McIntyre MathsMathematics Extension 2
Extension 2 › Complex Numbers I › Lesson 1

Lesson 1 · The arithmetic of complex numbers

Mathematics Extension 2 · Complex Numbers I · Outcome ME2-12-03 · Term 4 · Week 6
Learning intentionTo extend the number system to include ii and the complex numbers, and to add, subtract and multiply complex numbers.
Success criteria · I can…
  • define ii by i2=−1i^2=-1 and simplify powers of ii
  • solve x2+k=0x^2+k=0 where k>0k>0
  • classify numbers as belonging to N\mathbb{N}, Z\mathbb{Z}, Q\mathbb{Q}, R\mathbb{R} or C\mathbb{C}
  • identify the real part Re⁡(z)\Re(z) and imaginary part Im⁡(z)\Im(z) of a complex number
  • add, subtract and multiply complex numbers

Lesson flow

LaunchWarm-up · Why do we need a new number?Slide 2ThinkThink · Multiplying on the number lineSlide 3DefinitionsThe number iSlide 4ExamplesExample 2 · Powers of iSlide 5Key ideaThe cycle of fourSlide 6DefinitionsThe complex numbersSlide 7ExamplesExample 4 · Classifying numbersSlide 8DefinitionsAdding, subtracting and multiplyingSlide 9ExamplesExample 6 · MultiplyingSlide 10HSC-styleHSC practice · Expanding a powerSlide 11Exit ticketExit ticketSlide 12

Warm-up · Why do we need a new number?

Slide 2

Consider the equation x2+2x+4=0x^2+2x+4=0.

(a) Find the sum and the product of the roots.
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Answer
Sum =−2=-2, product =4=4.
Why
For ax2+bx+c=0ax^2+bx+c=0 the sum of the roots is −ba-\frac{b}{a} and the product is ca\frac{c}{a}. These formulae still make sense even if the roots are not real.
(b) Use the quadratic formula. What goes wrong?
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Answer
x=−2±4−162=−2±−122x=\dfrac{-2\pm\sqrt{4-16}}{2}=\dfrac{-2\pm\sqrt{-12}}{2}. The discriminant is Δ=−12<0\Delta=-12<0, so there are no real roots.
Why
We need a number whose square is negative. Once we have one, the roots are x=−1±i3x=-1\pm i\sqrt3. Check: the sum is −2-2 and the product is 1+3=41+3=4.

Think · Multiplying on the number line

Slide 3

Multiplying by −1-1 turns a number through 180∘180^\circ about 00 (for example 3→−33\to-3). What operation, done twice, gives a half-turn? Where does that operation take the number 11?

Discuss

If multiplying by a number ii twice has the same effect as multiplying by −1-1, what must i2i^2 equal? What does multiplying by ii do to the number line?
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Answer
i2=−1i^2=-1. Multiplying by ii is a quarter-turn (90∘90^\circ): it takes 11 off the real number line to a new vertical axis.
Why
This is only a preview. In Lesson 4 we plot complex numbers on the Argand diagram, and in Lesson 8 we use multiplication by ii as a rotation.

The number i

Slide 4
Definition
The number ii

ii is defined by i2=−1i^2=-1. So x2=−1x^2=-1 has the two solutions x=±ix=\pm i.

For k>0k>0:  x2+k=0 ⇒ x=±ik\ x^2+k=0\ \Rightarrow\ x=\pm i\sqrt{k}.

Watch out

We define ii by i2=−1i^2=-1 and think of it as “−1\sqrt{-1}” only as a reading aid. The rule ab=ab\sqrt{a}\sqrt{b}=\sqrt{ab} holds only for a,b≥0a,b\ge0. For example −4×−9=2i×3i=−6\sqrt{-4}\times\sqrt{-9}=2i\times3i=-6, not 36=6\sqrt{36}=6.

Example 1 · Solve

(a) x2+9=0x^2+9=0
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Answer
x=±3ix=\pm3i
Why
x2=−9=9×(−1)x^2=-9=9\times(-1), so x=±3ix=\pm3i.
(b) x2+5=0x^2+5=0
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Answer
x=±i5x=\pm i\sqrt5
Why
x2=−5x^2=-5, so x=±5 ix=\pm\sqrt5\,i.
(c) 4x2+49=04x^2+49=0
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Answer
x=±7i2x=\pm\dfrac{7i}{2}
Why
x2=−494x^2=-\frac{49}{4}, so x=±72ix=\pm\frac{7}{2}i.

Example 2 · Powers of i

Slide 5

Simplify each power of ii.

(a) i2i^2
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Answer
−1-1
Why
This is the definition of ii.
(b) i3i^3
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Answer
−i-i
Why
i3=i2×i=−ii^3=i^2\times i=-i.
(c) i4i^4
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Answer
11
Why
i4=(i2)2=(−1)2=1i^4=(i^2)^2=(-1)^2=1.
(d) i5i^5
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Answer
ii
Why
i5=i4×i=ii^5=i^4\times i=i.
(e) i6i^6
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Answer
−1-1
Why
i6=i4×i2=−1i^6=i^4\times i^2=-1.
(f) i7i^7
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Answer
−i-i
Why
i7=i4×i3=−ii^7=i^4\times i^3=-i. Can you spot the cycle of four before the next slide?
(g) i8i^8
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Answer
11
Why
i8=(i4)2=1i^8=(i^4)^2=1.
(h) i40i^{40}
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Answer
11
Why
i40=(i4)10=110=1i^{40}=(i^4)^{10}=1^{10}=1.
(i) i50i^{50}
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Answer
−1-1
Why
i50=(i4)12×i2=−1i^{50}=(i^4)^{12}\times i^2=-1.
(j) i231i^{231}
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Answer
−i-i
Why
231=4×57+3231=4\times57+3, so i231=(i4)57×i3=−ii^{231}=(i^4)^{57}\times i^3=-i.

The cycle of four

Slide 6
Key idea
Powers of ii repeat every fouri4k=1,i4k+1=i,i4k+2=−1,i4k+3=−i(k∈Z)i^{4k}=1,\qquad i^{4k+1}=i,\qquad i^{4k+2}=-1,\qquad i^{4k+3}=-i\qquad(k\in\mathbb{Z})

Divide the index by 4: the remainder decides the value.

HSC question

2021 HSC Q11(b)Evaluate i+i2+i3+i4+i5i+i^2+i^3+i^4+i^5.
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Answer
ii
Why
i+(−1)+(−i)+1+i=ii+(-1)+(-i)+1+i=i. Any four consecutive powers of ii add to 00, so only the fifth term survives. Extension: i+i2+⋯+i2027=−1i+i^2+\dots+i^{2027}=-1, since 2027=4×506+32027=4\times506+3 leaves i+i2+i3=−1i+i^2+i^3=-1.

The complex numbers

Slide 7
Definition
Complex numbers C={a+ib: a,b∈R}\mathbb{C}=\{a+ib:\ a,b\in\mathbb{R}\}

A complex number has the form z=a+ibz=a+ib, where aa and bb are real.

  • a=Re⁡(z)a=\Re(z) is the real part.
  • b=Im⁡(z)b=\Im(z) is the imaginary part. It is a real number, not bibi.
  • If Re⁡(z)=0\Re(z)=0 and b≠0b\ne0, then zz is purely imaginary.

Example 3 · Real and imaginary parts

Write down Re⁡(z)\Re(z) and Im⁡(z)\Im(z).

(a) 3+4i3+4i
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Answer
Re⁡=3, Im⁡=4\Re=3,\ \Im=4
Why
Im⁡(z)\Im(z) is 44, not 4i4i.
(b) −2i-2i
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Answer
Re⁡=0, Im⁡=−2\Re=0,\ \Im=-2
Why
−2i=0+(−2)i-2i=0+(-2)i, so zz is purely imaginary.
(c) 77
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Answer
Re⁡=7, Im⁡=0\Re=7,\ \Im=0
Why
7=7+0i7=7+0i. Every real number is a complex number.
(d) 1−i52\dfrac{1-i\sqrt5}{2}
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Answer
Re⁡=12, Im⁡=−52\Re=\dfrac12,\ \Im=-\dfrac{\sqrt5}{2}
Why
Split it as 12−52i\frac12-\frac{\sqrt5}{2}i.

Example 4 · Classifying numbers

Slide 8

Each set is an extension of the previous one: N⊂Z⊂Q⊂R⊂C\mathbb{N}\subset\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}\subset\mathbb{C}. Place each number in the smallest set it belongs to.

ℂ complexℝ realℚ rationalℤ integerℕ natural
The number sets are nested. The part of ℝ outside ℚ holds the irrational numbers, and the part of ℂ outside ℝ holds the non-real numbers.
(a) 5,1865,\quad \dfrac{18}{6}
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Answer
5∈N5\in\mathbb{N} and 186=3∈N\frac{18}{6}=3\in\mathbb{N}
Why
Whole numbers 1, 2, 3, ... (some texts include 0 in N\mathbb{N}; the syllabus does not insist either way).
(b) −8-8
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Answer
Z\mathbb{Z}
Why
A negative whole number is an integer but not a natural number.
(c) sin⁡π6,227\sin\dfrac{\pi}{6},\quad \dfrac{22}{7}
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Answer
sin⁡π6=12∈Q\sin\frac{\pi}{6}=\frac12\in\mathbb{Q} and 227∈Q\frac{22}{7}\in\mathbb{Q}
Why
Both are ratios of integers. Note 227\frac{22}{7} is only an approximation of π\pi.
(d) π,5\pi,\quad\sqrt5
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Answer
R\mathbb{R} (irrational)
Why
They are real but cannot be written as a ratio of integers.
(e) 49\sqrt{49}
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Answer
7∈N7\in\mathbb{N}
Why
49=7\sqrt{49}=7 exactly, so it is a natural number.
(f) 2+3i,4i2+3i,\quad 4i
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Answer
C\mathbb{C}
Why
Both are non-real, so C\mathbb{C} is the smallest set containing them.

Adding, subtracting and multiplying

Slide 9
Key idea
Arithmetic in C\mathbb{C}

Treat ii like a pronumeral (a special one) and replace i2i^2 with −1-1.

(a+ib)±(c+id)=(a±c)+i(b±d)(a+ib)\pm(c+id)=(a\pm c)+i(b\pm d)(a+ib)(c+id)=(ac−bd)+i(ad+bc)(a+ib)(c+id)=(ac-bd)+i(ad+bc)

Check your answers on a calculator in complex mode (the syllabus says “with and without digital tools”).

Example 5

Express in the form a+iba+ib.

(a) (3+4i)+(2+6i)(3+4i)+(2+6i)
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Answer
5+10i5+10i
Why
Add real parts and add imaginary parts.
(b) (4−3i)−(2−4i)(4-3i)-(2-4i)
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Answer
2+i2+i
Why
(4−2)+(−3+4)i=2+i(4-2)+(-3+4)i=2+i. Take care with the signs when subtracting.
(c) (3+2i)(4+2i)(3+2i)(4+2i)
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Answer
8+14i8+14i
Why
12+6i+8i+4i2=12+14i−4=8+14i12+6i+8i+4i^2=12+14i-4=8+14i.

Example 6 · Multiplying

Slide 10

Express in the form a+iba+ib.

(a) (2+5i)(3+6i)(2+5i)(3+6i)
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Answer
−24+27i-24+27i
Why
6+12i+15i+30i2=6+27i−30=−24+27i6+12i+15i+30i^2=6+27i-30=-24+27i.
(b) (3+2i)(3−2i)(3+2i)(3-2i)
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Answer
1313
Why
9−6i+6i−4i2=9+4=139-6i+6i-4i^2=9+4=13. The middle terms cancel.
(c) (a+ib)(a−ib)(a+ib)(a-ib)
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Answer
a2+b2a^2+b^2
Why
The product is always real. The numbers a+iba+ib and a−iba-ib are called conjugates (Lesson 2).

HSC practice · Expanding a power

Slide 11
Adapted from 2023 HSC Q1Expand (a+ib)3(a+ib)^3, where a,b∈Ra,b\in\mathbb{R}, and write it in the form x+iyx+iy. Hence find (2−i)3(2-i)^3.
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Answer
(a+ib)3=(a3−3ab2)+i(3a2b−b3)(a+ib)^3=(a^3-3ab^2)+i(3a^2b-b^3), and (2−i)3=2−11i(2-i)^3=2-11i.
Why
(a+ib)3=a3+3a2(ib)+3a(ib)2+(ib)3=a3+3a2b i−3ab2−b3i(a+ib)^3=a^3+3a^2(ib)+3a(ib)^2+(ib)^3=a^3+3a^2b\,i-3ab^2-b^3i, using i2=−1i^2=-1 and i3=−ii^3=-i. For (2−i)3(2-i)^3 put a=2a=2, b=−1b=-1: real part 8−6=28-6=2, imaginary part −12+1=−11-12+1=-11. A common slip is forgetting that i3=−ii^3=-i.

Exit ticket

Slide 12
1. Simplify i2027i^{2027}.
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Answer
−i-i
Why
2027=4×506+32027=4\times506+3, so i2027=(i4)506×i3=−ii^{2027}=(i^4)^{506}\times i^3=-i.
2. Express (1+i)4(1+i)^4 in the form a+iba+ib.
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Answer
−4-4
Why
(1+i)2=1+2i+i2=2i(1+i)^2=1+2i+i^2=2i, so (1+i)4=(2i)2=4i2=−4(1+i)^4=(2i)^2=4i^2=-4.
3. Find Re⁡(z)\Re(z) and Im⁡(z)\Im(z) if z=(2−i)(3+4i)z=(2-i)(3+4i).
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Answer
Re⁡(z)=10\Re(z)=10, Im⁡(z)=5\Im(z)=5
Why
z=6+8i−3i−4i2=6+5i+4=10+5iz=6+8i-3i-4i^2=6+5i+4=10+5i.