1

∎ 𝑎 + 𝑏𝑖 , 𝑎𝜖𝑅, 𝑏𝜖𝑅 𝑖 = √−1

∎ 𝑧 = 𝑎 + 𝑏𝑖 𝑎 = 𝑅𝑒(𝑧), 𝑏 = 𝐼𝑚(𝑧)

∎ 𝑔𝑖𝑣𝑒𝑛 𝑧 = 𝑎 + 𝑏𝑖 & 𝑤 = 𝑐 + 𝑑𝑖, 𝑎 + 𝑏𝑖 = 𝑐 + 𝑑𝑖 ⟺ 𝑎 = 𝑐 & 𝑏 = 𝑑

∎ 𝑇ℎ𝑒 𝑐𝑜𝑚𝑝𝑙𝑒𝑥 𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑜𝑓 𝑧 = 𝑎 + 𝑏𝑖 𝑖𝑠 𝑧∗ = 𝑎 − 𝑏𝑖

∎ 𝑇ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑐𝑜𝑚𝑝𝑙𝑒𝑥 𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒𝑠 𝑖𝑠 𝒓𝒆𝒂𝒍: 𝑧 + 𝑧∗ = 2𝑎

∎ 𝑇ℎ𝑒 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑐𝑜𝑚𝑝𝑙𝑒𝑥 𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒𝑠 𝑖𝑠 𝒓𝒆𝒂𝒍: 𝑧 𝑧∗ = 𝑎2 + 𝑏2

Properties of complex conjugates:

● 𝑅𝑒(𝑧) =1

2(𝑧 + 𝑧∗) 𝐼𝑚(𝑧) =

1

2𝑖(𝑧 − 𝑧∗)

● (𝑧1 ± 𝑧2)∗ = 𝑧1

∗ ± 𝑧2∗

● (𝑧∗)∗ = 𝑧

● (𝑧1𝑧2)∗ = 𝑧1

∗ × 𝑧2∗ & (

𝑧1𝑧2)∗

=𝑧1∗

𝑧2∗ , 𝑧2 ≠ 0

● (𝑧𝑛)∗ = (𝑧∗)𝑛

Presentation of complex number in Cartesian and polar coordinate system

𝑧 = 𝑥 + 𝑦𝑖⏟ 𝐶𝑎𝑟𝑡𝑒𝑠𝑖𝑎𝑛 𝑓𝑜𝑟𝑚

= 𝑟(𝑐𝑜𝑠 𝜃 + 𝑖 𝑠𝑖𝑛 𝜃)⏟ 𝑝𝑜𝑙𝑎𝑟 𝑓𝑜𝑟𝑚

𝑚𝑜𝑑𝑢𝑙𝑢𝑠−𝑎𝑟𝑔𝑢𝑚𝑒𝑛𝑡 𝑓𝑜𝑟𝑚

= 𝑟𝑒𝑖𝜃⏟𝐸𝑢𝑙𝑒𝑟 𝑓𝑜𝑟𝑚

= 𝑟 𝑐𝑖𝑠 𝜃

Argan plane is the complex plane

▪ 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑟 𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑎𝑙𝑢𝑒: |𝑧| = 𝑟 = √𝑥2 + 𝑦2

▪ 𝐴𝑟𝑔𝑢𝑚𝑒𝑛𝑡: 𝑎𝑟𝑔 𝑧 = 𝜃 = 𝑎𝑟𝑐 𝑡𝑎𝑛 𝑦

𝑥 𝑏𝑒 𝑐𝑎𝑟𝑒𝑓𝑢𝑙

▪ 𝑥 = 𝑟𝑐𝑜𝑠 𝜃, 𝑦 = 𝑟𝑠𝑖𝑛 𝜃

Recall that the argument should be

measured in radians

Very useful for fast conversions from

Cartesian into Euler form

it will give you visually position of the point,

and therefore quadrant for the angle

𝑧 = −𝑖 → 𝑧 = 𝑒𝑖3𝜋2

– 2 𝑖 = 2𝑒𝑖3𝜋2

𝑖 = 𝑒𝑖𝜋2

1 = 𝑒𝑖0

2

Practicality of Euler’s form

𝑧1 = 𝑥1 + 𝑖𝑦1 = 𝑟1(cos 𝜃1 + 𝑖 sin 𝜃1) & 𝑧2 = 𝑥2 + 𝑖𝑦2 = 𝑟2(cos 𝜃2 + 𝑖 sin 𝜃2)

Product is:

𝑧1𝑧2 = (𝑥1 + 𝑖𝑦1)(𝑥2 + 𝑖𝑦2) = 𝑟1𝑟2(cos 𝜃1 + 𝑖 sin 𝜃1)(cos 𝜃2 + 𝑖 sin 𝜃2)

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

= 𝑟1𝑟2[cos(𝜃1 + 𝜃2) + 𝑖 sin(𝜃1 + 𝜃2)]

∴ |𝒛𝟏𝒛𝟐| = |𝒛𝟏||𝒛𝟐| 𝒎𝒐𝒅𝒖𝒍𝒖𝒔 𝒐𝒇 𝒑𝒓𝒐𝒅𝒖𝒄𝒕 𝒊𝒔 𝒑𝒓𝒐𝒅𝒖𝒄𝒕 𝒐𝒇 𝒎𝒐𝒅𝒖𝒍𝒖𝒔

& 𝒂𝒓𝒈 (𝒛𝟏𝒛𝟐) = 𝒂𝒓𝒈 (𝒛𝟏) + 𝒂𝒓𝒈 (𝒛𝟐) 𝒂𝒓𝒈𝒎𝒆𝒏𝒕 𝒐𝒇 𝒑𝒓𝒐𝒅𝒖𝒄𝒕 𝒊𝒔 𝒔𝒖𝒎 𝒐𝒇 𝒂𝒓𝒈𝒖𝒎𝒆𝒏𝒕𝒔

▪ 𝒛𝟏𝒛𝟐 = [|𝒛𝟏|𝒆𝒊𝜽𝟏] [|𝒛𝟐|𝒆

𝒊𝜽𝟐] = |𝒛𝟏||𝒛𝟐|𝒆𝒊(𝜽𝟏+𝜽𝟐)

Quotient is:

𝑧1 𝑧2= (𝑥1 + 𝑖𝑦1)

(𝑥2 + 𝑖𝑦2)=𝑟1𝑟2

cos 𝜃1 + 𝑖 sin 𝜃1cos 𝜃2 + 𝑖 sin 𝜃2

=𝑟1𝑟2

cos 𝜃1 + 𝑖 sin 𝜃1cos 𝜃2 + 𝑖 sin 𝜃2

×cos 𝜃2 − 𝑖 sin 𝜃2cos 𝜃2 − 𝑖 sin 𝜃2

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

cos2 𝜃2 + sin2 𝜃2

=𝑟1𝑟2 [cos( 𝜃1−𝜃2) + 𝑖 sin( 𝜃1−𝜃2)]

∴ |𝒛𝟏𝒛𝟐| =

|𝒛𝟏|

|𝒛𝟐| 𝒎𝒐𝒅𝒖𝒍𝒖𝒔 𝒐𝒇 𝒒𝒖𝒐𝒕𝒊𝒆𝒏𝒕 𝒊𝒔 𝒒𝒖𝒐𝒕𝒊𝒆𝒏𝒕 𝒐𝒇 𝒎𝒐𝒅𝒖𝒍𝒖𝒔

& 𝒂𝒓𝒈 (𝒛𝟏𝒛𝟐) = 𝒂𝒓𝒈 (𝒛𝟏) − 𝒂𝒓𝒈 (𝒛𝟐) 𝒂𝒓𝒈𝒎𝒆𝒏𝒕 𝒐𝒇 𝒒𝒖𝒐𝒕𝒊𝒆𝒏𝒕 𝒊𝒔 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝒐𝒇 𝒂𝒓𝒈𝒖𝒎𝒆𝒏𝒕𝒔

▪ 𝒛𝟏 𝒛𝟐

= [|𝒛𝟏|𝒆

𝒊𝜽𝟏]

[|𝒛𝟐|𝒆𝒊𝜽𝟐]

=| 𝒛𝟏|

| 𝒛𝟐|𝒆𝒊(𝜽𝟏−𝜽𝟐)

Conclusion: Euler form of complex numbers follows ordinary algebra: 𝑎𝑛𝑎𝑚 = 𝑎𝑛+𝑚

𝐼𝑡 𝑖𝑠 𝑒𝑎𝑠𝑦 𝑡𝑜 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑡𝑤𝑜 𝑐𝑜𝑚𝑝𝑙𝑒𝑥 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑛 𝑎𝑛𝑦 𝑓𝑜𝑟𝑚, 𝑏𝑢𝑡 𝑤ℎ𝑎𝑡 𝑖𝑓 𝑦𝑜𝑢 ℎ𝑎𝑣𝑒 10 𝑓𝑎𝑐𝑡𝑜𝑟𝑠 𝑜𝑟 𝑧25 𝑜𝑟 √𝑧25

Properties of modulus and argument

▪ |𝑧∗| = |𝑧| & 𝑎𝑟𝑔 (𝑧∗) = −𝑎𝑟𝑔 𝑧

▪ 𝑧𝑧∗ = |𝑧|2

▪ 𝑒𝑖𝜃 = 𝑒𝑖(𝜃+𝑘2𝜋) 𝑘𝜖𝑍

▪ 𝑧1 = 𝑧2 ⇔ |𝑧1| = |𝑧2| & 𝜃2 = 𝜃1 + 2𝑘𝜋 𝑘𝜖𝑍

3 De Moivre’s Theorem

𝑧𝑛 = [𝑟(𝑐𝑜𝑠𝜃 + 𝑖 𝑠𝑖𝑛𝜃)]𝑛 = 𝑟𝑛(𝑐𝑜𝑠𝜃 + 𝑖 𝑠𝑖𝑛𝜃)𝑛

𝑧𝑛 = [𝑟𝑒𝑖𝜃]𝑛= 𝑟𝑛𝑒𝑖𝑛𝜃 = 𝑟𝑛(𝑐𝑜𝑠 𝑛𝜃 + 𝑖 𝑠𝑖𝑛 𝑛𝜃)

∴ (𝒄𝒐𝒔 𝒏𝜽 + 𝒊 𝒔𝒊𝒏 𝒏𝜽) = (𝒄𝒐𝒔𝜽 + 𝒊 𝒔𝒊𝒏𝜽)𝒏

Application of DeMoivre’s Theorem

(𝑎 + 𝑏)𝑛 = ∑(𝑛𝑟) 𝑎𝑛−𝑟𝑏𝑟 = 𝑎𝑛 + (

𝑛1) 𝑎𝑛−1𝑏 +⋯+ (

𝑛𝑟) 𝑎𝑛−𝑟𝑏𝑟 +⋯+ 𝑏𝑛

𝑛

𝑟=0

(𝑛𝑟) =

𝑛(𝑛 − 1)(𝑛 − 2)⋯ (𝑛 − 𝑟 + 1)

𝑟!

= 𝑛!

𝑟! (𝑛 − 𝑟)!=

𝑛!

(𝑛 − 𝑟)! 𝑟!= (

𝑛𝑛 − 𝑟

) (𝑛0) ≡ 1

Certain trigonometric identities can be derived using DeMoivre’s theorem. We can for instance express cosn , sin n

and tan n in terms of cos , sin and tan .

Example:

We can find an expression for cos5 Re(cos5 sin5 )i

= 5

Re cos sini (using DeMoivre’s theorem)

Then: cos 5𝜃 = 𝑅𝑒 (𝑐𝑜𝑠𝜃 + 𝑖 𝑠𝑖𝑛𝜃)5

= 𝑅𝑒(𝑐𝑜𝑠5𝜃 + 5𝑖 𝑐𝑜𝑠4𝜃 𝑠𝑖𝑛𝜃 + 10𝑖2 𝑐𝑜𝑠3𝜃 𝑠𝑖𝑛2𝜃 + 10𝑖3 𝑐𝑜𝑠2𝜃 𝑠𝑖𝑛3𝜃 + 5𝑖4 cos 𝜃 𝑠𝑖𝑛4𝜃 + 𝑖5𝑠𝑖𝑛5𝜃)

= 𝑅𝑒(𝑐𝑜𝑠5𝜃 + 5𝑖 𝑐𝑜𝑠4𝜃 𝑠𝑖𝑛𝜃 + 10𝑖2 𝑐𝑜𝑠3𝜃 𝑠𝑖𝑛2𝜃 + 10𝑖3 𝑐𝑜𝑠2𝜃 𝑠𝑖𝑛3𝜃 + 5𝑖4 cos 𝜃 𝑠𝑖𝑛4𝜃 + 𝑖5𝑠𝑖𝑛5𝜃)

= 𝑐𝑜𝑠5𝜃 − 10 𝑐𝑜𝑠3𝜃 𝑠𝑖𝑛2𝜃 + 5 cos 𝜃 𝑠𝑖𝑛4𝜃

cos5 5 3 2 4cos 10cos sin 5cos sin . (*)

If required, the right hand side can be expressed entirely in terms of cos only. We get:

cos5 5 3 2 2 2cos 10cos (1 cos ) 5cos (1 cos )

= 5 3 5 3 5cos 10cos 10cos 5cos 10cos 5cos

cos5 5 316cos 20cos 5cos

Note 1: We can also get an identity for sin5 :

sin5𝜃 = 𝐼𝑚 (𝑐𝑜𝑠𝜃 + 𝑖 𝑠𝑖𝑛𝜃)5

4 = 𝐼𝑚(𝑐𝑜𝑠5𝜃 + 5𝑖 𝑐𝑜𝑠4𝜃 𝑠𝑖𝑛𝜃 + 10𝑖2 𝑐𝑜𝑠3𝜃 𝑠𝑖𝑛2𝜃 + 10𝑖3 𝑐𝑜𝑠2𝜃 𝑠𝑖𝑛3𝜃 + 5𝑖4 cos 𝜃 𝑠𝑖𝑛4𝜃 + 𝑖5𝑠𝑖𝑛5𝜃)

= 𝐼𝑚(𝑐𝑜𝑠5𝜃 + 5𝑖 𝑐𝑜𝑠4𝜃 𝑠𝑖𝑛𝜃 + 10𝑖2 𝑐𝑜𝑠3𝜃 𝑠𝑖𝑛2𝜃 + 10𝑖3 𝑐𝑜𝑠2𝜃 𝑠𝑖𝑛3𝜃 + 5𝑖4 cos 𝜃 𝑠𝑖𝑛4𝜃 + 𝑖5𝑠𝑖𝑛5𝜃)

= 5 𝑐𝑜𝑠4𝜃 𝑠𝑖𝑛𝜃 − 10 𝑐𝑜𝑠2𝜃 𝑠𝑖𝑛3𝜃 + 𝑠𝑖𝑛5𝜃

sin 5𝜃 = 5 𝑐𝑜𝑠4𝜃 𝑠𝑖𝑛𝜃 − 10 𝑐𝑜𝑠2𝜃 𝑠𝑖𝑛3𝜃 + 𝑠𝑖𝑛5𝜃 (**)

If required, the right hand side can be expressed entirely in terms of sin

sin5 2 2 2 3 55(1 sin ) sin 10(1 sin )sin sin

= 3 5 3 5 55sin 10sin 5sin 10sin 10sin sin

sin5 5 316sin 20sin 5sin

Note 2: We can also get an expression for tan5 by dividing equation (*) by equation (**):

4 2 3 5

5 3 2 4

sin5 5cos sin 10cos sin sintan5

cos5 cos 10cos sin 5cos sin

Dividing every term on the top and bottom by 5cos gives:

4 2 3 5

5 5 5

5 3 2 4

5 5 5

5cos sin 10cos sin sin

cos cos costan 5cos 10cos sin 5cos sin

cos cos cos

= 3 5

2 4

tan 10 tan tan

1 10 tan 5tan

Question:

a) Find an expression for cos4 in terms of cos only.

b) Find an expression for sin 4 in terms of sin only.

c) Show that

3

2 4

4 4tan 4

1 6

t t

t t

, where t = tanθ.

Questions:

1. Find an expression for cos6 in terms of cosc and sins

2. Find an expression for sin 7 in terms of sin only.

3. Find an expression for tan 7 in terms of tant .

5

Finding n-th root of a complex number z 𝑧 = |𝒛|𝑒𝑖𝜃 = |𝒛|𝑒𝑖(𝜃+𝑘2𝜋)

√𝑧𝑛

= √|𝑧|𝑛

𝑒𝑖 (𝜃+𝑘2𝜋)

𝑛 = √|𝑧|𝑛

{𝑐𝑜𝑠 (𝜃 + 𝑘2𝜋

𝑛) + 𝑖 𝑠𝑖𝑛 (

𝜃 + 𝑘2𝜋

𝑛) } 𝑘 = 0, 1, 2, 3… , 𝑛 − 1

There are exactly n nth roots of z.

Only the values k = 0, 1, …, n - 1 give different values of √𝑧𝑛

.

Geometrically, the n th roots are the vertices

of a regular polygon with n sides in Argan plane.

√17

Example: Find 5th root of 1. Or show that if 𝜔 = 𝑒2

5𝜋𝑖 , then the 5th roots of unity can be expressed as 2 3 41, , , ,

1 = 𝑒𝑖(0+𝑘2𝜋) 𝑘𝜖𝑁

√15

= 𝑒𝑖(0+𝑘2𝜋)

5 𝑘 = 0, 1, 2, 3, 4

So the 5th roots of unity are 𝑒0, 𝑒𝑖 2𝜋/5, 𝑒𝑖 4𝜋/5, 𝑒𝑖 6𝜋/5, 𝑒𝑖 8𝜋/5 (𝑟 = 1)

Example: Find the cube roots of 8 – 8i, i.e. find z such that z3 = 8 – 8i.

8 − 8𝑖 = [√128,−𝜋

4] = [√128,−

𝜋

4+ 2𝑘𝜋] (from Argan plane straight forward)

The cube roots of 8 – 8i are:

612 12

128(cos( ) sin( ))i

6 7 712 12

128(cos( ) sin( ))i

6 15 1512 12

128(cos( ) sin( ))i

The cube roots (to 3 sf) are:

2.17 - 0.581i

-0.581 + 2.17i

-1.59 – 1.59i

Note: The cube roots of 8 – 8i can be shown on an Argand diagram: -2

-1

0

1

2

3

1 2 3-1-2

1

1

-1

-1

6 REAL POLYNOMIALS are polynomials with real coefficients. Two polynomials are equal if and only if they have the same degree (order) and corresponding terms have equal coefficients. if 2x3 + 3x2 – 4x + 6 = ax3 + bx2 – cx + d then

a = 2, b = 3, c = – 4, d = 6

Remainder If P(x) is divided by D(x) until a remainder R is obtain:

𝑃(𝑥)

𝐷(𝑥)= 𝑄(𝑥) +

𝑅(𝑥)

𝐷(𝑥)

𝑃(𝑥) = 𝐷(𝑥)𝑄(𝑥) + 𝑅(𝑥)

When we divide by a polynomial of degree 1 ("ax+b") the remainder will have degree 0 (a constant)

The Remainder Theorem

When a polynomial P(x) is divided by (x – k) until a remainder R is obtain, then R = P(k):

proof:

𝑃(𝑥)

𝑥 − 𝑘= 𝑄(𝑥) +

𝑅

𝑥 − 𝑘

𝑃(𝑥) = 𝑄(𝑥)(𝑥 − 𝑘) + 𝑅

𝑃(𝑘) = 𝑅

The Factor Theorem

D(x) is the divisor

Q(x) is the quotient

R(x) is the remainder

7 k is a zero of P(x) ⇔ (x – k) is a factor of P(x)

proof: k is a zero of P(x) ⇔ P(k) = 0

⇔ R = 0

as 𝑃(𝑥) = 𝑄(𝑥)(𝑥 − 𝑘) + 𝑅 = 𝑄(𝑥)(𝑥 − 𝑘) =

⇔ (x – k) is a factor of P(x)

Corollar

(x – k) is a factor of P(x) ⇔ there exists a polynomial Q(x) such that P(x) = (x – k) Q(x)

8

P(x) is real and – 3 + i is zero. ∴ – 3 – i is zero

[𝑎𝑥3 + 9𝑥2 + 𝑎𝑥 − 30] = [𝑥 − (−3 + 𝑖)][𝑥 − (−3 − 𝑖)](𝑎𝑥 + 𝑏)

[𝑎𝑥3 + 9𝑥2 + 𝑎𝑥 − 30] = (𝑥2 + 6𝑥 + 10)(𝑎𝑥 + 𝑏) = 𝑎𝑥3 + (6𝑎 + 𝑏)𝑥2 + (10𝑎 + 6𝑏)𝑥 + 10𝑏

∴ 6𝑎 + 𝑏 = 9 & 10𝑏 = −30 10𝑎 + 6𝑏 = 𝑎

10𝑏 = −30 → 𝑏 = −3 10𝑎 + 6𝑏 = 𝑎 → 9𝑎 + 6𝑏 = 0 → 𝑎 = 2

𝑜𝑟 6𝑎 + 𝑏 = 9 → 𝑎 = 2

linear factor (ax+b) = 2x – 3 so zeroes are:

−3 ± 𝑖 𝑎𝑛𝑑 3

2

Polynomials: Sums and Products of Roots

𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 = 𝟎 → 𝒑, 𝒒 𝒓𝒐𝒐𝒕𝒔 𝒕𝒉𝒆𝒏 𝒑 + 𝒒 = −𝒃

𝒂 𝒑𝒒 =

𝒄

𝒂

(𝑥 − 𝑝)(𝑥 − 𝑞) = 0 → 𝑥2 − (𝑝 + 𝑞)𝑥 + 𝑝𝑞 = 0

𝑥2 +𝑏

𝑎𝑥 +

𝑐

𝑎= 0

𝒂 𝒙𝟑 + 𝒃 𝒙𝟐 + 𝒄𝒙 + 𝒅 = 𝟎 → 𝒑, 𝒒, 𝒓 𝒓𝒐𝒐𝒕𝒔 𝒑 + 𝒒 + 𝒓 = −𝒃

𝒂 𝒑𝒒𝒓 = −

𝒅

𝒂

(𝑥 − 𝑝)(𝑥 − 𝑞)(𝑥 − 𝑟) = 0 →

(𝑥 − 𝑝)(𝑥 − 𝑞)(𝑥 − 𝑟) = 0 → [ 𝑥2 − (𝑝 + 𝑞)𝑥 + 𝑝𝑞](𝑥 − 𝑟) = 0

𝑥3 − (𝑝 + 𝑞 + 𝑟) 𝑥2 + (𝑝𝑞 + 𝑝𝑟 + 𝑞𝑟)𝑥 − 𝑝𝑞𝑟(𝑥 − 𝑟) = 0

𝑥3 +𝑏

𝑎 𝑥2 +

𝑐

𝑎𝑥 +

𝑑

𝑎= 0

in general:

𝛽1, 𝛽2, 𝛽3, ⋯ , 𝛽𝑛 are roots of the equation

𝑎𝑛 𝑥𝑛 + 𝑎𝑛−1 𝑥

𝑛−1 + ⋯+ 𝑎1 𝑥1 + 𝑎0 = 0 (𝑎0 ≠ 0)

𝐬𝐮𝐦 𝐨𝐟 𝐭𝐡𝐞 𝐫𝐨𝐨𝐭𝐬 = −𝒔𝒆𝒄𝒐𝒏𝒅 𝒄𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕

𝒇𝒊𝒓𝒔𝒕 𝒄𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕= −

𝒂𝒏−𝟏

𝒂𝒏

𝐩𝐫𝐨𝐝𝐮𝐜𝐭 𝐨𝐟 𝐭𝐡𝐞 𝐫𝐨𝐨𝐭𝐬 = (−𝟏)𝒏𝒇𝒓𝒆𝒆 𝒕𝒆𝒓𝒎

𝒇𝒊𝒓𝒔𝒕 𝒄𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕= (−𝟏)𝒏

𝒂𝟎

𝒂𝒏