Lecture 4/12: Polar Form and Euler’s Formulamaths.nuigalway.ie/~niall/teaching/Archive/0607/MA154/04PolarForm.pdf · Lecture 4/12: Polar Form and Euler’s Formula MA154: Algebra

Lecture 4/12: Polar Form and Euler’sFormula

MA154: Algebra for 1st Year IT

Niall Madden

[email protected]

25 Jan 2007

CS457 — Lecture 4/12: Polar Form and Euler’s Formula 1/17

Outline

1 Recall... Polar coordinates

2 Euler’s formula.A little Historycos x = ...

3 Inverse, Products, and Quotients

4 Binomial Coefficients.

5 Some combinatorics

6 De Moivre’s Theorem

6 De Moivre’s Theorem

6 Applications of de Moirve’s Theorem:


Recall... Polar coordinates

Given z ∈ C then r = |z | and the angle θ that the line fromthe origin (0, 0) to the complex number z makes with thepositive x-axis are called the Polar Coordinates of z .

z = (r cos θ, r sin θ) = r (cos θ + i sin θ) is called the polarform of the complex number z = x + i y .

Define arg z = {θ : z = r (cos θ + i sin θ)} for z 6= 0.Any θ ∈ arg z is called an argument of z .

If θ1, θ2 ∈ arg z then θ2 = θ1 + 2nπ for some integer n ∈ Z.

The principal value Arg z of arg z is defined as the uniquevalue θ ∈ arg z with −π < θ 6 π.So arg z = {Arg z + 2nπ : n ∈ Z}.

Arg z = π ⇔ z is a negative real number.



















































Example:

z = 1 + i in polar form is:



To find θ in general: If z 6= 0, then

y

x=

r sin θ

r cos θ=

sin θ

cos θ= tanθ.

Hence arg z ⊆ tan−1 yx where tan−1 u = {v ∈ R : tan v = u},

u ∈ R.



Example:

Express z = 2 − 2i in polar form.


Euler’s formula.

For θ ∈ R define Euler’s1 Formula:

exp(i θ) = e i θ = cos θ + i sin θ

So can write the polar form of z 6= 0 as

z = r (cos θ + i sin θ) = r e i θ.

Exercise: Show that ddθ(e i θ) = d

dθ(cos θ + i sin θ)

1Leonhard Euler, 1707–1783


Euler’s formula. A little History

“Euler’s” formula is actually due to Roger Cotes in 1714. His formwas

log(cos x + i sin x) = ix .

Euler published the equation in its current form in 1748.Neither of them used the geometrical interpretation of the formulathat we have used. That was first given 50 years later by CasparWessel.Cotes is also famous for the “Newton-Cotes” methods forestimating definite integrals, the most famous examples of whichare the Trapezium Rule and Simpson’s Rule.


Euler’s formula. cos x = ...

We can use Euler’s formula to show that

cos(x) =1

2

(e i x + e−ix

),

as follows:

Exer: Show that

sin x =1

2

(e i x − e−ix

).


Euler’s formula. cos x = ...

We can use Euler’s formula to show that

cos(x) =1

2

(e i x + e−ix

),

as follows:

Exer: Show that

sin x =1

2

(e i x − e−ix

).


Inverse, Products, and Quotients

Suppose z = re i θ, z1 = r1ei θ1 and z2 = r2e

i θ2 . Then

z−1 =1

z=

1

re i θ=

1

re− i θ,

z1z2 = r1r2 e i (θ1+θ2)

and

z1

z2=

r1r2

e i (θ1−θ2).

Multiplication in polar form is...:


Inverse, Products, and Quotients

Suppose z = re i θ, z1 = r1ei θ1 and z2 = r2e

i θ2 . Then

z−1 =1

z=

1

re i θ=

1

re− i θ,

z1z2 = r1r2 e i (θ1+θ2)

and

z1

z2=

r1r2

e i (θ1−θ2).

Multiplication in polar form is...:


Binomial Coefficients.

The Binomial Theorem:

(a+b)n = an+

(n

1

)an−1b+

(n

2

)an−2b2+· · ·+

(n

n − 1

)abn−1+bn

More compactly:

(a + b)n =

n∑k=0

(n

k

)an−kbk ,

Here the Binomial Coefficient(nk

)(“n choose k”) is(

n

k

)=

n!

k! (n − k)!,

and n! (“n factorial”) is n! = n · (n − 1) . . . 3 · 2 · 1.


Binomial Coefficients.

The Binomial Theorem:

(a+b)n = an+

(n

1

)an−1b+

(n

2

)an−2b2+· · ·+

(n

n − 1

)abn−1+bn

More compactly:

(a + b)n =

n∑k=0

(n

k

)an−kbk ,

Here the Binomial Coefficient(nk

)(“n choose k”) is(

n

k

)=

n!

k! (n − k)!,

and n! (“n factorial”) is n! = n · (n − 1) . . . 3 · 2 · 1.


Some combinatorics

Permutations: There are n!(n−k)! ways of picking and ordered

collection of k objects from n

Combinations: There are(nk

)= n!

k!(n−k)! ways of choosing,without order, k objects from n

Pascal’s Triangle:

(n

k

)=

(n − 1

k − 1

)+

(n − 1

k

).

Sum (a = b = 1):n∑

k=0

(n

k

)= 2n.


Some combinatorics




)= n!



(n

k

)=

(n − 1

k − 1

)+

(n − 1

k

).

Sum (a = b = 1):n∑

k=0

(n

k

)= 2n.


Some combinatorics




)= n!



(n

k

)=

(n − 1

k − 1

)+

(n − 1

k

).

Sum (a = b = 1):n∑

k=0

(n

k

)= 2n.


Some combinatorics




)= n!



(n

k

)=

(n − 1

k − 1

)+

(n − 1

k

).

Sum (a = b = 1):n∑

k=0

(n

k

)= 2n.


Some combinatorics

De Moivre’s Theorem2 is:

(cos θ + i sin θ)n = cos nθ + i sin nθ for all n ∈ Z.

Proof (1): Use Euler’s formula:

2Abraham de Moivre, 1667-1754


Some combinatorics

De Moivre’s Theorem2 is:

(cos θ + i sin θ)n = cos nθ + i sin nθ for all n ∈ Z.

Proof (1): Use Euler’s formula:

2Abraham de Moivre, 1667-1754


Some combinatorics

Proof (2): Exercise Use an inductive proof and the fact that

cosα cosβ − sinα sinβ = cos(α + β)

and

sinα cosβ + cosα sinβ = sin(α + β)


Applications of de Moirve’s Theorem:

De Moivre’s Theorem can be used to express cos or sin ofmultiple angles in terms of powers of cos θ and sin θ.Example: Show that

cos 3θ = 4 cos3 θ − 3 cos θ.

(Note: cos2 θ is short-hand for (cos θ)2)(Hint: Also need to use that cos2 θ = 1 − sin2 θ)



The theorem can be used to express powers of cos θ and sin θ interms of cos or sin of multiple angles.Example: Show that cos2 θ = 1

2(cos 2θ + 1).



Exercise: Show that 16 cos4 θ = 2 cos 4θ + 2 cos 2θ + 6.(Hint: use the previous identity for cos2 θ.


Documents

Lecture 4/12: Polar Form and Euler’s Formulamaths.nuigalway.ie/~niall/teaching/Archive/0607/MA154/04PolarForm.pdf · Lecture 4/12: Polar Form and Euler’s Formula MA154: Algebra