McIntyre MathsMathematics Extension 2
Extension 2 › Complex Numbers I › Lesson 9

Lesson 9 · Lines and circles in the Argand diagram

Mathematics Extension 2 · Complex Numbers I · Outcome ME2-12-03 · Term 1 · Week 2
Learning intentionTo sketch and describe loci defined by modulus conditions: circles and perpendicular bisectors.
Success criteria · I can…
  • read ∣z−z1∣|z-z_1| as the distance from zz to z1z_1
  • graph ∣z−z1∣=r|z-z_1|=r and the regions ∣z−z1∣<r|z-z_1|<r and ∣z−z1∣>r|z-z_1|>r
  • graph ∣z−z1∣=∣z−z2∣|z-z_1|=|z-z_2| and give its equation and a geometric description
  • find a Cartesian equation by letting z=x+iyz=x+iy
  • find a locus of ww given a condition on zz

Lesson flow

Warm-upRetrieval and coordinate geometrySlide 2Key ideasModulus is distanceSlide 3ExamplesExample 1 · Sketching circlesSlide 4ExamplesExample 2 · Regions bounded by circlesSlide 5Key ideasPerpendicular bisectorsSlide 6ExamplesExample 3 · A perpendicular bisectorSlide 7Exam practiceExample 4 · A circle and a lineSlide 8Exam practiceExample 5 · Touching circlesSlide 9ExamplesExample 6 · The locus of wSlide 10ExamplesExample 7 · Purely imaginarySlide 11ExtensionExample 8 · Extension: not a straight lineSlide 12Spot the errorSpot the error · Circles and bisectorsSlide 13Exit ticketExit ticketSlide 14

Retrieval and coordinate geometry

Slide 2
(a) Write the equation of the circle with centre (1,−2)(1,-2) and radius 33.
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Answer
(x−1)2+(y+2)2=9(x-1)^2+(y+2)^2=9
Why
A circle with centre (a,b)(a,b) and radius rr has equation (x−a)2+(y−b)2=r2(x-a)^2+(y-b)^2=r^2.
(b) Find the perpendicular bisector of the interval from (0,0)(0,0) to (4,2)(4,2).
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Answer
y=−2x+5y=-2x+5
Why
The midpoint is (2,1)(2,1) and the interval has gradient 12\frac12, so the bisector has gradient −2-2: y−1=−2(x−2)y-1=-2(x-2).
(c) Find ∣3−4i∣|3-4i|.
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Answer
55
Why
32+(−4)2=25=5\sqrt{3^2+(-4)^2}=\sqrt{25}=5.
(d) Write z+2−iz+2-i in the form z−(   )z-(\ \ \ ).
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Answer
z−(−2+i)z-(-2+i)
Why
z+2−i=z−(−2+i)z+2-i=z-(-2+i). Parts (a), (b) and (d) are the three skills this lesson needs.

Modulus is distance

Slide 3
Key idea
Circles

∣z−z1∣|z-z_1| is the distance from ZZ to Z1Z_1. So ∣z−z1∣=r|z-z_1|=r is the circle with centre z1z_1 and radius rr.

∣z−z1∣<r|z-z_1|<r is the inside of the circle (dashed boundary). ∣z−z1∣≥r|z-z_1|\ge r is the outside, including the circle (solid boundary).

Write ∣z+3−i∣|z+3-i| as ∣z−(−3+i)∣|z-(-3+i)| to read off the centre.

ReIm-112345-112340z1rz
|z − z1| = r: every point z is a distance r from the centre z1.
Algebraic check
∣z−(a+ib)∣=r ⇔ (x−a)2+(y−b)2=r2|z-(a+ib)|=r\ \Leftrightarrow\ (x-a)^2+(y-b)^2=r^2 for z=x+iyz=x+iy.

Example 1 · Sketching circles

Slide 4

Sketch each locus and write its Cartesian equation.

(a) ∣z∣=3|z|=3
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Answer
x2+y2=9x^2+y^2=9: centre OO, radius 33.
Why
∣z−0∣=3|z-0|=3: the distance from OO is always 3.
(b) ∣z−(1−2i)∣=2|z-(1-2i)|=2
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Answer
(x−1)2+(y+2)2=4(x-1)^2+(y+2)^2=4: centre (1,−2)(1,-2), radius 22.
Why
The centre is 1 unit from the imaginary axis and the radius is 2, so the circle cuts the imaginary axis.
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ReIm-4-3-2-11234-4-3-2-1123401 − 2i
(a) |z| = 3 and (b) |z − (1 − 2i)| = 2.

Example 2 · Regions bounded by circles

Slide 5

Sketch each region. Take care with the boundary.

(a) ∣z+3−i∣≤2|z+3-i|\le2
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Answer
The closed disc with centre (−3,1)(-3,1) and radius 22, solid boundary, shaded inside.
Why
Rewrite as ∣z−(−3+i)∣≤2|z-(-3+i)|\le2. The inequality is “≤\le”, so the circle itself is included.
(b) ∣z−2i∣>1|z-2i|>1
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Answer
Everything outside the circle with centre (0,2)(0,2) and radius 11, with a dashed boundary.
Why
∣z−2i∣>1|z-2i|>1 means the distance from 2i2i is more than 1. The inequality is strict, so the circle is not included.
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ReIm-6-5-4-3-2-1123-1123450−3 + i
(a) The closed disc |z + 3 − i| ≤ 2.
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ReIm-3-2-1123-11234502ishade outside
(b) The boundary of |z − 2i| > 1 is dashed; the region is everything outside.

Perpendicular bisectors

Slide 6
Key idea
Equidistant from two points: ∣z−z1∣=∣z−z2∣|z-z_1|=|z-z_2|

ZZ is the same distance from Z1Z_1 and Z2Z_2, so the locus is the perpendicular bisector of the interval joining z1z_1 and z2z_2.

Find its equation by either method: (1) the midpoint and the perpendicular gradient, or (2) let z=x+iyz=x+iy, square both sides and simplify.

ReIm-112345-1123450z1z2M
The perpendicular bisector of Z1Z2 passes through the midpoint M.

Example 3 · A perpendicular bisector

Slide 7

Example 3

Sketch ∣z−(1+i)∣=∣z−(2−i)∣|z-(1+i)|=|z-(2-i)| and find its Cartesian equation using two methods.
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Answer
2x−4y−3=02x-4y-3=0
Why
Method A. Midpoint (32,0)\left(\frac32,0\right). Gradient of Z1Z2Z_1Z_2 is −1−12−1=−2\frac{-1-1}{2-1}=-2, so the bisector has gradient 12\frac12: y=12(x−32)y=\frac12\left(x-\frac32\right), giving 2x−4y−3=02x-4y-3=0.
Method B. (x−1)2+(y−1)2=(x−2)2+(y+1)2(x-1)^2+(y-1)^2=(x-2)^2+(y+1)^2 gives −2x−2y+2=−4x+2y+5-2x-2y+2=-4x+2y+5, so 2x−4y−3=02x-4y-3=0.
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ReIm-2-112345-3-2-112301 + i2 − i
The perpendicular bisector 2x − 4y − 3 = 0.

Example 4 · A circle and a line

Slide 8
Exam practice3 marks(i) On the same diagram sketch ∣z−(3+2i)∣=2|z-(3+2i)|=2 and ∣z+3∣=∣z−5∣|z+3|=|z-5|. (ii) Find the complex number(s) satisfying both equations.
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Answer
(i) A circle with centre (3,2)(3,2) and radius 22, and the vertical line x=1x=1 (the perpendicular bisector of −3-3 and 55). (ii) The line is tangent to the circle: z=1+2iz=1+2i.
Why
The leftmost point of the circle is x=3−2=1x=3-2=1, so the line x=1x=1 touches the circle at exactly one point, (1,2)(1,2).
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ReIm-4-3-2-11234567-11234501 + 2i3 + 2i
The line x = 1 is tangent to the circle.

Example 5 · Touching circles

Slide 9
Exam practice2 marksFor what values of k>0k>0 does the circle ∣z−2i∣=k|z-2i|=k meet ∣z−(3+2i)∣=2|z-(3+2i)|=2 at exactly one point?
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Answer
k=1k=1 or k=5k=5.
Why
The centres (0,2)(0,2) and (3,2)(3,2) are 33 apart. External tangency: k+2=3k+2=3, so k=1k=1. Internal tangency (the smaller circle inside the larger): k−2=3k-2=3, so k=5k=5.
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ReIm-6-5-4-3-2-112345678-4-3-2-11234567802i3 + 2i
The solid circle is tangent to the dashed circles with k = 1 and k = 5.

Example 6 · The locus of w

Slide 10

Example 6

Find the locus of ww if w=z−1zw=\dfrac{z-1}{z} and ∣z∣=2|z|=2.
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Answer
The circle ∣w−1∣=12|w-1|=\dfrac12: centre 11, radius 12\dfrac12.
Why
Make zz the subject: wz=z−1wz=z-1, so z(w−1)=−1z(w-1)=-1 and z=−1w−1z=\dfrac{-1}{w-1}. Then ∣z∣=2|z|=2 gives 1∣w−1∣=2\dfrac{1}{|w-1|}=2, so ∣w−1∣=12|w-1|=\frac12. In Cartesian form, (u−1)2+v2=14(u-1)^2+v^2=\frac14 for w=u+ivw=u+iv. Strategy: if ww is given in terms of zz, make zz the subject, then apply the condition on zz.
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ReIm-1123-2-11201
The locus of w: a circle of radius 1/2 centred at 1.

Example 7 · Purely imaginary

Slide 11

Example 7

Find the locus of zz if w=z−2zw=\dfrac{z-2}{z} is purely imaginary.
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Answer
The circle (x−1)2+y2=1(x-1)^2+y^2=1, with the points z=0z=0 and z=2z=2 excluded.
Why
w=(x−2+iy)(x−iy)x2+y2=x2+y2−2x+2iyx2+y2w=\dfrac{(x-2+iy)(x-iy)}{x^2+y^2}=\dfrac{x^2+y^2-2x+2iy}{x^2+y^2}. Purely imaginary means Re⁡(w)=0\Re(w)=0 (and w≠0w\ne0), so x2+y2−2x=0x^2+y^2-2x=0, that is (x−1)2+y2=1(x-1)^2+y^2=1. Exclude z=0z=0 (where ww is undefined) and z=2z=2 (where w=0w=0, which is real). Mark both with open circles.
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ReIm-1123-2-112002
The locus with two excluded points.

Example 8 · Extension: not a straight line

Slide 12

This locus is not one of the named syllabus loci, but it shows why the “equal distance” condition matters.

Example 8

ExtensionDescribe the locus given by ∣z+2∣=2∣z−2+i∣|z+2|=2|z-2+i|.
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Answer
A circle with centre (103,−43)\left(\dfrac{10}{3},-\dfrac43\right) and radius 2173\dfrac{2\sqrt{17}}{3}.
Why
(x+2)2+y2=4[(x−2)2+(y+1)2](x+2)^2+y^2=4\left[(x-2)^2+(y+1)^2\right] gives 3x2+3y2−20x+8y+16=03x^2+3y^2-20x+8y+16=0, so (x−103)2+(y+43)2=689\left(x-\frac{10}{3}\right)^2+\left(y+\frac43\right)^2=\frac{68}{9}. When the ratio of distances is not 1, the locus is a circle, not a line.

Spot the error · Circles and bisectors

Slide 13

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

(a) ∣z+3−i∣=2|z+3-i|=2 is a circle with centre (3,−1)(3,-1).
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Answer
The centre is (−3,1)(-3,1).
Why
Write ∣z+3−i∣=∣z−(−3+i)∣|z+3-i|=|z-(-3+i)|, so the centre is −3+i-3+i, that is (−3,1)(-3,1). The sign flips when you read off the centre.
(b) ∣z−2i∣<1|z-2i|<1 is drawn with a solid boundary.
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Answer
The boundary should be dashed.
Why
The inequality is strict, so the points on the circle are not in the region.
(c) ∣2z−4i∣=6|2z-4i|=6 is a circle with radius 66.
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Answer
The radius is 33 and the centre is 2i2i.
Why
∣2z−4i∣=2∣z−2i∣=6|2z-4i|=2|z-2i|=6, so ∣z−2i∣=3|z-2i|=3. Always factor out the coefficient of zz first.
(d) ∣z−1∣=∣z−i∣|z-1|=|z-i| is the circle through 11 and ii.
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Answer
It is a line, not a circle: the line y=xy=x.
Why
Equal distances from 11 and ii give the perpendicular bisector of the interval joining them, which is y=xy=x.

Exit ticket

Slide 14
1. Sketch ∣z−1−i∣=∣z+1+i∣|z-1-i|=|z+1+i| and give its equation.
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Answer
The line y=−xy=-x.
Why
This is the perpendicular bisector of (1,1)(1,1) and (−1,−1)(-1,-1). The midpoint is OO and the interval has gradient 11, so the bisector passes through OO with gradient −1-1.
2. Write the Cartesian equation of ∣2z−4i∣=6|2z-4i|=6 and sketch it.
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Answer
x2+(y−2)2=9x^2+(y-2)^2=9.
Why
Divide by 2: ∣z−2i∣=3|z-2i|=3, a circle with centre (0,2)(0,2) and radius 33.
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ReIm-4-3-2-11234-4-3-2-112345602i
Answers to the exit ticket: y = −x and the circle |z − 2i| = 3.