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How do equations and inequalities in z carve out lines, circles and regions in the Argand plane?

Describe and sketch subsets of the complex plane defined by equations and inequalities in modulus and argument

WACE Specialist Unit 3 loci in the complex plane: circles from modulus conditions, perpendicular bisectors from equal distances, rays from argument conditions, and shaded regions from inequalities, translated to Cartesian form, with a worked example.

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  1. What this dot point is asking
  2. Modulus conditions give circles and distances
  3. Modulus of a quotient or product condition
  4. Argument conditions give rays
  5. More loci: ratios and sums of distances
  6. Converting an argument condition carefully
  7. Combining conditions

What this dot point is asking

SCSA wants you to interpret modulus and argument conditions geometrically, sketch the resulting curves and regions, and where needed convert to Cartesian equations.

Modulus conditions give circles and distances

Since za|z - a| is the distance from zz to the fixed point aa, the equation za=r|z - a| = r describes all points at distance rr from aa, a circle of radius rr centred at aa. The inequality zar|z - a| \le r shades the closed disc; za>r|z - a| > r shades the outside.

The condition za=zb|z - a| = |z - b| says zz is equidistant from aa and bb, which is the perpendicular bisector of the segment joining them. The inequality za<zb|z - a| < |z - b| shades the half-plane nearer to aa.

Modulus of a quotient or product condition

Some conditions involve the modulus of a quotient. Because zazb=k\left|\dfrac{z - a}{z - b}\right| = k can be rewritten as za=kzb|z - a| = k\,|z - b|, it describes the set of points whose distance to aa is a fixed multiple kk of the distance to bb. When k=1k = 1 this is the perpendicular bisector, but for k1k \neq 1 it is a circle, known as a circle of Apollonius. Squaring za2=k2zb2|z - a|^2 = k^2 |z - b|^2 and expanding with z=x+iyz = x + iy produces a Cartesian equation that, after completing the square, reveals the centre and radius. Recognising the family the condition belongs to before grinding through the algebra is the efficient approach.

Argument conditions give rays

The condition arg(za)=α\arg(z - a) = \alpha describes all points zz for which the displacement from aa points in the fixed direction α\alpha. This is a ray (half-line) starting at aa (open, since z=az = a has no argument) at angle α\alpha to the horizontal. A condition like π6arg(za)π3\tfrac{\pi}{6} \le \arg(z - a) \le \tfrac{\pi}{3} shades a wedge between two rays.

More loci: ratios and sums of distances

Beyond circles, bisectors and rays, SCSA can set a few richer conditions. A condition of the form za+zb=k|z - a| + |z - b| = k (a constant sum of distances to two fixed points) is an ellipse with foci aa and bb, provided kk exceeds the distance between them. A condition Re(z)=c\operatorname{Re}(z) = c is the vertical line x=cx = c, and Im(z)=c\operatorname{Im}(z) = c is the horizontal line y=cy = c. Recognising the geometric object before converting to Cartesian form saves time and avoids algebra errors, because the shape tells you what equation to expect.

Converting an argument condition carefully

An argument condition needs more care than a modulus condition when converted to Cartesian form. The equation arg(za)=α\arg(z - a) = \alpha gives the relationship ya2xa1=tanα\dfrac{y - a_2}{x - a_1} = \tan\alpha, but only on the correct half-line: the ray points in the direction α\alpha, so the signs of xa1x - a_1 and ya2y - a_2 must match that direction. Writing the Cartesian relation as a line and then restricting to the correct ray (and excluding the start point aa) is the rigorous way to handle it.

Combining conditions

Many SCSA questions intersect two conditions, for example a disc and a wedge. Sketch each boundary, decide which side each inequality shades, and the answer is the overlap. Always state whether boundaries are included (solid) or excluded (dashed).

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20226 marksCalculator-assumed. On an Argand diagram sketch the region {z:z1+2i2}{z:0arg(z1+2i)π2}\{z : |z - 1 + 2i| \le 2\} \cap \{z : 0 \le \arg(z - 1 + 2i) \le \dfrac{\pi}{2}\}.
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An intersection of a disc and a wedge.

Rewrite z1+2i=z(12i)z - 1 + 2i = z - (1 - 2i), so both conditions are centred at the point 12i1 - 2i, i.e. (1,2)(1, -2).

The first condition is the closed disc of radius 22 centred at (1,2)(1, -2). The second is the wedge of directions from (1,2)(1, -2) with angle between 00 (the positive real direction) and π2\dfrac{\pi}{2} (the positive imaginary direction), i.e. the first-quadrant wedge measured from (1,2)(1, -2).

The region is the quarter-disc: that part of the disc lying up and to the right of the centre, bounded by the two radii along the positive horizontal and vertical directions and the arc. Boundaries are solid (inequalities are weak). Markers reward the correct centre (1,2)(1, -2), the disc, the quarter-wedge, and the overlap shaded with solid boundaries.

WACE 20245 marksCalculator-free. The locus of zz satisfies z4=z2i|z - 4| = |z - 2i|. Find its Cartesian equation and describe it geometrically.
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A perpendicular-bisector locus.

The condition says zz is equidistant from 44 (the point (4,0)(4, 0)) and 2i2i (the point (0,2)(0, 2)), so the locus is the perpendicular bisector of the segment joining them.

Algebraically, put z=x+iyz = x + iy. Then z42=(x4)2+y2|z - 4|^2 = (x - 4)^2 + y^2 and z2i2=x2+(y2)2|z - 2i|^2 = x^2 + (y - 2)^2. Setting equal: (x4)2+y2=x2+(y2)2(x - 4)^2 + y^2 = x^2 + (y - 2)^2, so x28x+16+y2=x2+y24y+4x^2 - 8x + 16 + y^2 = x^2 + y^2 - 4y + 4, giving 8x+16=4y+4-8x + 16 = -4y + 4, hence 4y=8x124y = 8x - 12, that is y=2x3y = 2x - 3.

Markers reward recognising the perpendicular bisector, the squared-modulus expansion, and the line y=2x3y = 2x - 3.

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