McIntyre MathsMathematics Extension 2
Extension 2 › Complex Numbers I › Lesson 4

Lesson 4 · The Argand diagram

Mathematics Extension 2 · Complex Numbers I · Outcome ME2-12-03 · Term 4 · Week 7
Learning intentionTo represent complex numbers as points in the complex plane and describe simple sets of points.
Success criteria · I can…
  • explain how the complex plane is labelled and where its name comes from
  • plot zz, z‾\conj{z}, −z-z and iziz and describe each transformation
  • explain why multiplying by ii is a rotation of 90∘90^\circ about the origin
  • sketch Re⁡(z)=c\Re(z)=c, Im⁡(z)=k\Im(z)=k and simple regions
  • find a locus by letting z=x+iyz=x+iy

Lesson flow

LaunchWarm-up · Retrieval from Lessons 1 to 3Slide 2BackgroundWhy a plane? · Where the name comes fromSlide 3DefinitionsThe complex plane · How it is labelledSlide 4CheckCheck · Reading the planeSlide 5DefinitionsExample 1 · The complex planeSlide 6ExamplesExample 2 · Powers of i on the planeSlide 7ExamplesExample 3 · Why multiplying by i is a rotationSlide 8ExamplesExample 4 · Lines and regionsSlide 9DefinitionsLociSlide 10ExamplesExample 6 · A locus with fractionsSlide 11ExamplesExample 7 · Show that it is a circleSlide 12Exam practiceExam practice · Multiple choice and the unit circleSlide 13Exit ticketExit ticketSlide 14

Warm-up · Retrieval from Lessons 1 to 3

Slide 2
(a) Find real aa, bb with (a+ib)(1+i)=3+i(a+ib)(1+i)=3+i.
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Answer
a=2a=2, b=−1b=-1
Why
a+ib=3+i1+i=(3+i)(1−i)2=4−2i2=2−ia+ib=\dfrac{3+i}{1+i}=\dfrac{(3+i)(1-i)}{2}=\dfrac{4-2i}{2}=2-i. Check: (2−i)(1+i)=2+2i−i+1=3+i(2-i)(1+i)=2+2i-i+1=3+i.
(b) Find the square roots of 2i2i.
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Answer
±(1+i)\pm(1+i)
Why
(1+i)2=1+2i−1=2i(1+i)^2=1+2i-1=2i, so the square roots are ±(1+i)\pm(1+i).
(c) Find a real quadratic with root 1+2i1+2i.
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Answer
z2−2z+5=0z^2-2z+5=0
Why
The other root is 1−2i1-2i. Sum =2=2, product =1+4=5=1+4=5.
(d) Find the distance between (1,2)(1,2) and (4,6)(4,6).
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Answer
55
Why
32+42=5\sqrt{3^2+4^2}=5. This previews distance on the Argand diagram.

Why a plane? · Where the name comes from

Slide 3
Key idea
The number line is full

The real numbers fill every point of the number line, so there is nowhere on it for ii. A complex number x+iyx+iy is made of two real numbers, so we need two dimensions to show it.

One idea, several names: the Argand diagram, the complex plane, the Argand plane or the zz-plane.

YearWhat happened
1545Cardano meets square roots of negative numbers while solving cubics
1572Bombelli writes down the rules for calculating with them
1637Descartes calls them “imaginary”, and it was not a compliment
1777Euler introduces the symbol ii for −1\sqrt{-1}
1797Wessel, a Norwegian–Danish surveyor, first draws them as points in a plane
1806Argand, a Swiss bookkeeper in Paris, publishes the same idea and his name sticks
1831Gauss makes the plane widely accepted and names them “complex numbers”

Wessel wrote in Danish and was overlooked for about a century.

The complex plane · How it is labelled

Slide 4
Definition
Reading the plane
  • Horizontal: the real axis, labelled Re.
  • Vertical: the imaginary axis, labelled Im.
  • z=x+iyz=x+iy is the point Z(x,y)Z(x,y): across Re⁡(z)\Re(z), up Im⁡(z)\Im(z).
  • Real numbers lie on the real axis; purely imaginary numbers lie on the imaginary axis.
  • Every complex number is exactly one point, and every point is exactly one complex number.
  • Capital letter for the point (ZZ), lower case for the number (zz): “the point ZZ represents zz”.

Check · Reading the plane

Slide 5

Quick-fire: write the point for each number, then the number for each point.

(a) Write the point for −3+5i-3+5i.
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Answer
(−3,5)(-3,5).
Why
The real part is the xx-coordinate and the imaginary part is the yy-coordinate.
(b) Write the point for 2i2i.
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Answer
(0,2)(0,2).
Why
2i=0+2i2i=0+2i, so it lies on the imaginary axis.
(c) Write the point for 77.
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Answer
(7,0)(7,0).
Why
7=7+0i7=7+0i, so it lies on the real axis.
(d) Write the point for (2+i) i(2+i)\,i.
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Answer
(−1,2)(-1,2).
Why
(2+i)i=2i+i2=−1+2i(2+i)i=2i+i^2=-1+2i. This previews multiplying by ii.
(e) Write the number for (−3,0)(-3,0).
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Answer
−3-3.
Why
A point on the real axis is a real number.
(f) Write the number for (0,3)(0,3).
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Answer
3i3i.
Why
A point on the imaginary axis is purely imaginary.
(g) Write the number for (7,−5)(7,-5).
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Answer
7−5i7-5i.
Why
x=7x=7 and y=−5y=-5, so z=7+(−5)iz=7+(-5)i.
(h) Write the number for (a,b)(a,b).
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Answer
a+iba+ib.
Why
This is the general rule: (x,y)(x,y) represents x+iyx+iy.

Example 1 · The complex plane

Slide 6

The Argand diagram (also called the complex plane or Argand plane) plots z=x+iyz=x+iy at the point (x,y)(x,y): the real axis runs across and the imaginary axis runs up.

Example 1

Let z=3+4iz=3+4i. Plot (a) zz, (b) z‾\conj{z}, (c) −z-z, (d) iziz. Describe each as a transformation of zz.
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Answer
z=(3,4)z=(3,4); z‾=3−4i\conj{z}=3-4i at (3,−4)(3,-4); −z=−3−4i-z=-3-4i at (−3,−4)(-3,-4); iz=−4+3iiz=-4+3i at (−4,3)(-4,3).
Why
z‾\conj{z} is the reflection of zz in the real axis. −z-z is a rotation of 180∘180^\circ about the origin. iz=i(3+4i)=−4+3iiz=i(3+4i)=-4+3i is a rotation of 90∘90^\circ anticlockwise about the origin. In general iz=i(x+iy)=−y+ixiz=i(x+iy)=-y+ix. The gradient of OzOz is 43\frac43 and the gradient of O(iz)O(iz) is −34-\frac34: the product is −1-1, so they are perpendicular.
Show diagramHide diagram
ReIm-5-4-3-2-112345-5-4-3-2-1123450zz̄−ziz
z = 3 + 4i with its conjugate, its negative and iz.

Example 2 · Powers of i on the plane

Slide 7

Let z=1+3iz=1+3i. The points AA, BB, CC, DD represent zz, iziz, i2zi^2z and i3zi^3z.

(a) Plot the four points.
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Answer
A=1+3iA=1+3i, B=−3+iB=-3+i, C=−1−3iC=-1-3i, D=3−iD=3-i.
Why
iz=i+3i2=−3+iiz=i+3i^2=-3+i, i2z=−z=−1−3ii^2z=-z=-1-3i, i3z=−iz=3−ii^3z=-iz=3-i.
(b) What type of quadrilateral is ABCDABCD?
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Answer
A square centred at the origin.
Why
All sides have length 20\sqrt{20} (for example ABAB: (4,2)(4,2) has length 20\sqrt{20}) and the diagonals ACAC and BDBD are equal, both of length 2102\sqrt{10}.
(c) What does multiplying by ii appear to do?
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Answer
It rotates the point 90∘90^\circ anticlockwise about the origin.
Why
Each step A→B→C→D→AA\to B\to C\to D\to A is a quarter-turn. Four quarter-turns return to the start, which is why i4=1i^4=1.
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ReIm-4-3-2-11234-4-3-2-112340ABCD
ABCD is a square: each multiplication by i is a quarter-turn.

Example 3 · Why multiplying by i is a rotation

Slide 8

The numbers z=a+ibz=a+ib and w=izw=iz (with a,b≠0a,b\ne0) are plotted at AA and BB.

(a) By considering gradients, show that OA⊥OBOA\perp OB.
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Answer
w=i(a+ib)=−b+iaw=i(a+ib)=-b+ia, so A=(a,b)A=(a,b) and B=(−b,a)B=(-b,a). The gradient of OAOA is ba\frac ba and of OBOB is a−b\frac{a}{-b}. The product is ba×(−ab)=−1\frac ba\times\left(-\frac ab\right)=-1, so OA⊥OBOA\perp OB.
Why
Two lines are perpendicular when the product of their gradients is −1-1. (If aa or bb is 00, the point lies on an axis and the result is clear from a sketch.)
(b) Use the distance formula to show that OA=OBOA=OB. What type of triangle is △OAB\triangle OAB?
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Answer
OA=a2+b2OA=\sqrt{a^2+b^2} and OB=b2+a2OB=\sqrt{b^2+a^2}, so OA=OBOA=OB. The triangle OABOAB is right-angled and isosceles.
Why
So BB is AA rotated 90∘90^\circ anticlockwise about OO. This is the geometric fact behind rotations in Lesson 8.

Example 4 · Lines and regions

Slide 9
Key idea
Conventions

Re⁡(z)=c\Re(z)=c is the vertical line x=cx=c. Im⁡(z)=k\Im(z)=k is the horizontal line y=ky=k.

Use a dashed boundary when it is excluded (<< or >>) and a solid boundary when it is included (≤\le or ≥\ge).

(a) Sketch Re⁡(z)=5\Re(z)=5 and Im⁡(z)=−2\Im(z)=-2.
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Answer
A vertical line through (5,0)(5,0) and a horizontal line through (0,−2)(0,-2).
Why
They meet at the point 5−2i5-2i.
(b) Shade the region where Im⁡(z)<1\Im(z)<1 and Re⁡(z)≥−2\Re(z)\ge-2.
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Answer
The region below the dashed line y=1y=1 and to the right of the solid line x=−2x=-2. The corner (−2,1)(-2,1) is not included.
Why
Both conditions must hold, so shade the overlap. The corner is excluded because Im⁡(z)<1\Im(z)<1 is strict there.
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ReIm-4-3-2-1123456-4-3-2-112305 − 2i
(a) Re(z) = 5 and Im(z) = −2.
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ReIm-4-3-2-1123456-4-3-2-11230
(b) Im(z) < 1 and Re(z) ≥ −2.

Loci

Slide 10
Definition
Locus

A locus is the set of all points zz satisfying a condition. Method: let z=x+iyz=x+iy, simplify to a Cartesian equation, then recognise and sketch the curve.

Example 5

Given that zz‾=4z\conj{z}=4, find the locus of zz and sketch it.
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Answer
The circle with centre OO and radius 22: x2+y2=4x^2+y^2=4.
Why
(x+iy)(x−iy)=x2+y2=4(x+iy)(x-iy)=x^2+y^2=4. Since zz‾=∣z∣2z\conj{z}=|z|^2, this is the same as ∣z∣=2|z|=2.
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ReIm-3-2-1123-3-2-11230
|z| = 2

Example 6 · A locus with fractions

Slide 11
Given that 1z+1z‾=1\dfrac1z+\dfrac1{\conj{z}}=1, find the locus of zz and sketch it.
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Answer
The circle (x−1)2+y2=1(x-1)^2+y^2=1, centre (1,0)(1,0) and radius 11, excluding the origin.
Why
The left side is undefined at z=0z=0, so z≠0z\ne0. Multiply by zz‾z\conj{z}: z‾+z=zz‾\conj{z}+z=z\conj{z}, so 2x=x2+y22x=x^2+y^2 and (x−1)2+y2=1(x-1)^2+y^2=1. Draw an open circle at the origin. (Alternatively, 1z+1z‾=2Re⁡(1z)\frac1z+\frac1{\conj z}=2\Re\left(\frac1z\right) by Lesson 2.)
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ReIm-1123-2-1120
The circle with the origin removed.

Example 7 · Show that it is a circle

Slide 12
Show that (z−2)(z‾−2)=9(z-2)(\conj{z}-2)=9 represents a circle in the Argand diagram, and sketch it.
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Answer
It is the circle with centre (2,0)(2,0) and radius 33.
Why
Let z=x+iyz=x+iy: (x−2+iy)(x−2−iy)=(x−2)2+y2=9(x-2+iy)(x-2-iy)=(x-2)^2+y^2=9. Neater: z‾−2=z−2‾\conj{z}-2=\conj{z-2}, so the left side is ∣z−2∣2|z-2|^2 and the equation says ∣z−2∣=3|z-2|=3: the distance from zz to 22 is 33.
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ReIm-2-1123456-4-3-2-1123402
|z − 2| = 3

Exam practice · Multiple choice and the unit circle

Slide 13
Exam practiceWhich equation describes the circle with centre OO and radius 22?
(A) zz‾=2z\conj{z}=2(B) (z+z‾)2−(z−z‾)2=16(z+\conj{z})^2-(z-\conj{z})^2=16(C) Re⁡(z)+Im⁡(z)=2\Re(z)+\Im(z)=2(D) ∣z∣2=2|z|^2=2
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Answer
(B)
Why
(z+z‾)2−(z−z‾)2=(2x)2−(2iy)2=4x2+4y2=16(z+\conj{z})^2-(z-\conj{z})^2=(2x)^2-(2iy)^2=4x^2+4y^2=16, so x2+y2=4x^2+y^2=4. (A) and (D) both give radius 2\sqrt2. (C) is a straight line.
Exam practiceThe point ZZ lies on the unit circle in the first quadrant, as shown. Where does W=1ZW=\dfrac1Z lie?
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Answer
On the unit circle in the fourth quadrant: WW is the reflection of ZZ in the real axis.
Why
∣Z∣=1|Z|=1, so ZZ‾=1Z\conj{Z}=1 and 1Z=Z‾\frac1Z=\conj{Z}. This uses 1z=z‾∣z∣2\frac1z=\frac{\conj{z}}{|z|^2} from Lesson 2.
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ReIm-2-112-2-1120Z
Z on the unit circle.

Exit ticket

Slide 14
1. Let z=−2+iz=-2+i. Plot zz, z‾\conj{z} and iziz on one Argand diagram.
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Answer
z=(−2,1)z=(-2,1), z‾=(−2,−1)\conj{z}=(-2,-1) and iz=(−1,−2)iz=(-1,-2).
Why
iz=i(−2+i)=−2i+i2=−1−2iiz=i(-2+i)=-2i+i^2=-1-2i, which is zz rotated 90∘90^\circ anticlockwise about the origin.
2. Find and sketch the locus of zz if Im⁡(z2)=0\Im(z^2)=0.
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Answer
Both axes: the lines x=0x=0 and y=0y=0 (the origin is included).
Why
z2=x2−y2+2ixyz^2=x^2-y^2+2ixy, so Im⁡(z2)=2xy=0\Im(z^2)=2xy=0, giving x=0x=0 or y=0y=0.