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

Lesson 2 · Conjugates, division and equality

Mathematics Extension 2 · Complex Numbers I · Outcome ME2-12-03 · Term 4 · Week 6
Learning intentionTo use the complex conjugate to divide complex numbers, and to solve equations by equating real and imaginary parts.
Success criteria · I can…
  • find z‾\conj{z} and use zz‾=∣z∣2z\conj{z}=|z|^2
  • prove and use z1+z2‾=z1‾+z2‾\conj{z_1+z_2}=\conj{z_1}+\conj{z_2} and z1z2‾=z1‾ z2‾\conj{z_1z_2}=\conj{z_1}\,\conj{z_2}
  • divide complex numbers by realising the denominator
  • use Re⁡(z)=z+z‾2\Re(z)=\frac{z+\conj{z}}{2}, Im⁡(z)=z−z‾2i\Im(z)=\frac{z-\conj{z}}{2i} and 1z=z‾∣z∣2\frac1z=\frac{\conj{z}}{|z|^2}
  • equate real and imaginary parts to solve for unknowns

Lesson flow

LaunchWarm-up · Real productsSlide 2DefinitionsThe conjugate and the modulusSlide 3ExamplesExample 2 · Proving the conjugate rulesSlide 4ExamplesExample 3 · Using the conjugate rulesSlide 5DefinitionsExample 4 · Dividing: realise the denominatorSlide 6ExamplesExample 5 · DivisionSlide 7Key ideaExample 6 · Re, Im and the reciprocalSlide 8Key ideaExample 7 · Equality of complex numbersSlide 9HSC-styleHSC practice · Conjugates and divisionSlide 10HSC-styleHSC practice · Equating partsSlide 11Exit ticketExit ticketSlide 12

Warm-up · Real products

Slide 2

Expand and simplify.

(a) (5+i)(5−i)(5+i)(5-i)
Show answerHide answer
Answer
2626
Why
25−i2=25+1=2625-i^2=25+1=26. The middle terms cancel.
(b) (a+ib)(a−ib)(a+ib)(a-ib)
Show answerHide answer
Answer
a2+b2a^2+b^2
Why
The same pattern as (a): a2−i2b2=a2+b2a^2-i^2b^2=a^2+b^2, which is always real.
(c) Rationalise the denominator of 13−2\dfrac{1}{3-\sqrt2}.
Show answerHide answer
Answer
3+27\dfrac{3+\sqrt2}{7}
Why
Multiply top and bottom by 3+23+\sqrt2: 3+29−2\frac{3+\sqrt2}{9-2}. This is the template for dividing complex numbers.

The conjugate and the modulus

Slide 3
Definition
Conjugate and modulus

The conjugate of z=a+ibz=a+ib is z‾=a−ib\conj{z}=a-ib (the reflection of zz in the real axis).

The modulus of zz is ∣z∣=zz‾=a2+b2|z|=\sqrt{z\conj{z}}=\sqrt{a^2+b^2}, so zz‾=∣z∣2z\conj{z}=|z|^2 is always real.

Example 1

(a) −1+4i‾\conj{-1+4i}
Show answerHide answer
Answer
−1−4i-1-4i
Why
Change the sign of the imaginary part.
(b) −2i‾\conj{-2i}
Show answerHide answer
Answer
2i2i
Why
−2i=0−2i-2i=0-2i, so its conjugate is 0+2i0+2i.
(c) If z=3+4iz=3+4i, find zz‾z\conj{z} and ∣z∣|z|.
Show answerHide answer
Answer
zz‾=25z\conj{z}=25 and ∣z∣=5|z|=5.
Why
(3+4i)(3−4i)=9+16=25(3+4i)(3-4i)=9+16=25, and ∣z∣=25=5|z|=\sqrt{25}=5.
(d) 2+3i‾‾\conj{\conj{2+3i}}
Show answerHide answer
Answer
2+3i2+3i
Why
Conjugating twice returns the original number.

Example 2 · Proving the conjugate rules

Slide 4

Let z1=a+ibz_1=a+ib and z2=c+idz_2=c+id, where a,b,c,d∈Ra,b,c,d\in\mathbb{R}.

(a) Prove that z1+z2‾=z1‾+z2‾\conj{z_1+z_2}=\conj{z_1}+\conj{z_2}.
Show answerHide answer
Answer
z1+z2=(a+c)+i(b+d)z1+z2‾=(a+c)−i(b+d)=(a−ib)+(c−id)=z1‾+z2‾\begin{aligned}z_1+z_2&=(a+c)+i(b+d)\\ \conj{z_1+z_2}&=(a+c)-i(b+d)=(a-ib)+(c-id)=\conj{z_1}+\conj{z_2}\end{aligned}
Why
Add first, then conjugate; the result matches conjugating first, then adding.
(b) Prove that z1z2‾=z1‾ z2‾\conj{z_1z_2}=\conj{z_1}\,\conj{z_2}.
Show answerHide answer
Answer
z1z2=(ac−bd)+i(ad+bc)z1z2‾=(ac−bd)−i(ad+bc)z1‾ z2‾=(a−ib)(c−id)=ac−iad−ibc+i2bd=(ac−bd)−i(ad+bc)\begin{aligned}z_1z_2&=(ac-bd)+i(ad+bc)\\ \conj{z_1z_2}&=(ac-bd)-i(ad+bc)\\ \conj{z_1}\,\conj{z_2}&=(a-ib)(c-id)=ac-iad-ibc+i^2bd=(ac-bd)-i(ad+bc)\end{aligned} The two sides are equal.
Why
Expand both sides and compare real and imaginary parts. These identities are part of the syllabus: you must be able to prove and use them.

Example 3 · Using the conjugate rules

Slide 5

Evaluate in two ways: directly, and using the conjugate rules.

(a) (5+i)+(3−2i)‾\conj{(5+i)+(3-2i)}
Show answerHide answer
Answer
8+i8+i
Why
Directly: (8−i)(8-i) conjugated is 8+i8+i. By the rule: (5−i)+(3+2i)=8+i(5-i)+(3+2i)=8+i.
(b) (5+i)(3−2i)‾\conj{(5+i)(3-2i)}
Show answerHide answer
Answer
17+7i17+7i
Why
By the rule: (5−i)(3+2i)=15+10i−3i+2=17+7i(5-i)(3+2i)=15+10i-3i+2=17+7i. Directly: (5+i)(3−2i)=17−7i(5+i)(3-2i)=17-7i, whose conjugate is 17+7i17+7i.

Example 4 · Dividing: realise the denominator

Slide 6
Key idea
Dividing complex numbers

Multiply the numerator and denominator by the conjugate of the denominator:

zw=z w‾w w‾=z w‾∣w∣2,w≠0\frac{z}{w}=\frac{z\,\conj{w}}{w\,\conj{w}}=\frac{z\,\conj{w}}{|w|^2},\qquad w\ne0

Example 4

Express in the form a+iba+ib.

(a) 32i\dfrac{3}{2i}
Show answerHide answer
Answer
−32 i-\dfrac32\,i
Why
Multiply by ii\frac{i}{i}: 3i2i2=−3i2\frac{3i}{2i^2}=-\frac{3i}{2}. Multiplying by −2i−2i\frac{-2i}{-2i} works too, but is slower.
(b) 13+2i\dfrac{1}{3+2i}
Show answerHide answer
Answer
313−213 i\dfrac{3}{13}-\dfrac{2}{13}\,i
Why
13+2i×3−2i3−2i=3−2i9+4\frac{1}{3+2i}\times\frac{3-2i}{3-2i}=\frac{3-2i}{9+4}.

Example 5 · Division

Slide 7
Express 3+4i2−3i\dfrac{3+4i}{2-3i} in the form a+iba+ib.
Show answerHide answer
Answer
−613+1713 i-\dfrac{6}{13}+\dfrac{17}{13}\,i
Why
(3+4i)(2+3i)(2−3i)(2+3i)=6+9i+8i−124+9=−6+17i13\frac{(3+4i)(2+3i)}{(2-3i)(2+3i)}=\frac{6+9i+8i-12}{4+9}=\frac{-6+17i}{13} Check the answer on a calculator in complex mode (the syllabus says “with and without digital tools”).

Example 6 · Re, Im and the reciprocal

Slide 8
Theorem
Three useful relationshipsRe⁡(z)=z+z‾2,Im⁡(z)=z−z‾2i,1z=z‾∣z∣2(z≠0)\Re(z)=\frac{z+\conj{z}}{2},\qquad \Im(z)=\frac{z-\conj{z}}{2i},\qquad \frac1z=\frac{\conj{z}}{|z|^2}\quad(z\ne0)

Example 6

(a) Prove each relationship.
Show answerHide answer
Answer
Let z=a+ibz=a+ib. Then z+z‾=2az+\conj{z}=2a, so a=z+z‾2a=\frac{z+\conj{z}}{2}. Also z−z‾=2ibz-\conj{z}=2ib, so b=z−z‾2ib=\frac{z-\conj{z}}{2i}. Finally 1z=z‾zz‾=z‾∣z∣2\frac1z=\frac{\conj{z}}{z\conj{z}}=\frac{\conj{z}}{|z|^2}.
Why
Add and subtract zz and its conjugate, then use zz‾=∣z∣2z\conj{z}=|z|^2.
(b) Hence show that zz is real if and only if z=z‾z=\conj{z}.
Show answerHide answer
Answer
zz is real ⇔Im⁡(z)=0⇔z−z‾2i=0⇔z=z‾\Leftrightarrow\Im(z)=0\Leftrightarrow\frac{z-\conj{z}}{2i}=0\Leftrightarrow z=\conj{z}.
Why
Every step is reversible, so the argument proves both directions.
(c) What can you say about z≠0z\ne0 if z‾=−z\conj{z}=-z?
Show answerHide answer
Answer
zz is purely imaginary.
Why
z‾=−z⇔z+z‾=0⇔2Re⁡(z)=0⇔Re⁡(z)=0\conj{z}=-z\Leftrightarrow z+\conj{z}=0\Leftrightarrow 2\Re(z)=0\Leftrightarrow\Re(z)=0.

Example 7 · Equality of complex numbers

Slide 9
Key idea
Equal complex numbers

For real a,b,c,da,b,c,d:

a+ib=c+id  ⟺  a=c  and  b=da+ib=c+id\iff a=c\ \text{ and }\ b=d

Equate real parts and imaginary parts to get two equations.

Example 7

Find real aa and bb such that z=a+ibz=a+ib satisfies z+3iz‾=−1+13iz+3i\conj{z}=-1+13i.
Show answerHide answer
Answer
a=5a=5, b=−2b=-2, so z=5−2iz=5-2i.
Why
z+3iz‾=(a+ib)+3i(a−ib)=(a+3b)+i(3a+b)z+3i\conj{z}=(a+ib)+3i(a-ib)=(a+3b)+i(3a+b). Real parts: a+3b=−1a+3b=-1. Imaginary parts: 3a+b=133a+b=13. Eliminating: 3×(first)−(second)3\times(\text{first})-(\text{second}) gives 8b=−168b=-16, so b=−2b=-2 and a=5a=5. Check: 5−2i+3i(5+2i)=5−2i+15i−6=−1+13i5-2i+3i(5+2i)=5-2i+15i-6=-1+13i.

HSC practice · Conjugates and division

Slide 10
2021 HSC Q11(e)Let z=5+iz=5+i and w=2−4iw=2-4i. Express z‾w\dfrac{\conj{z}}{w} in the form x+iyx+iy.
Show answerHide answer
Answer
710+910 i\dfrac{7}{10}+\dfrac{9}{10}\,i
Why
5−i2−4i×2+4i2+4i=10+20i−2i+420=14+18i20\frac{5-i}{2-4i}\times\frac{2+4i}{2+4i}=\frac{10+20i-2i+4}{20}=\frac{14+18i}{20}.
Exam practiceLet z=1+3iz=1+3i and w=2−iw=2-i. Find, in the form x+iyx+iy: (i) z‾ w\conj{z}\,w; (ii) zw\dfrac{z}{w}.
Show answerHide answer
Answer
(i) −1−7i-1-7i. (ii) −15+75 i-\dfrac15+\dfrac75\,i.
Why
(i) (1−3i)(2−i)=2−i−6i+3i2=−1−7i(1-3i)(2-i)=2-i-6i+3i^2=-1-7i. (ii) (1+3i)(2+i)5=2+i+6i−35=−1+7i5\frac{(1+3i)(2+i)}{5}=\frac{2+i+6i-3}{5}=\frac{-1+7i}{5}.

HSC practice · Equating parts

Slide 11
Adapted from 2024 HSC Q11(b)Let z=2+3iz=2+3i and w=1−5iw=1-5i. Find in the form x+iyx+iy: (i) z+w‾z+\conj{w}; (ii) z2z^2; (iii) zw\dfrac{z}{w}.
Show answerHide answer
Answer
(i) 3+8i3+8i. (ii) −5+12i-5+12i. (iii) −12+12 i-\dfrac12+\dfrac12\,i.
Why
(i) w‾=1+5i\conj{w}=1+5i, so z+w‾=3+8iz+\conj{w}=3+8i. (ii) z2=4+12i+9i2=−5+12iz^2=4+12i+9i^2=-5+12i. (iii) (2+3i)(1+5i)∣w∣2=2+10i+3i−1526=−13+13i26\frac{(2+3i)(1+5i)}{|w|^2}=\frac{2+10i+3i-15}{26}=\frac{-13+13i}{26}.
Adapted from 2024 HSC Q12(c)Find the complex number zz such that ∣z∣=z+8+12i|z|=z+8+12i.
Show answerHide answer
Answer
z=5−12iz=5-12i
Why
∣z∣|z| is real, so the imaginary part of the right-hand side must be zero. Let z=a+ibz=a+ib: then b+12=0b+12=0, so b=−12b=-12. The real parts give a2+144=a+8\sqrt{a^2+144}=a+8, so a2+144=a2+16a+64a^2+144=a^2+16a+64 and a=5a=5. Check: a+8=13>0a+8=13>0 and ∣5−12i∣=13|5-12i|=13.

Exit ticket

Slide 12
1. Express 51−2i\dfrac{5}{1-2i} in the form a+iba+ib.
Show answerHide answer
Answer
1+2i1+2i
Why
5(1+2i)(1−2i)(1+2i)=5(1+2i)5=1+2i\frac{5(1+2i)}{(1-2i)(1+2i)}=\frac{5(1+2i)}{5}=1+2i.
2. Find real xx and yy such that (x+iy)(2−3i)=−13i(x+iy)(2-3i)=-13i.
Show answerHide answer
Answer
x=3x=3, y=−2y=-2
Why
Expanding: (2x+3y)+i(2y−3x)=−13i(2x+3y)+i(2y-3x)=-13i, so 2x+3y=02x+3y=0 and 2y−3x=−132y-3x=-13, giving x=3x=3, y=−2y=-2. Alternatively divide: x+iy=−13i(2+3i)13=−i(2+3i)=3−2ix+iy=\frac{-13i(2+3i)}{13}=-i(2+3i)=3-2i.