt{2}}{2} + \frac{5\sqrt{2}}{2}i, \qquad q = \frac{5\sqrt{2}}{2} - \frac{5\sqrt{2}}{2}i.

Answer

p=−522+522 ip = -\dfrac{5\sqrt{2}}{2} + \dfrac{5\sqrt{2}}{2}\,i and q=522−522 iq = \dfrac{5\sqrt{2}}{2} - \dfrac{5\sqrt{2}}{2}\,i

3((iv))

1. Use the geometric properties of the points on the Argand diagram to deduce the argument.

Since pp and qq are the intersections of the line y=−xy = -x (passing through the origin) and the circle ∣z∣=5|z| = 5, they form the endpoints of a diameter of the circle.
The point z=5z = 5 also lies on the circle ∣z∣=5|z| = 5 since its distance from the origin is 55.
By the geometric property of an angle in a semicircle, the angle subtended by the diameter pqpq at the point 55 is a right angle (π2\frac{\pi}{2} rad).
Thus, the vector 5−p5-p is perpendicular to the vector 5−q5-q, which implies that 5−p5−q\frac{5-p}{5-q} is purely imaginary, and its argument is either π2\frac{\pi}{2} or −π2-\frac{\pi}{2}.

From the earlier parts, pp is in the second quadrant and qq is in the fourth quadrant.
The vector 5−q5-q (pointing from qq to 55) lies in the first quadrant (positive real and positive imaginary parts), while the vector 5−p5-p (pointing from pp to 55) lies in the fourth quadrant (positive real and negative imaginary parts).
Since the angle measured from 5−q5-q to 5−p5-p is clockwise, the argument difference is negative.
arg⁡(5−p5−q)=−π2\arg\left(\frac{5-p}{5-q}\right) = -\frac{\pi}{2}

Answer

−π2-\frac{\pi}{2}

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