McIntyre MathsMathematics Extension 2
Extension 2 › Complex Numbers I › Lesson 10

Lesson 10 · Rays and regions

Mathematics Extension 2 · Complex Numbers I · Outcome ME2-12-03 · Term 1 · Week 2
Learning intentionTo sketch rays and regions defined by argument and modulus conditions, and find extreme values on them.
Success criteria · I can…
  • graph arg⁡(z−z1)=θ\arg(z-z_1)=\theta as a ray from z1z_1, with z1z_1 excluded
  • sketch regions that combine circles, lines and rays, with correct boundaries and endpoints
  • find the greatest and least ∣z∣|z| and arg⁡z\arg z in a region
  • (extension) use circle geometry for arc loci

Lesson flow

Warm-upRetrieval from Lessons 5 and 9Slide 2Key ideasArgument fixes a directionSlide 3ExamplesExample 1 · RaysSlide 4ExamplesExample 2 · More raysSlide 5ExamplesExample 3 · RegionsSlide 6HSC-styleHSC practice · RegionsSlide 7Exam practiceExample 4 · Greatest and least valuesSlide 8ExamplesExample 5 · A region and its extremesSlide 9ExtensionExtension · ArcsSlides 10-11Spot the errorSpot the error · Rays and regionsSlide 12Exit ticketExit ticketSlide 13

Retrieval from Lessons 5 and 9

Slide 2
(a) Find Arg⁡(−1−i)\Arg(-1-i).
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Answer
−3π4-\dfrac{3\pi}{4}
Why
−1−i-1-i is in the third quadrant with reference angle π4\frac{\pi}{4}. The principal argument lies in (−π,π](-\pi,\pi], so it is −π+π4=−3π4-\pi+\frac{\pi}{4}=-\frac{3\pi}{4}.
(b) Describe the region ∣z−1∣≤2|z-1|\le2.
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Answer
The closed disc with centre 11 and radius 22 (solid boundary).
Why
∣z−1∣|z-1| is the distance from 11, and “≤\le” includes the circle itself.
(c) A circle has centre (2,0)(2,0) and radius 11. A tangent from OO touches it at TT. Find ∠TOX\angle TOX, where XX is on the positive real axis.
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Answer
π6\dfrac{\pi}{6}
Why
The radius to TT is perpendicular to the tangent, so in the right-angled triangle sin⁡∠TOX=12\sin\angle TOX=\frac12 and ∠TOX=π6\angle TOX=\frac{\pi}{6}. This is used again in Example 4.
(d) Write tan⁡π3\tan\dfrac{\pi}{3} exactly.
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Answer
3\sqrt3
Why
From the 30∘30^\circ-60∘60^\circ-90∘90^\circ triangle, tan⁡60∘=31\tan60^\circ=\frac{\sqrt3}{1}.

Argument fixes a direction

Slide 3
Key idea
Rays: arg⁡(z−z1)=θ\arg(z-z_1)=\theta

The vector from Z1Z_1 to ZZ makes the angle θ\theta with the positive real direction.

The locus is a ray starting at z1z_1. The point z1z_1 itself is excluded (open circle), because arg⁡0\arg0 is undefined.

ReIm-1123456-1123450θz1
arg(z − z1) = θ: a ray from z1 at angle θ to the horizontal.

Example 1 · Rays

Slide 4

Sketch each locus.

(a) arg⁡z=π6\arg z=\dfrac{\pi}{6}
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Answer
A ray from OO at angle π6\frac{\pi}{6} to the positive real axis, with an open circle at OO.
Why
arg⁡z=π6\arg z=\frac{\pi}{6} means z=r(cos⁡π6+isin⁡π6)z=r\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right) with r>0r>0. OO is excluded.
(b) arg⁡(z−(1+i))=π6\arg\left(z-(1+i)\right)=\dfrac{\pi}{6}
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Answer
A ray from (1,1)(1,1) at angle π6\frac{\pi}{6} (parallel to the ray in (a)), with an open circle at 1+i1+i.
Why
Common errors: drawing a full line, or starting the ray at OO instead of at 1+i1+i.
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ReIm-1123456-11234501 + i
(a) the ray from O and (b) the parallel ray from 1 + i.

Example 2 · More rays

Slide 5

Sketch each locus.

(a) arg⁡(z+2)=3π4\arg(z+2)=\dfrac{3\pi}{4}
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Answer
A ray from −2-2 going up and to the left at 135∘135^\circ, with an open circle at −2-2.
Why
z+2=z−(−2)z+2=z-(-2), so the ray starts at −2-2.
(b) arg⁡(z−1+i)=arg⁡(z+1−i)\arg(z-1+i)=\arg(z+1-i)
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Answer
Two rays on the line y=−xy=-x: one starting at (1,−1)(1,-1) going down-right, and one starting at (−1,1)(-1,1) going up-left. Both endpoints are excluded, and the segment between them is not part of the locus.
Why
The directions from (1,−1)(1,-1) and from (−1,1)(-1,1) to ZZ are equal. This happens only on the line through the two points, and only when ZZ is outside the interval between them (inside it the two directions are opposite).
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ReIm-5-4-3-2-112345-4-3-2-112340
(a) the ray from −2 at 135°. (b) the two rays of y = −x beyond (1, −1) and (−1, 1) (drawn on the same axes).

Example 3 · Regions

Slide 6

Sketch each region.

(a) 1≤∣z∣≤31\le|z|\le3
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Answer
The annulus between the circles of radius 11 and 33 about OO, both boundaries solid.
Why
∣z∣≥1|z|\ge1 is outside or on the unit circle and ∣z∣≤3|z|\le3 is inside or on the circle of radius 3.
(b) π6≤arg⁡z≤π3\dfrac{\pi}{6}\le\arg z\le\dfrac{\pi}{3}
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Answer
The sector between the rays at π6\frac{\pi}{6} and π3\frac{\pi}{3} from OO, rays solid, with an open circle at OO.
Why
OO is excluded because arg⁡0\arg0 is undefined.
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ReIm-4-3-2-11234-4-3-2-112340shade between
(a) The annulus 1 ≤ |z| ≤ 3.
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ReIm-112345-1123450
(b) The sector π/6 ≤ arg z ≤ π/3.

HSC practice · Regions

Slide 7
Adapted from 2022 HSC Q12 marksSketch the region where Re⁡(z)>1\Re(z)>1 and Arg⁡z<π3\Arg z<\dfrac{\pi}{3}.
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Answer
The region to the right of the dashed line x=1x=1 and below the dashed ray y=3 xy=\sqrt3\,x (the ray from OO at π3\frac{\pi}{3}). The corner (1,3)(1,\sqrt3) is not included.
Why
Points with Re⁡(z)>1\Re(z)>1 already have Arg⁡z>−π2\Arg z>-\frac{\pi}{2}, so only the upper bound matters. The two boundaries meet at (1,3)(1,\sqrt3).
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ReIm-112345-4-3-2-11234560
Re(z) > 1 and Arg z < π/3.
Adapted from 2024 HSC Q11(f)4 marksSketch the region where ∣z∣<3|z|<3 and 0≤arg⁡(z−i)≤π20\le\arg(z-i)\le\dfrac{\pi}{2}.
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Answer
Inside the dashed circle of radius 33 about OO, in the quarter-plane x≥0x\ge0, y≥1y\ge1 with corner at ii. The two straight edges (from ii) are solid, and the point ii itself is excluded (open circle).
Why
0≤arg⁡(z−i)≤π20\le\arg(z-i)\le\frac{\pi}{2} says that z−iz-i lies in the first quadrant, including its two axes but not the origin. The edges end where they meet the circle, at (8,1)\left(\sqrt8,1\right) and (0,3)(0,3). Marks are usually for the dashed circle, the quarter-plane from ii, the open point at ii and the correct shading.
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ReIm-4-3-2-11234-112340i
|z| < 3 and 0 ≤ arg(z − i) ≤ π/2.

Example 4 · Greatest and least values

Slide 8
Exam practice5 marksPP represents zz, where ∣z−2∣=1|z-2|=1. (i) Sketch the locus. (ii) Find the greatest and least values of ∣z∣|z| and of arg⁡z\arg z. (iii) OP1OP_1 and OP2OP_2 are tangents to the locus. Write z1z_1 and z2z_2 in polar form.
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Answer
(i) Circle with centre 22 and radius 11. (ii) 1≤∣z∣≤31\le|z|\le3 and −π6≤arg⁡z≤π6-\frac{\pi}{6}\le\arg z\le\frac{\pi}{6}. (iii) z1=3(cos⁡π6+isin⁡π6)z_1=\sqrt3\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right) and z2=3(cos⁡π6−isin⁡π6)z_2=\sqrt3\left(\cos\frac{\pi}{6}-i\sin\frac{\pi}{6}\right).
Why
(ii) Along the real axis the nearest and farthest points are 11 and 33. A tangent from OO is perpendicular to the radius, so in the right-angled triangle sin⁡α=12\sin\alpha=\frac12 and α=π6\alpha=\frac{\pi}{6}. (iii) The tangent length is 22−12=3\sqrt{2^2-1^2}=\sqrt3. As a check, z12+z22=3cis⁡π3+3cis⁡(−π3)=3z_1^2+z_2^2=3\cis\frac{\pi}{3}+3\cis\left(-\frac{\pi}{3}\right)=3.
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ReIm-112345-3-2-112302P1P2
The tangents from O to the circle touch it at angles ±π/6.

Example 5 · A region and its extremes

Slide 9

Example 5

Sketch the region where ∣z−1−i∣≤2|z-1-i|\le\sqrt2 and 0≤arg⁡z≤π40\le\arg z\le\dfrac{\pi}{4}. Find the greatest value of ∣z∣|z| in the region and the point where it occurs.
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Answer
The greatest value is ∣z∣=22|z|=2\sqrt2, at z=2+2iz=2+2i.
Why
The circle with centre (1,1)(1,1) and radius 2\sqrt2 passes through OO and meets the real axis at OO and 22. The diagonal y=xy=x passes through the centre, so the region is the half-disc below y=xy=x minus the small segment below the real axis. It is bounded by the chord from OO to 22 on the real axis, the arc from 22 to 2+2i2+2i, and the diameter from 2+2i2+2i back to OO. OO is excluded because arg⁡0\arg0 is undefined. The farthest point from OO is the other end of the diameter through OO, 2+2i2+2i, which lies on arg⁡z=π4\arg z=\frac{\pi}{4} (included).
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ReIm-11234-112340O2 + 2i
The region (shaded) and the point where |z| is greatest.

Extension · Arcs

Slides 10-11
Key idea
Angle subtended by a fixed interval

arg⁡(z−z1z−z2)=arg⁡(z−z1)−arg⁡(z−z2)=θ\arg\left(\dfrac{z-z_1}{z-z_2}\right)=\arg(z-z_1)-\arg(z-z_2)=\theta means the interval Z2Z1Z_2Z_1 subtends a constant angle θ\theta at ZZ.

So ZZ lies on an arc of a circle (angles in the same segment are equal), and z1z_1 and z2z_2 are excluded. This is not a named syllabus locus, but it follows from “using circle geometry theorems where necessary”.

Example 6

ExtensionSketch the locus given by arg⁡(z−2)−arg⁡(z+2)=π3\arg(z-2)-\arg(z+2)=\dfrac{\pi}{3}, and find the centre and radius of the circle.
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Answer
The major arc, above the real axis, of the circle with centre (0,23)\left(0,\dfrac{2}{\sqrt3}\right) and radius 43\dfrac{4}{\sqrt3}, with ±2\pm2 excluded.
Why
arg⁡(z−2)−arg⁡(z+2)\arg(z-2)-\arg(z+2) is the angle at ZZ in the triangle with vertices −2-2, 22 and ZZ. The chord of length 44 subtends π3\frac{\pi}{3} at the circumference, so 2Rsin⁡π3=42R\sin\frac{\pi}{3}=4 and R=43R=\frac{4}{\sqrt3}. The centre is on the imaginary axis at height Rcos⁡π3=23R\cos\frac{\pi}{3}=\frac{2}{\sqrt3}. The circle is x2+(y−23)2=163x^2+\left(y-\frac{2}{\sqrt3}\right)^2=\frac{16}{3} with y>0y>0. (Algebraically, 4y=3(x2+y2−4)4y=\sqrt3(x^2+y^2-4) gives the same circle.)
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ReIm-4-3-2-11234-2-1123450−22centre
The arc locus (drawn in teal): the major arc above the real axis with the end points excluded.

Spot the error · Rays and regions

Slide 12

Each statement is wrong. Find the error and correct it.

(a) arg⁡z=π4\arg z=\dfrac{\pi}{4} is the line y=xy=x.
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Answer
It is only the ray y=xy=x, x>0x>0, with OO excluded.
Why
A fixed argument gives a ray, not a full line. arg⁡0\arg0 is undefined, so OO is an open circle.
(b) arg⁡(z−1)=π2\arg(z-1)=\dfrac{\pi}{2} is a ray starting at OO.
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Answer
The ray starts at 11: it is the vertical ray x=1x=1, y>0y>0, with the point 11 excluded.
Why
z−1=z−(1)z-1=z-(1), so the ray starts at 11, not at OO.
(c) In 1≤∣z∣≤31\le|z|\le3 both circles are dashed.
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Answer
Both circles are solid.
Why
“≤\le” includes the boundary, so both circles are part of the region.
(d) The region π6≤arg⁡z≤π3\dfrac{\pi}{6}\le\arg z\le\dfrac{\pi}{3} includes the origin.
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Answer
The origin is excluded (open circle).
Why
arg⁡0\arg0 is undefined, so OO cannot satisfy the condition.

Exit ticket

Slide 13
1. Sketch the region ∣z−2∣≤2|z-2|\le2 and Im⁡(z)≥1\Im(z)\ge1.
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Answer
The part of the closed disc with centre 22 and radius 22 on or above the solid line y=1y=1 (a segment of the disc).
Why
The line y=1y=1 cuts the circle where (x−2)2+1=4(x-2)^2+1=4, that is x=2±3x=2\pm\sqrt3.
2. Find the least value of ∣z∣|z| if arg⁡(z−2i)=−π4\arg(z-2i)=-\dfrac{\pi}{4}.
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Answer
2\sqrt2, at z=1+iz=1+i.
Why
The locus is the ray from 2i2i with points (s,2−s)(s,2-s), s>0s>0. Then ∣z∣2=s2+(2−s)2=2(s−1)2+2|z|^2=s^2+(2-s)^2=2(s-1)^2+2 is least at s=1s=1. (This is the foot of the perpendicular from OO to the line, and it lies on the ray.)
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ReIm-112345-11234501 + i
Exit ticket answers: the disc is bounded by the circle; the ray from 2i meets the line y = 1 at 1 + i.