Lesson 6 · Multiplying and dividing in polar form

Learning intentionTo multiply, divide and find powers of complex numbers in polar form, and to prove the modulus and argument identities.

Success criteria · I can…

prove that moduli multiply and arguments add
multiply, divide and find powers in polar form, giving the principal argument
prove and use $|z_1z_2|=|z_1||z_2|$ , $\arg(z_1z_2)=\arg z_1+\arg z_2$ and the quotient and power rules
use two forms of the same number to find exact trig values

ResourcesLesson slides
Practice worksheet
Worked solutions

Lesson flow

LaunchWarm-up · Retrieval from Lesson 5Slide 2 ExamplesExample 1 · Proving the multiplication ruleSlide 3 Key ideasMultiply moduli, add argumentsSlide 4 ExamplesExample 2 · Multiplying in polar formSlide 5 ExamplesExample 3 · Dividing in polar formSlide 6 ExamplesExample 4 · Proving the identitiesSlide 7 ExamplesExample 5 · Two forms, exact valuesSlide 8 HSC-styleHSC practice · Powers in polar formSlide 9 HSC-styleHSC practice · Modulus and argument of a new numberSlide 10 Exit ticketExit ticketSlide 11

Warm-up · Retrieval from Lesson 5

Slide 2

(a) Write

1-i

in polar form.

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Answer

\sqrt2\left(\cos\left(-\dfrac\pi4\right)+i\sin\left(-\dfrac\pi4\right)\right)

Why

r=\sqrt2

. The point is in the fourth quadrant with reference angle

\frac\pi4

, so

\theta=-\frac\pi4

(b) State

\Arg(-2)

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Answer

\pi

(not

-\pi

)

Why

The principal argument lies in

(-\pi,\pi]

, which includes

\pi

and excludes

-\pi

(c) Expand

\cos(A+B)

and

\sin(A+B)

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Answer

\cos(A+B)=\cos A\cos B-\sin A\sin B

and

\sin(A+B)=\sin A\cos B+\cos A\sin B

Why

Keep these on the board: they are exactly what Example 1 needs.

(d) Write

2\left(\cos\dfrac{5\pi}{6}+i\sin\dfrac{5\pi}{6}\right)

x+iy

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Answer

-\sqrt3+i

Why

2\left(-\frac{\sqrt3}{2}+\frac12 i\right)=-\sqrt3+i

Example 1 · Proving the multiplication rule

Slide 3

Let $z_1=r_1(\cos\theta_1+i\sin\theta_1)$ and $z_2=r_2(\cos\theta_2+i\sin\theta_2)$ .

Prove that

z_1z_2=r_1r_2\big(\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)\big).

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Answer

\begin{aligned}z_1z_2&=r_1r_2(\cos\theta_1+i\sin\theta_1)(\cos\theta_2+i\sin\theta_2)\\&=r_1r_2\big[(\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2)+i(\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2)\big]\\&=r_1r_2\big(\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)\big).\end{aligned}

Why

Expand using

i^2=-1

, collect real and imaginary parts, then recognise the compound-angle formulas for

\cos(\theta_1+\theta_2)

and

\sin(\theta_1+\theta_2)

. So

|z_1z_2|=|z_1||z_2|

and

\arg(z_1z_2)=\arg z_1+\arg z_2

, up to a multiple of

2\pi

: multiply the moduli and add the arguments.

Multiply moduli, add arguments

Slide 4

Theorem

Modulus and argument rules

For non-zero complex numbers $z_1$ , $z_2$ , $z$ and an integer $n$ :

|z_1z_2|=|z_1||z_2|\qquad \arg(z_1z_2)=\arg z_1+\arg z_2

\left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}\qquad \arg\left(\frac{z_1}{z_2}\right)=\arg z_1-\arg z_2

|z^n|=|z|^n\qquad \arg(z^n)=n\arg z

The argument rules hold for $\arg$ (any argument). For the principal argument $\Arg$ the answer may need $\pm2\pi$ to land in $(-\pi,\pi]$ (Example 2 shows this). For $n=-1$ the quotient rule with $z_1=1$ gives $\left|\frac1z\right|=\frac1{|z|}$ and $\arg\frac1z=-\arg z$ . The full proof for all integers $n$ is de Moivre's theorem, later in this unit.

Think

In words: what does multiplying

z

r\cis\theta

do to the point

z

on the Argand diagram?

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Answer

It multiplies the distance from

O

r

(a dilation about

O

) and adds

\theta

to the argument (a rotation about

O

through

\theta

Why

This is the geometric meaning of “multiply the moduli, add the arguments”. It is developed fully in Lesson 8.

Example 2 · Multiplying in polar form

Slide 5

Express in polar form with the principal argument.

(a)

4\cis\dfrac{3\pi}{4}\times 3\cis\dfrac{\pi}{2}

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Answer

12\cis\left(-\dfrac{3\pi}{4}\right)

Why

Multiply moduli:

4\times3=12

. Add arguments:

\frac{3\pi}{4}+\frac{\pi}{2}=\frac{5\pi}{4}

. This is more than

\pi

, so subtract

2\pi

to get

-\frac{3\pi}{4}

(b)

\left(2\cis\dfrac{2\pi}{3}\right)^2

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Answer

4\cis\left(-\dfrac{2\pi}{3}\right)

Why

2^2=4

and

\frac{2\pi}{3}+\frac{2\pi}{3}=\frac{4\pi}{3}

, which is outside

(-\pi,\pi]

, so subtract

2\pi

to get

-\frac{2\pi}{3}

Show diagramHide diagram

Part (a): the two factors. Their product has modulus 12 and argument 5π/4, which is off this scale (lower left, at −3π/4).

Example 3 · Dividing in polar form

Slide 6

Express in polar form with the principal argument.

(a)

\dfrac{12\cis 0}{6\cis\frac{2\pi}{3}}

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Answer

2\cis\left(-\dfrac{2\pi}{3}\right)=2\left(\cos\dfrac{2\pi}{3}-i\sin\dfrac{2\pi}{3}\right)=-1-\sqrt3\,i

Why

Divide moduli:

12\div6=2

. Subtract arguments:

0-\frac{2\pi}{3}=-\frac{2\pi}{3}

, already in

(-\pi,\pi]

(b)

\dfrac{6\cis\left(-\frac{\pi}{3}\right)}{2\cis\frac{2\pi}{3}}

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Answer

3\cis\pi=-3

Why

6\div2=3

and

-\frac{\pi}{3}-\frac{2\pi}{3}=-\pi

. The principal argument lies in

(-\pi,\pi]

, so write

\pi

instead of

-\pi

. A negative real number has principal argument

\pi

Example 4 · Proving the identities

Slide 7

Let $z_1=r_1\cis\theta_1$ and $z_2=r_2\cis\theta_2$ , with $z_2\ne0$ .

(a) Prove that

\left|\dfrac{z_1}{z_2}\right|=\dfrac{|z_1|}{|z_2|}

and

\arg\left(\dfrac{z_1}{z_2}\right)=\arg z_1-\arg z_2

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Answer

Let

w=\dfrac{z_1}{z_2}

, so

z_1=wz_2

. By the product rule,

|z_1|=|w||z_2|

and

\arg z_1=\arg w+\arg z_2

. Rearranging gives

|w|=\dfrac{|z_1|}{|z_2|}

and

\arg w=\arg z_1-\arg z_2

Why

Use the product rule you have already proved, applied to

z_1=wz_2

. (Alternatively, multiply top and bottom by

\cos\theta_2-i\sin\theta_2

and simplify.)

(b) Prove that

\arg\conj{z}=-\arg z

and

\conj{z_1z_2}=\conj{z_1}\,\conj{z_2}

using polar form.

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Answer

z=r\cis\theta

then

\conj{z}=r(\cos\theta-i\sin\theta)=r(\cos(-\theta)+i\sin(-\theta))

, so

\arg\conj{z}=-\theta

. Then

\conj{z_1}\,\conj{z_2}=r_1\cis(-\theta_1)\cdot r_2\cis(-\theta_2)=r_1r_2\cis(-(\theta_1+\theta_2))=\conj{z_1z_2}

Why

Cosine is even and sine is odd. The conjugate of a product has modulus

r_1r_2

and argument

-(\theta_1+\theta_2)

, which is exactly the product of the conjugates.

Example 5 · Two forms, exact values

Slide 8

Let $z_1=1+\sqrt3\,i$ and $z_2=1+i$ .

(a) Write

z_1

and

z_2

in polar form.

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Answer

z_1=2\cis\dfrac{\pi}{3}

and

z_2=\sqrt2\cis\dfrac{\pi}{4}

Why

|z_1|=\sqrt{1+3}=2

with

\tan\theta=\sqrt3

|z_2|=\sqrt2

with

\tan\theta=1

. Both are in the first quadrant.

(b) Find

z_1z_2

and

\dfrac{z_1}{z_2}

in polar form and in Cartesian form.

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Answer

z_1z_2=2\sqrt2\cis\dfrac{7\pi}{12}=(1-\sqrt3)+(1+\sqrt3)i

\dfrac{z_1}{z_2}=\sqrt2\cis\dfrac{\pi}{12}=\dfrac{(1+\sqrt3)+(\sqrt3-1)i}{2}

Why

Polar: multiply moduli and add arguments (

\frac{\pi}{3}+\frac{\pi}{4}=\frac{7\pi}{12}

), or divide moduli and subtract (

\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}

). Cartesian: expand

(1+\sqrt3 i)(1+i)

, and for the quotient multiply top and bottom by

1-i

(c) Hence find the exact values of

\sin\dfrac{7\pi}{12}

and

\cos\dfrac{\pi}{12}

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Answer

\sin\dfrac{7\pi}{12}=\cos\dfrac{\pi}{12}=\dfrac{\sqrt2+\sqrt6}{4}

Why

Equate imaginary parts of the two forms of

z_1z_2

2\sqrt2\sin\frac{7\pi}{12}=1+\sqrt3

, so

\sin\frac{7\pi}{12}=\frac{1+\sqrt3}{2\sqrt2}=\frac{\sqrt2+\sqrt6}{4}

. Equate real parts of

\frac{z_1}{z_2}

\sqrt2\cos\frac{\pi}{12}=\frac{1+\sqrt3}{2}

, so

\cos\frac{\pi}{12}=\frac{\sqrt2+\sqrt6}{4}

Show diagramHide diagram

The product z1z2 sits at angle 7π/12 (105°), the sum of the arguments 60° and 45°.

HSC practice · Powers in polar form

Slide 9

Adapted from 2024 HSC Q11(e)Let

z=\sqrt3+i

. (i) Write

z

in polar form. (ii) Hence express

z^7

in the form

x+iy

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Answer

(i)

z=2\cis\dfrac{\pi}{6}

. (ii)

z^7=128\cis\dfrac{7\pi}{6}=-64\sqrt3-64i

Why

\arg(z^n)=n\arg z

and

|z^n|=|z|^n

, so

z^7=2^7\cis\frac{7\pi}{6}=128\left(-\frac{\sqrt3}{2}-\frac{1}{2}i\right)

Adapted from 2022 HSC Q11(c)(ii)Let

z=-\sqrt3+i

. Use polar form to find

z^{10}

in the form

x+iy

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Answer

z=2\cis\dfrac{5\pi}{6}

and

z^{10}=1024\cis\dfrac{25\pi}{3}=1024\cis\dfrac{\pi}{3}=512+512\sqrt3\,i

Why

10\times\frac{5\pi}{6}=\frac{25\pi}{3}=8\pi+\frac{\pi}{3}

. Remove whole turns of

2\pi

, then use

\cos\frac{\pi}{3}=\frac12

and

\sin\frac{\pi}{3}=\frac{\sqrt3}{2}

. The rule

\arg(z^n)=n\arg z

is enough here; de Moivre's theorem is proved by induction later in this unit.

HSC practice · Modulus and argument of a new number

Slide 10

Adapted from 2020 HSC Q4Let

z

be a non-zero complex number and

w=\dfrac{z^2}{|z|}

. Find

|w|

and

\arg w

in terms of

|z|

and

\arg z

. Describe where

w

lies compared with

z

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Answer

|w|=|z|

and

\arg w=2\arg z

. So

w

is on the same circle

|w|=|z|

about

O

z

, at double the angle.

Why

|w|=\dfrac{|z|^2}{|z|}=|z|

. Dividing by the positive real number

|z|

does not change the argument (a positive real has argument

0

), so

\arg w=\arg(z^2)-0=2\arg z

Exit ticket

Slide 11

1. Use polar form to express

\dfrac{(1+i)^5}{(1-i)^3}

in the form

x+iy

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Answer

2

Why

1+i=\sqrt2\cis\frac{\pi}{4}

and

1-i=\sqrt2\cis\left(-\frac{\pi}{4}\right)

. The quotient is

\dfrac{(\sqrt2)^5\cis\frac{5\pi}{4}}{(\sqrt2)^3\cis\left(-\frac{3\pi}{4}\right)}=2\cis\left(\frac{5\pi}{4}+\frac{3\pi}{4}\right)=2\cis2\pi=2

2. Prove that

|z^n|=|z|^n

and

\arg(z^n)=n\arg z

for

n=2

and

n=3

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Answer

With

z=r\cis\theta

z^2=z\cdot z=r^2\cis2\theta

, and

z^3=z^2\cdot z=r^3\cis3\theta

Why

Apply the product rule once for

n=2

and again for

n=3

. The general case is proved by induction in Complex Numbers II (de Moivre's theorem).