CorePure2 Chapter 1 :: Complex Numbers jfrost@tiffin.kingston.sch.uk

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Last modified: 6th August 2018

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OVERVIEW

𝑐𝑜𝑠 2𝜃 + 𝑖 𝑠𝑖𝑛 2𝜃 → 𝑒2𝑖𝜃

1 :: Exponential form of a complex number

2 :: Multiplying and dividing complex numbers

If 𝑧1 and 𝑧2 are two complex numbers, what happens to their moduli when we find 𝑧1𝑧2. What happens to their

arguments when we find 𝑧1

𝑧2?

3 :: De Moivre’s Theorem

If 𝑧 = 𝑟(cos 𝜃 + 𝑖 sin 𝜃), 𝑧𝑛 = 𝑟𝑛(cos 𝑛𝜃 + 𝑖 sin 𝑛𝜃)

4 :: De Moivre’s for Trigonometric Identities

“Express cos 3𝜃 in terms of powers of cos 𝜃”

6 :: Roots

“Solve 𝑧4 = 3 + 2𝑖”

5 :: Sums of series

“Given that 𝑧 = cos𝜋

𝑛+ 𝑖 sin

𝜋

𝑛,

where 𝑛 is a positive integer, show that 1 + 𝑧 + 𝑧2 +⋯+ 𝑧𝑛−1

= 1 + 𝑖 cot𝜋

2𝑛

”

Note for teachers: Most of this content is the old FP2, but sums of series is new.

Im[z]

Re[z]

𝜃

RECAP :: Modulus-Argument Form

If 𝒛 = 𝒙 + 𝒊𝒚 (and suppose in this case 𝑧 is in the first quadrant), what was:

𝑟 = 𝑧 = 𝒙𝟐 + 𝒚𝟐

𝜃 = arg 𝑧 = 𝐭𝐚𝐧−𝟏𝒚

𝒙

(𝑥, 𝑦)

Then in terms of 𝑟 and 𝜃: 𝒙 = 𝒓 𝐜𝐨𝐬𝜽 𝒚 = 𝒓 𝐬𝐢𝐧𝜽 𝒛 = 𝒙 + 𝒊𝒚 = 𝒓 𝒄𝒐𝒔 𝜽 + 𝒊 𝒓𝒔𝒊𝒏 𝜽 = 𝒓 𝐜𝐨𝐬𝜽 + 𝒊 𝐬𝐢𝐧𝜽

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This is known as the modulus-argument form of 𝑧. ?

modulus

argument

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? When −𝜋 < 𝜃 < 𝜋 it is known as the principal argument. ?

RECAP :: Modulus-Argument Form

𝒙 + 𝒊𝒚 𝒓 𝜽 Mod-arg form

−1 1 𝜋 𝑧 = cos 𝜋 + 𝑖 sin 𝜋

𝑖 1 𝜋

2 𝑧 = cos

𝜋

2+ 𝑖 sin

𝜋

2

1 + 𝑖 2 𝜋

4 𝑧 = 2 cos

𝜋

4+ 𝑖 sin

𝜋

4

− 3 + 𝑖 2 5𝜋

6 𝑧 = 2 cos

5𝜋

6+ 𝑖 sin

5𝜋

6

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Exponential Form

We’ve seen the Cartesian form a complex number 𝑧 = 𝑥 + 𝑦𝑖 and the modulus-argument form 𝑧 = 𝑟(cos 𝜃 + 𝑖 sin 𝜃). But wait, there’s a third form! In the later chapter on Taylor expansions, you’ll see that you that you can write functions as an infinitely long polynomial:

cos 𝑥 = 1 −𝑥2

2! +

𝑥4

4! −

𝑥6

6! + ⋯

sin 𝑥 = 𝑥 −𝑥3

3! +

𝑥5

5! − ⋯

𝑒𝑥 = 1 + 𝑥 +𝑥2

2!+𝑥3

3!+𝑥4

4!+𝑥5

5!+𝑥6

6!+

It looks like the cos 𝑥 and sin 𝑥 somehow add to give 𝑒𝑥. The one problem is that the signs don’t quite match up. But 𝑖 changes sign as we raise it to higher powers.

𝑒i𝜃 = 1 + 𝑖𝜃 +𝑖𝜃 2

2!+𝑖𝜃 3

3!+𝑖𝜃 4

4!+ ⋯

= 1 + 𝑖𝜃 −𝜃2

2!−𝑖𝜃3

3!+𝜃4

4!+ ⋯

= 1 −𝜃2

2!+𝜃4

4!−𝜃6

6!+ ⋯ + 𝑖 𝜃 −

𝜃3

3!+𝜃5

5!− ⋯

= cos 𝜃 + 𝑖 sin 𝜃 Holy smokes Batman!

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Sub i𝜃 into 𝑒𝑥

Exponential Form

! Exponential form

𝑧 = 𝑟𝑒𝑖𝜃

You need to be able to convert to and from exponential form.

𝒙 + 𝒊𝒚 Mod-arg form Exp Form

−1 𝑧 = cos 𝜋 + 𝑖 sin 𝜋 𝑧 = 𝑒𝑖𝜋

2 − 3𝑖 𝑧 = 13𝑒−0.983𝑖

2 cos𝜋

10+ 𝑖 sin

𝜋

10 𝑧 = 2𝑒

𝜋𝑖10

−1 + 𝑖 2 cos

3𝜋

4+ 𝑖 sin

3𝜋

4 𝑧 = 2𝑒

3𝜋𝑖4

2 cos3𝜋

5+ 𝑖 sin

3𝜋

5 𝑧 = 2𝑒

23𝜋𝑖5

𝒆𝒊𝝅 + 𝟏 = 𝟎 This is Euler’s identity. It relates the five most fundamental constants in maths!

Notice this is not a principal argument.

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? ? To get Cartesian form, put in modulus-argument form first.

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A Final Example

Use 𝑒𝑖𝜃 = cos 𝜃 + 𝑖 sin 𝜃 to show that cos 𝜃 =1

2(𝑒𝑖𝜃 + 𝑒−𝑖𝜃)

𝑒𝑖𝜃 = cos 𝜃 + 𝑖 sin 𝜃 𝑒−𝑖𝜃 = cos −𝜃 + 𝑖 sin −𝜃 = cos 𝜃 − 𝑖 sin 𝜃 ∴ 𝑒𝑖𝜃 + 𝑒−𝑖𝜃 = 2 cos 𝜃

∴ cos 𝜃 =1

2(𝑒𝑖𝜃 + 𝑒−𝑖𝜃)

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Exercise 1A

Pearson Core Pure Year 2 Page 5

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Multiplying and Dividing Complex Numbers

If 𝑧1 = 𝑟1 cos 𝜃1 + 𝑖 sin 𝜃1 and 𝑧2 = 𝑟2 cos 𝜃2 + 𝑖 sin 𝜃2 Then:

𝑧1𝑧2 = 𝒓𝟏𝒓𝟐(𝐜𝐨𝐬 𝜽𝟏 + 𝜽𝟐 + 𝒊 𝐬𝐢𝐧(𝜽𝟏 + 𝜽𝟐))

If you multiply two complex numbers, you multiply the moduli and add the arguments, and if you divide them, you divide the moduli and subtract the arguments.

Similarly if 𝑧1 = 𝑟1𝑒𝑖𝜃1 and 𝑧2 = 𝑟2𝑒

𝑖𝜃2 Then:

𝑧1𝑧2 = 𝒓𝟏𝒓𝟐𝒆𝒊 𝜽𝟏+𝜽𝟐

𝑧1𝑧2=𝒓𝟏𝒓𝟐𝒆𝒊 𝜽𝟏−𝜽𝟐

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Examples

3 cos5𝜋

12+ 𝑖 sin

5𝜋

12× 4 cos

𝜋

12+ 𝑖 sin

𝜋

12

= 𝟏𝟐 𝐜𝐨𝐬𝝅

𝟐+ 𝒊 𝒔𝒊𝒏

𝝅

𝟐

= 𝟏𝟐 𝟎 + 𝒊 𝟏 = 𝟏𝟐𝒊

2 cos𝜋

15+ 𝑖 sin

𝜋

15× 3 cos

2𝜋

5− 𝑖 sin

2𝜋

5

= 2 cos𝜋

15+ 𝑖 sin

𝜋

15× 3 cos −

2𝜋

5+ 𝑖 sin −

2𝜋

5

= 𝟔 𝐜𝐨𝐬 −𝝅

𝟑+ 𝒊 𝒔𝒊𝒏 −

𝝅

𝟑

= 𝟑 − 𝟑 𝟑 𝒊

2 cos𝜋12 + 𝑖 sin

𝜋12

2 cos5𝜋6 + 𝑖 sin

5𝜋6

= 𝟐 𝐜𝐨𝐬 −𝟑𝝅

𝟒+ 𝒊 𝐬𝐢𝐧 −

𝟑𝝅

𝟒

= 𝟐 𝒆−𝟑𝝅𝒊𝟒

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Write in the form 𝑟𝑒𝑖𝜃:

1

2 cos𝜃 − 𝑖 sin 𝜃 can be written as cos −𝜃 + 𝑖 sin(−𝜃)

3

Test Your Understanding

𝑧 = 5 3 − 5𝑖 Find (a) |𝑧| (1) (b) arg (𝑧) in terms of 𝜋 (2)

𝑤 = 2 cos𝜋

4+ 𝑖 sin

𝜋

4

Find

(c) 𝑤

𝑧 (1)

(d) arg𝑤

𝑧 (2)

Edexcel FP2(Old) June 2013 Q2

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Exercise 1B

Pearson Core Pure Year 2 Pages 7-8

De Moivre’s Theorem

We saw that: 𝑧1𝑧2 = 𝒓𝟏𝒓𝟐(𝐜𝐨𝐬 𝜽𝟏 + 𝜽𝟐 + 𝒊 𝐬𝐢𝐧(𝜽𝟏 + 𝜽𝟐))

Can you think therefore what 𝑧𝑛 is going to be?

Prove by induction that 𝑧𝑛 = 𝑟𝑛 cos 𝑛𝜃 + 𝑖 sin 𝑛𝜃 They so went there...

Edexcel FP2(Old) June 2013 Q4

? Secret Alternative Way ?

! If 𝑧 = 𝑟 𝑐𝑜𝑠 𝜃 + 𝑖 𝑠𝑖𝑛 𝜃 𝒛𝒏 = 𝒓𝒏 𝐜𝐨𝐬𝒏𝜽 + 𝒊 𝐬𝐢𝐧𝒏𝜽 This is known as De Moivre’s Theorem. ?

De Moivre’s Theorem for Exponential Form

If 𝑧 = 𝑟𝑒𝑖𝜃 then 𝒛𝒏 = 𝒓𝒆𝒊𝜽𝒏= 𝒓𝒏𝒆𝒊𝒏𝜽 ?

Examples

Simplify cos

9𝜋

17+𝑖 sin

9𝜋

17

5

cos2𝜋

17 − 𝑖 sin

2𝜋

17

3

=𝒄𝒐𝒔

𝟗𝝅𝟏𝟕+ 𝒊 𝒔𝒊𝒏

𝟗𝝅𝟏𝟕

𝟓

𝒄𝒐𝒔 −𝟐𝝅𝟏𝟕

+ 𝒊 𝒔𝒊𝒏 −𝟐𝝅𝟏𝟕

𝟑

=𝐜𝐨𝐬

𝟒𝟓𝝅𝟏𝟕

+ 𝒊 𝐬𝐢𝐧𝟒𝟓𝝅𝟏𝟕

𝒄𝒐𝒔 −𝟔𝝅𝟏𝟕

+ 𝒊 𝒔𝒊𝒏 −𝟔𝝅𝟏𝟕

= 𝐜𝐨𝐬𝟑𝝅 + 𝒊 𝐬𝐢𝐧𝟑𝝅

= 𝐜𝐨𝐬𝝅 + 𝒊 𝐬𝐢𝐧𝝅 = −𝟏

Apply De Moivres

- sinx = sin (-x)

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𝑧1𝑧2=𝒓𝟏𝒓𝟐 (𝐜𝐨𝐬 𝜽𝟏 − 𝜽𝟐 + 𝒊 𝐬𝐢𝐧(𝜽𝟏 − 𝜽𝟐)) ?

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Examples

Express 1 + 3 𝑖7

in the form 𝑥 + 𝑖𝑦 where 𝑥, 𝑦 ∈ ℝ.

𝟏 + 𝟑 𝒊 = 𝟐(𝐜𝐨𝐬𝝅

𝟑+ 𝒊 𝐬𝐢𝐧

𝝅

𝟑)

Therefore 𝟏 + 𝟑 𝒊𝟕= 𝟐𝟕 𝐜𝐨𝐬

𝟕𝝅

𝟑+ 𝒊 𝐬𝐢𝐧

𝟕𝝅

𝟑

= 𝟏𝟐𝟖 𝐜𝐨𝐬𝝅

𝟑+ 𝒊 𝐬𝐢𝐧

𝝅

𝟑

= 𝟔𝟒 + 𝟔𝟒 𝟑 𝒊

𝟏

𝟑 ?

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Edexcel FP2(Old) June 2010 Q4

Test Your Understanding

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Exercise 1C

Pearson Core Pure Year 2 Pages 10-11

Applications of de Moivre #1: Trig identities

Express cos 3𝜃 in terms of powers of cos 𝜃

Is there someway we could use de Moivre’s theorem to get a cos 3𝜃 term?

𝐜𝐨𝐬𝜽 + 𝒊 𝐬𝐢𝐧𝜽 𝟑 = 𝐜𝐨𝐬𝟑𝜽 + 𝒊 𝐬𝐢𝐧𝟑𝜽 Could we use the LHS to get another expression? Use a Binomial expansion!

cos 𝜃 + 𝑖 sin 𝜃 3 = 𝟏𝐜𝐨𝐬𝟑 𝜽 + 𝟑𝐜𝐨𝐬𝟐 𝜽 𝒊 𝐬𝐢𝐧𝜽 + 𝟑 𝐜𝐨𝐬𝜽 𝒊 𝐬𝐢𝐧𝜽 𝟐 + 𝒊 𝐬𝐢𝐧𝜽 𝟑 = 𝐜𝐨𝐬𝟑 𝜽 + 𝟑𝒊 𝐜𝐨𝐬𝟐 𝜽 𝐬𝐢𝐧𝜽 − 𝟑 𝐜𝐨𝐬𝜽 𝐬𝐢𝐧𝟐 𝜽 − 𝒊 𝐬𝐢𝐧𝟑 𝜽

How could we then compare the two expressions? Equating real parts:

𝐜𝐨𝐬𝟑𝜽 = 𝐜𝐨𝐬𝟑 𝜽 − 𝟑𝐜𝐨𝐬𝜽 𝐬𝐢𝐧𝟐 𝜽 = ⋯ = 𝟒𝐜𝐨𝐬𝟑 𝜽 − 𝟑𝐜𝐨𝐬𝜽

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Applications of de Moivre #1: Trig identities

Express cos 6𝜃 in terms of cos 𝜃.

cos 𝜃 + 𝑖 sin 𝜃 6 = cos 6𝜃 + 𝑖 sin 6𝜃 = cos6 𝜃 + 6 cos5 𝜃 𝑖 sin 𝜃 + ⋯ = cos6 𝜃 + 6𝑖 cos5 𝜃 sin 𝜃 − 15 cos4 𝜃 sin2 𝜃 − 20𝑖 cos3 𝜃 sin3 𝜃 + 15 cos2 𝜃 sin4 𝜃

+ 6𝑖 cos 𝜃 sin5 𝜃 − sin6 𝜃 Equating real parts:

cos 6𝜃 = cos6 𝜃 − 15 cos4 𝜃 sin2 𝜃 + 15 cos2 𝜃 sin4 𝜃 − sin6 𝜃 = 32 cos6 𝜃 − 48 cos4 𝜃 + 18 cos2 𝜃 − 1

Equating imaginary parts: (and simplifying)

sin 6𝜃 = 32 cos5 𝜃 − 32 cos3 𝜃 + 6 cos 𝜃

Test Your Understanding

(a) Use de Moivre’s theorem to show that (5) sin 5𝜃 = 16 sin5 𝜃 − 20 sin3 𝜃 + 5 sin 𝜃

Hence, given also that sin 3𝜃 = 3 sin 𝜃 − 4 sin3 𝜃

(b) Find all the solutions of sin 5𝜃 = 5 sin 3𝜃

in the interval 0 ≤ 𝜃 < 2𝜋. Give your answers to 3 decimal places. (6)

Edexcel FP2(Old) June 2011 Q7

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Applications of de Moivre #1: Trig identities

Finding identities for sin𝑛 𝜃 and cos𝑛 𝜃

The technique we’ve seen allows us to write say cos 3𝜃 in terms of powers of cos 𝜃 (e.g. cos3 𝜃). Is it possible to do the opposite, to say express cos3 𝜃 in terms of a linear combination of cos 3𝜃 and cos 𝜃 (with no powers)?

If 𝑧 = cos 𝜃 + 𝑖 sin 𝜃,

𝒛𝒏 = 𝐜𝐨𝐬𝒏𝜽 + 𝒊 𝐬𝐢𝐧𝒏𝜽 𝒛−𝒏 = 𝐜𝐨𝐬𝒏𝜽 − 𝒊 𝐬𝐢𝐧𝒏𝜽

∴ 𝒛𝒏 + 𝒛−𝒏 = 𝟐𝐜𝐨𝐬𝒏𝜽 𝒛𝒏− 𝒛−𝒏 = 𝟐𝒊𝒔𝒊𝒏𝒏𝜽

! Results you need to use (but you don’t need to memorise)

𝑧 +1

𝑧= 2 cos 𝜃 𝑧𝑛 +

1

𝑧𝑛= 2 cos 𝑛𝜃

𝑧 −1

𝑧= 2𝑖 sin 𝜃 𝑧𝑛 −

1

𝑧𝑛= 2𝑖 sin 𝑛𝜃

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Adding or subtracting gives:

In exponential form: 𝑐𝑜𝑠𝑛𝜃 =1

2𝑒𝑖𝑛𝜃 + 𝑒−𝑖𝑛𝜃

sin𝑛𝜃 =1

2𝑖𝑒𝑖𝑛𝜃 − 𝑒−𝑖𝑛𝜃

Finding identities for sin𝑛 𝜃 and cos𝑛 𝜃

It follows that, for example

(𝑧 +1

𝑧)𝑛 = 2 cos 𝜃 𝑛

Hence:

cosn 𝜃 =1

2n( 𝑧 +

1

𝑧)𝑛

! Results you need to use (but you don’t need to memorise)

𝑧 +1

𝑧= 2 cos 𝜃 𝑧𝑛 +

1

𝑧𝑛= 2 cos 𝑛𝜃

𝑧 −1

𝑧= 2𝑖 sin 𝜃 𝑧𝑛 −

1

𝑧𝑛= 2𝑖 sin 𝑛𝜃

We now use binomial to expand the RHS

Finding identities for sin𝑛 𝜃 and cos𝑛 𝜃

Express cos5 𝜃 in the form 𝑎 cos 5𝜃 + 𝑏 cos 3𝜃 + 𝑐 cos 𝜃

𝑧 +1

𝑧= 2 cos 𝜃 𝑧𝑛 +

1

𝑧𝑛= 2 cos 𝑛𝜃

𝑧 −1

𝑧= 2𝑖 sin 𝜃 𝑧𝑛 −

1

𝑧𝑛= 2𝑖 sin 𝑛𝜃

2 cos 𝜃 5 = 𝑧 +1

𝑧

5

32 cos5 𝜃 = 𝑧5 + 5𝑧3 + 10𝑧 +10

𝑧+5

𝑧3+1

𝑧5

32 cos5 𝜃 = 𝑧5 +1

𝑧5+ 5 𝑧3 +

1

𝑧3+ 10 𝑧 +

1

𝑧

= 2 cos 5𝜃 + 5(2 cos 3𝜃) + 10(2 cos 𝜃)

Therefore cos5 𝜃 =1

16cos 5𝜃 +

5

16cos 3𝜃 +

5

8cos 𝜃

Notice how the powers descend by 2 each time.

Starting point?

Using identities below.

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Finding identities for sin𝑛 𝜃 and cos𝑛 𝜃

Prove that sin3 𝜃 = −1

4sin 3𝜃 +

3

4sin 𝜃

𝑧 +1

𝑧= 2 cos 𝜃 𝑧𝑛 +

1

𝑧𝑛= 2 cos 𝑛𝜃

𝑧 −1

𝑧= 2𝑖 sin 𝜃 𝑧𝑛 −

1

𝑧𝑛= 2𝑖 sin 𝑛𝜃

2𝑖 sin 𝜃 3 = 𝑧 −1

𝑧

3

−8𝑖 sin3 𝜃 = 𝑧3 − 3𝑧 +3

𝑧−1

𝑧3

= 𝑧3 −1

𝑧3− 3 𝑧 −

1

𝑧

= 2𝑖 sin 3𝜃 − 3(2𝑖 sin 𝜃) Therefore dividing by −8𝑖:

sin3 𝜃 = −1

4sin 3𝜃 +

3

4sin 𝜃

Speed Tip: Remember that terms in such an expansion oscillate between positive and negative.

Again using identities.

Starting point?

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Test Your Understanding

(a) Express sin4 𝜃 in the form 𝑎 cos 4𝜃 + 𝑏 cos 2𝜃 + 𝑐

(b) Hence find the exact value of sin4 𝜃 𝑑𝜃𝜋

20

2𝑖 sin 𝜃 4 = 𝑧 −1

𝑧

4

16 sin4 𝜃 = 𝑧4 − 4𝑧2 + 6 −4

𝑧2+1

𝑧4

= 𝑧4 +1

𝑧4− 4 𝑧2 +

1

𝑧2+ 6

= 2 cos 4𝜃 − 4 2 cos 2𝜃 + 6

∴ sin4 𝜃 =1

8cos 4𝜃 −

1

2cos 2𝜃 +

3

8

sin4 𝜃 𝑑𝜃

𝜋2

0

=1

32sin 4𝜃 −

1

4sin 2𝜃 +

3

8𝜃0

𝜋2

=3𝜋

16

𝑧 +1

𝑧= 2 cos 𝜃 𝑧𝑛 +

1

𝑧𝑛= 2 cos 𝑛𝜃

𝑧 −1

𝑧= 2𝑖 sin 𝜃 𝑧𝑛 −

1

𝑧𝑛= 2𝑖 sin 𝑛𝜃

Starting point?

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Exercise 1D

Pearson Core Pure Year 2 Pages 14-15

Sums of Series

The formula for the sum of a geometric series also applies to complex numbers:

For 𝑤, 𝑧 ∈ ℂ,

Σ𝑟=0𝑛−1𝑤𝑧𝑟 = 𝑤 + 𝑤𝑧 + 𝑤𝑧2 +⋯+𝑤𝑧𝑛−1 =

𝑤 𝑧𝑛 − 1

𝑧 − 1

Σ𝑟=0∞ 𝑤𝑧𝑟 = 𝑤 + 𝑤𝑧 + 𝑤𝑧2 +⋯+𝑤𝑧𝑛−1 +⋯ =

𝑤

𝑧−1 provided 𝑧 < 1

No need to write these down. These are just 𝑆𝑛 and 𝑆∞ for geometric series.

Side Note: When we covered series before, | … | was understood to give the ‘unsigned value’ of a quantity. But this can just be thought of as the 1D version of finding the magnitude of a vector/complex number. e.g. −4 = 4 because -4 has a distance of 4 from 0 on a number line just as 34

= 5 because (3,4) has a distance of 5 from the origin.

Example

[Textbook] Given that 𝑧 = cos𝜋

𝑛+ 𝑖 sin

𝜋

𝑛, where 𝑛 is a positive integer, show that

1 + 𝑧 + 𝑧2 +⋯+ 𝑧𝑛−1 = 1 + 𝑖 cot𝜋

2𝑛

IMPORTANT: One of our earlier identities:

𝑧𝑛 − 𝑧−𝑛 = 2𝑖 sin 𝑛𝜃 Thus if we had an expression

of the form 𝑒𝑖𝜃 − 1, we could

cleverly factorise out 𝑒𝑖𝜃

2 (i.e. half the power) to get

𝑒𝑖𝜃

2 𝑒𝑖𝜃

2 − 𝑒−𝑖𝜃

2

→ 𝑒𝑖𝜃2 2𝑖 sin

𝜃

2

Thus where 𝑒𝑖𝜃 − 1 occurs in a fraction, multiply numerator

and denominator by 𝑒−𝑖𝜃

2 so

that we have just 𝑒𝑖𝜃

2 − 𝑒−𝑖𝜃

2

1 + 𝑧 + 𝑧2 +⋯+ 𝑧𝑛−1 =𝑧𝑛 − 1

𝑧 − 1

=𝑒𝜋𝑖𝑛

𝑛

− 1

𝑒𝜋𝑖𝑛 − 1

=𝑒𝜋𝑖 − 1

𝑒𝜋𝑖𝑛 − 1

=−2

𝑒𝜋𝑖𝑛 − 1

=−2

𝑒𝜋𝑖2𝑛 𝑒

𝜋𝑖2𝑛 − 𝑒−

𝜋𝑖2𝑛

=−2𝑒−

𝜋𝑖2𝑛

𝑒𝜋𝑖2𝑛 − 𝑒−

𝜋𝑖2𝑛

=−2𝑒−

𝜋𝑖2𝑛

2𝑖 sin𝜋2𝑛

=−2𝑖𝑒−

𝜋𝑖2𝑛

2𝑖2 sin𝜋2𝑛

=𝑖𝑒−

𝜋𝑖2𝑛

sin𝜋2𝑛

=𝑖 cos −

𝜋2𝑛

+ 𝑖 sin −𝜋2𝑛

sin𝜋2𝑛

=𝑖 cos

𝜋2𝑛 − 𝑖 sin

𝜋2𝑛

sin𝜋2𝑛

=sin

𝜋2𝑛 + 𝑖 cos

𝜋2𝑛

sin𝜋2𝑛

= 1 + 𝑖 cot𝜋

2𝑛

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𝑒𝜋𝑖

𝑛 since 𝑧 = 𝑟𝑒𝑖θ

Let’s practise that hard bit…

If 𝒛 = 𝒄𝒐𝒔 𝜽 + 𝒊 𝒔𝒊𝒏 𝜽 = 𝒆𝒊𝜽

𝑧𝑛 − 𝑧−𝑛 = 2𝑖 sin 𝑛𝜃 Thus if we had an expression

of the form 𝑒𝑖𝜃 − 1, we could

cleverly factorise out 𝑒𝑖𝜃

2 (i.e. half the power) to get

𝑒𝑖𝜃

2 𝑒𝑖𝜃

2 − 𝑒−𝑖𝜃

2

→ 𝑒𝑖𝜃2 2𝑖 sin

𝜃

2

Thus where 𝑒𝑖𝜃 − 1 occurs in a fraction, multiply numerator

and denominator by 𝑒−𝑖𝜃

2 so

that we have just 𝑒𝑖𝜃

2 − 𝑒−𝑖𝜃

2

3

𝑒2𝑖𝜃 − 1=

𝟑

𝒆𝒊𝜽(𝒆𝒊𝜽 − 𝒆−𝒊𝜽)=

𝟑𝒆−𝒊𝜽

𝒆𝒊𝜽 − 𝒆−𝒊𝜽=𝟑𝒆−𝒊𝜽

𝟐𝒊 𝐬𝐢𝐧𝜽

4𝑒𝑖𝜃

𝑒4𝑖𝜃 − 1=

𝟒𝒆−𝒊𝜽

𝒆𝟐𝒊𝜽 − 𝒆−𝟐𝒊𝜽=

𝟒𝒆−𝒊𝜽

𝟐𝒊 𝐬𝐢𝐧𝟐𝜽

1

𝑒𝑖𝜃3 − 1

=𝒆−𝒊𝜽𝟔

𝒆𝒊𝜽𝟔 − 𝒆−

𝒊𝜽𝟔

=𝒆−𝒊𝜽𝟔

𝟐𝒊 𝐬𝐢𝐧𝜽𝟔

The 𝑛 is 2 because if

𝑧 = 𝑒𝑖𝜃 then 𝑧2 = 𝑒2𝑖𝜃

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Using mod-arg form to split summation

𝑒𝑖𝜃 + 𝑒2𝑖𝜃 + 𝑒3𝑖𝜃 +⋯+ 𝑒𝑛𝑖𝜃 is a geometric series,

∴ 𝑒𝑖𝜃 + 𝑒2𝑖𝜃 + 𝑒3𝑖𝜃 +⋯+ 𝑒𝑛𝑖𝜃 =𝑒𝑖𝜃 𝑒𝑛𝑖𝜃 − 1

𝑒𝑖𝜃 − 1

Converting each exponential term to modulus-argument form would allow us to consider the real and imaginary parts of the series separately:

𝑒𝑖𝜃 + 𝑒2𝑖𝜃 + 𝑒3𝑖𝜃 +⋯+ 𝑒𝑛𝑖𝜃 = cos𝜃 + 𝑖 sin 𝜃 + cos 2𝜃 + 𝑖 sin 2𝜃 + ⋯ = cos𝜃 + cos 2𝜃 + ⋯ + 𝑖 sin 𝜃 + sin 2𝜃 + ⋯

Thus cos 𝜃 + cos 2𝜃 + ⋯ is the real part of 𝑒𝑖𝜃 𝑒𝑛𝑖𝜃−1

𝑒𝑖𝜃−1 and

sin 𝜃 + sin 2𝜃 +⋯ the imaginary part.

Example

[Textbook] 𝑆 = 𝑒𝑖𝜃 + 𝑒2𝑖𝜃 + 𝑒3𝑖𝜃 +⋯+ 𝑒8𝑖𝜃, for 𝜃 ≠ 2𝑛𝜋, where 𝑛 is an integer.

(a) Show that 𝑆 =𝑒9𝑖𝜃2 sin 4𝜃

sin𝜃

2

Let 𝑃 = cos 𝜃 + cos 2𝜃 + cos 3𝜃 +⋯+ cos8𝜃 and 𝑄 = sin 𝜃 + sin 2𝜃 + ⋯+ sin 8𝜃

(b) Use your answer to part a to show that 𝑃 = cos9𝜃

2sin 4𝜃 𝑐𝑜𝑠𝑒𝑐

𝜃

2 and find similar

expressions for 𝑄 and 𝑄

𝑃

𝑒𝑖𝜃 + 𝑒2𝑖𝜃 + 𝑒3𝑖𝜃 +⋯+ 𝑒8𝑖𝜃 =𝑒𝑖𝜃 𝑒𝑖𝜃

8− 1

𝑒𝑖𝜃 − 1

=𝑒𝑖𝜃 𝑒𝑖𝜃

8− 1

𝑒𝑖𝜃 − 1=𝑒𝑖𝜃 𝑒8𝑖𝜃 − 1

𝑒𝑖𝜃 − 1

=𝑒𝑖𝜃2 𝑒8𝑖𝜃 − 1

𝑒𝑖𝜃2 − 𝑒−

𝑖𝜃2

=𝑒𝑖𝜃2 𝑒8𝑖𝜃 − 1

2𝑖 sin𝜃2

=𝑖𝑒𝑖𝜃2 𝑒8𝑖𝜃 − 1

2𝑖2 sin𝜃2

=𝑒9𝑖𝜃2 sin 4𝜃

sin𝜃2

We can again use the trick of multiplying

𝑒𝑖𝜃 − 1 by 𝑒−𝑖𝜃

2

𝑧𝑛 − 𝑧−𝑛 = 2𝑖 sin 𝑛𝜃

a

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Example

[Textbook] 𝑆 = 𝑒𝑖𝜃 + 𝑒2𝑖𝜃 + 𝑒3𝑖𝜃 +⋯+ 𝑒8𝑖𝜃, for 𝜃 ≠ 2𝑛𝜋, where 𝑛 is an integer.

(a) Show that 𝑆 =𝑒9𝑖𝜃2 sin 4𝜃

sin𝜃

2

Let 𝑃 = cos 𝜃 + cos 2𝜃 + cos 3𝜃 +⋯+ cos8𝜃 and 𝑄 = sin 𝜃 + sin 2𝜃 + ⋯+ sin 8𝜃

(b) Use your answer to part a to show that 𝑃 = cos9𝜃

2sin 4𝜃 𝑐𝑜𝑠𝑒𝑐

𝜃

2 and find similar

expressions for 𝑄 and 𝑄

𝑃

First note that 𝑆 =𝑒9𝑖𝜃2 sin 4𝜃

sin𝜃

2

= 𝑒9𝑖𝜃

2 sin 4𝜃 𝑐𝑜𝑠𝑒𝑐𝜃

2

But also, by expressing each term within 𝑆 = 𝑒𝑖𝜃 + 𝑒2𝜃 … in modulus-argument form, we have 𝑆 = 𝑃 + 𝑖𝑄

∴ 𝑃 = 𝑅𝑒 𝑆 = 𝑅𝑒 𝑒9𝑖𝜃2 sin 4𝜃 𝑐𝑜𝑠𝑒𝑐

𝜃

2

= cos9𝜃

2sin 4𝜃 𝑐𝑜𝑠𝑒𝑐

𝜃

2

Similarly:

𝑄 = 𝐼𝑚 𝑆 = sin9𝜃

2sin 4𝜃 𝑐𝑜𝑠𝑒𝑐

𝜃

2

𝑄

𝑃=sin9𝜃2sin 4𝜃 𝑐𝑜𝑠𝑒𝑐

𝜃2

cos9𝜃2sin 4𝜃 𝑐𝑜𝑠𝑒𝑐

𝜃2

= tan9𝜃

2

b

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Example

Part b in notes

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Exercise 1E

Pearson Core Pure Year 2 Pages 18-19

Applications of de Moivre #2: Roots

𝑧𝑛 = 𝑟𝑛 cos 𝑛𝜃 + 𝑖 sin 𝑛𝜃

We have used de Moivre’s theorem when 𝑛 is an integer but it also hold true when 𝑛 is a rational number. We can use it solve equations of the form 𝑧𝑛 = 𝜔, by considering the fact that the argument of a number is not unique. This is equivalent to finding the nth roots of 𝜔. The fundamental theorem of algebra holds true for complex numbers: Hence 𝑧𝑛 = 𝜔 has n distinct roots If 𝜔 = 1 we call these the roots of unity. Since the argument of a complex number is not unique we can say: If then

𝑧𝑛 = 𝑟𝑛 cos 𝑛𝜃 + 𝑖 sin 𝑛𝜃 𝑧 = 𝑟 cos( 𝜃 + 2𝑘п) + 𝑖 sin(𝜃 +2𝑘п

Applications of de Moivre #2: Roots

Solve 𝑧3 = 1

Begin by writing 1 in mod- arg form. 𝑧3 = 1 cos 0 + 𝑖 sin 0 = 1(cos 0 + 2𝑘𝜋 + 𝑖 sin 0 + 2𝑘𝜋

𝑧 = cos 2𝑘𝜋 + 𝑖 sin 2𝑘𝜋13

= cos2𝑘𝜋

3+ 𝑖 sin

2𝑘𝜋

3

𝑘 = 0, 𝑧1 = cos 0 + 𝑖 sin 0 = 1

𝑘 = 1, 𝑧2 = cos2𝜋

3+ 𝑖 sin

2𝜋

3= −

1

2+ 𝑖

3

2

𝑘 = 2, 𝑧3 = cos4𝜋

3+ 𝑖 sin

4𝜋

3= −

1

2− 𝑖

3

2

These are known as the ‘cube roots of unity’ (more on this in a bit).

2𝜋

3

2𝜋

3

2𝜋

3

𝑧2

𝑧1

𝑧3

Plot these roots on an Argand diagram.

1 is always a root – the rest are spread out equally distributed. We’ll see why we get this pattern on the next slide.

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Fundamental Theorem of Algebra

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Roots of Unity

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More on Roots of Unity

The solutions to 𝑧𝑛 = 1 are known as the “roots of unity” (1 is a “unit”). Why are they nicely spread out? 1 is clearly a root as 1𝑛 = 1.

In the method on the previous slide, we

ended up adding 2𝑘𝜋

𝑛 to the argument, but

left the modulus untouched. If 𝑛 = 3 then

this rotates the line 2𝜋

3= 120° each time.

When 𝑘 = 𝑛 then we would have rotated 2𝑛𝜋

𝑛= 2𝜋 so end up where we started.

Suppose this root is 𝜔 (Greek letter “omega”)…

𝜔 = cos2𝜋

3+ 𝑖 sin

2𝜋

3

Then by De Moivre’s, then 𝜔2 would be:

𝜔2 = cos2𝜋

3+ 𝑖 sin

2𝜋

3

2

= cos4𝜋

3+ 𝑖 sin

4𝜋

3

i.e. the next root! Therefore, the roots can be represented as 1,𝜔,𝜔2 . Since the resultant ‘vector’ is 0, then 1 + 𝜔 + 𝜔2 = 0. Formal Proof: Geometric series where 𝑎 = 1 and 𝑟 = 𝜔 and 𝑛 = 𝑛.

𝑆𝑛 =𝑎 𝑟𝑛 − 1

𝑟 − 1=𝜔𝑛 − 1

𝜔 − 1=1 − 1

𝜔 − 1= 0

2𝜋

3

2𝜋

3

2𝜋

3

𝑧2

𝑧1

𝑧3

𝒛𝟑 = 𝟏

! If 𝑧𝑛 = 1 and 𝜔 = 𝑒2𝜋𝑖

𝑛 then 1,𝜔, 𝜔2, … , 𝜔𝑛−1 are the roots and 1 + 𝜔 + 𝜔2 +⋯+𝜔𝑛−1 = 0

Applications of de Moivre #2: Roots

𝑧𝑛 = 𝑤

We can use a similar method when w is not equal to 1. Again, our first step is to write w in mod-arg form and consider multiples of the argument.

Example

[Textbook] Solve 𝑧4 = 2 + 2 3 𝑖

𝑧4 = 4 cos𝜋

3+ 𝑖 sin

𝜋

3

= 4 cos𝜋

3+ 2𝑘𝜋 + 𝑖 sin(

𝜋

3+ 2𝑘𝜋)

𝑧 = 2 cos𝜋

12+2𝑘𝜋

4+ 𝑖 sin

𝜋

12+2𝑘𝜋

4

𝑘 = 0, 𝑧 = 2 cos𝜋

12+ 𝑖 sin

𝜋

12

𝑘 = 1, 𝑧 = 2 cos7𝜋

12+ 𝑖 sin

7𝜋

12

𝑘 = 2, 𝑧 = 2 cos13𝜋

12+ 𝑖 sin

13𝜋

12

= 2 cos −11𝜋

12+ 𝑖 sin −

11𝜋

12

𝑘 = 3, 𝑧 = ⋯ We need principal roots. Alternative is to use 𝑘 = 0, 1, −1,−2 instead of 0, 1, 2, 3, which will give your principal roots directly.

Note: W is not equal to 1 in this case.

1. Write in mod-arg form considering multiple arguments

2. Apply de Moivre’s

3. Consider consecutive values of k to generate solutions

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Test Your Understanding

a) Express the complex number −2 + 2 3 𝑖 in the form 𝑟 cos 𝜃 + 𝑖 sin 𝜃 ,

− 𝜋 < 𝜃 ≤ 𝜋. (3) b) Solve the equation

𝑧4 = −2 + 2 3 i

giving the roots in the form 𝑟 cos 𝜃 + 𝑖 sin 𝜃 , −𝜋 < 𝜃 ≤ 𝜋. (5)

Edexcel FP2(Old) June 2012 Q3

You can avoid the previous large amount of working by just dividing your angle by the power (4), and adding/subtracting increments of 2𝜋 again divided by the power.

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Exercise 1F

Pearson Core Pure Year 2 Pages 24-25

Solving Geometric Problems

𝑧6 = 7 + 24𝑖

We have already seen in the previous exercise that the roots of an equation such as that on the left give evenly spaced points in the Argand diagram, because the

modulus remained the same but we kept adding 2𝜋

𝑛 to

the argument. These points formed a regular hexagon, and in general for 𝑧𝑛 = 𝑠, form a regular 𝑛-agon. Recall that 𝜔 is the first root of unity of 𝑧𝑛 = 1:

𝜔 = cos2𝜋

𝑛+ 𝑖 sin

2𝜋

𝑛.

This has modulus 1 and argument 2𝜋

𝑛.

Suppose 𝒛𝟏 was the first root of 𝑧6 = 7 + 24𝑖. Then consider the product 𝑧1𝜔. What happens? Multiplying these multiplies the moduli. But 𝝎 = 𝟏 therefore the modulus of 𝒛𝟏 is unaffected.

We also add the arguments. 𝐚𝐫𝐠 𝝎 =𝟐𝝅

𝒏 so

multiplying by 𝝎 rotates 𝒛𝟏 by this. We can see therefore that it gives us the next root, i.e. 𝒛𝟐 = 𝒛𝟏𝝎.

! If 𝑧1 is one root of the equation 𝑧6 = 𝑠, and 1, 𝜔, 𝜔2, … , 𝜔𝑛−1 are the 𝑛th roots of unity, then the roots of 𝑧𝑛 = 𝑠 are given by 𝑧1, 𝑧1𝜔, 𝑧1𝜔

2, … , 𝑧1𝜔𝑛−1.

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Example

[Textbook] The point 𝑃 3, 1 lies at one vertex of an equilateral triangle. The

centre of the triangle is at the origin. (a) Find the coordinates of the other vertices of the triangle. (b) Find the area of the triangle.

Convert 3 + 𝑖 to either exponential of modulus-argument form. We can then rotate the point 120° around origin (and again) by multiplying by 𝜔.

3 + 𝑖 = 2𝑒𝜋

6𝑖 and 𝜔 = 𝑒

2𝜋

3𝑖

Next vertex:

2𝑒𝜋

6𝑖 × 𝑒

2𝜋

3𝑖 = 2𝑒

5𝜋

6𝑖 = − 3 + 𝑖

Next vertex:

2𝑒5𝜋

6𝑖 × 𝑒

2𝜋

3𝑖 = 2𝑒

3𝜋

2𝑖 = 2𝑒−

𝜋

2𝑖 = −2𝑖

Area =1

2× 𝑏𝑎𝑠𝑒 × ℎ𝑒𝑖𝑔ℎ𝑡

=1

2× 2 3 × 3

= 3 3

(− 3, 1) ( 3, 1)

(0,−2)

𝑥

𝑦

a

b

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? Argand Diagram

Test your understanding

An equilateral triangle has its centroid located at the origin and a vertex at (1,0). What are the coordinates of the other two vertices?

z = cos2𝑘𝜋

3+ 𝑖𝑠𝑖𝑛

2𝑘𝜋

3

𝑘 = 1,2,3

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Exercise 1G

Pearson Core Pure Year 2 Pages 26-27