Further Maths: Year 13 : Chapter 1 : Pure

How can you prove de Moivre's theorem using Euler's relation ?

(r(cosθ + i sinθ))^n = (re^iθ)^n = r^n x e^inθ = r^n (cosnθ + isinnθ) Valid only for any integer exponent, n

if z₁=r₁e^iθ₁ and z₂=r₂e^iθ₂, then z₁/z₂ =

(r₁/r₂)e^(θ₁-θ₂)

Show the proof by induction method to prove de Moivre's theorem for positive integer exponents: Assumption step

Assume that de Moivre's theorem is true for n=k (r(cosθ+isinθ))^k = r^k(coskθ+isinkθ)

What is the sum of the series e^iθ+e^2iθ+e^3iθ+...+e^niθ, Sn=

Sn= e^iθ((e^niθ)-1)/((e^iθ)-1)

show the final exponential form of the series e^iθ+e^2iθ+e^3iθ+...+e^niθ, into modulus argument form

(cosθ+cos2θ+cos3θ+...+cosnθ)+ i(sinθ+sin2θ+sin3θ+...+sin nθ) =e^iθ((e^niθ)-1)/((e^iθ)-1))

Show the proof by induction method to prove de Moivre's theorem for positive integer exponents: Basic step

1. Basic step: n=1; LHS= (r(cosθ+isinθ))¹=r(cosθ+isinθ) RHS = r¹(cos1θ+isin1θ) =r(cosθ+isinθ) As LHS=RHS, de Moivre's theorem is true for n=1

if z=cosθ+i sinθ, then what is 1/z=

1/z=(cosθ+i sinθ)⁻¹ 1/z=(cos(-θ)+i sin(-θ)) 1/z=cosθ-i sinθ

e^iθ = cosθ + i sinθ : what is this known as ?

Euler's relation

Can we automatically assume the laws of indices work the same way with complex numbers as with real numbers ?

NO

What is the equation for the sum of an infinite geometric series S∞=

S∞=a/(1-r) with first term a and common ratio is r

Since we cannot assume laws of indices work the same way with complex numbers, what does this mean?

We can only apply these two properties: 1. r₁r₂e^i(θ₁+θ₂)=z₁z₂ 2.(r₁/r₂)e^(θ₁-θ₂) =z₁/z₂

if z₁ is one root of the equation zⁿ=s, and 1,ω, ω², ..., wⁿ⁻¹ are the nth roots of unity, then the roots of zⁿ =s are given by...

are given by z₁, z₁ω, z₁ω², z₁ωⁿ⁻¹

how do you write cosθ as infinite series of powers of theta?

cosθ = 1-(θ²/2!) +(θ⁴/4!)-(θ⁶/6!) + ....+((-1^r)θ^2r/(2r)!)+....

equate the real parts in : (cosθ+cos2θ+cos3θ+...+cosnθ)+ i(sinθ+sin2θ+sin3θ+...+sin nθ)=e^iθ((e^niθ)-1)/((e^iθ)-1))

cosθ+cos2θ+cos3θ+...+cosnθ=Re(e^iθ((e^niθ)-1)/((e^iθ)-1))

Euler's relation

e^iθ = cosθ + i sinθ

e^iθ =

e^iθ = cosθ + i sinθ

Show how you would convert the exponential form of the series e^iθ+e^2iθ+e^3iθ+...+e^niθ, into modulus argument form

e^iθ+e^2iθ+e^3iθ+...+e^niθ = (cosθ+i sinθ)+(cos2θ+i sin2θ)+(cos3θ+i sin3θ)+...+(cosnθ+i sin nθ) =(cosθ+cos2θ+cos3θ+...+cosnθ)+ i(sinθ+sin2θ+sin3θ+...+sin nθ)

How do you write e^x as a series expansion in powers of x?

e^x = 1+x + (x²/2!)+(x³/3!)+(x⁴/4!)+(x⁵/5!)+....+(x^r/r!)+.....

How can you prove de Moivre's theorem for positive integer exponents?

from the modulus argument form of a complex number using the addition formula for sin and cos

Show the proof by induction method to prove de Moivre's theorem for positive integer exponents: conclusion step

if de Moivre's theorem is true for n=k, then it has been shown to be true for n=k+1 As de Moivre's theorem is true for n=1, it is now proven to be true for all n ∈ Z⁺ by mathematical induction

To prove de Moivre's theorem for positive integer exponents, what method is used?

proof by induction

if z₁=r₁e^iθ₁ and z₂=r₂e^iθ₂, then z₁z₂=

r₁r₂e^i(θ₁+θ₂)

how do you write sinθ as infinite series of powers of theta?

sinθ = θ-(θ³/3!) +(θ⁵/5!)-(θ⁷/7!) + ....+((-1^r)(θ^2r+1)/(2r+1)!)+......

equate the imaginary parts in :(cosθ+cos2θ+cos3θ+...+cosnθ)+ i(sinθ+sin2θ+sin3θ+...+sin nθ)=e^iθ((e^niθ)-1)/((e^iθ)-1))

sinθ+sin2θ+sin3θ+...+sin nθ=Im(e^iθ((e^niθ)-1)/((e^iθ)-1))

if z and w are non-zero complex numbers an n is a positive integer, then

the equation zⁿ =w has n distinct solutions

the nth roots of any complex numbers s lie

the nth roots of any complex numbers s lie on the vertices of a regular n-gon with its centre at the origin

Show the proof by induction method to prove de Moivre's theorem for positive integer exponents: Inductive step

when n=k+1, (r(cosθ+isinθ))^k+1 =(r(cosθ+isinθ))^k x r(cosθ+isinθ) =r^k(coskθ+isinkθ) x r(cosθ+isinθ) = r^k+1(coskθ+isinkθ)(cosθ+isinθ) =r^k+1((coskθcosθ-sinkθsinθ)+i(sinkθcosθ+coskθsinθ)) =r^k+1(cos(kθ+θ)+isin(kθ+θ)) =r^k+1(cos((k+1)θ)+isin((k+1)θ)) Therefore, de Moivre's theorem is true when n=k+1

in general, the solutions to Zⁿ =1 are

z = cos [2kπ /n] + i sin[2kπ/n] = e∧2πi/n such that the nth roots of unity are 1,ω, ω², ..., wⁿ⁻¹ 1,ω, ω², ..., wⁿ⁻¹ form the vertices of a regular n-gon 1+ ω, ω², ..., wⁿ⁻¹ = 0

for any complex number z = r [ cos θ + i sinθ ] you can write

z = r[cos [θ+2kπ] + i sin[θ+2kπ]] where k is an integer

if z=cosθ+i sinθ, what is z+(1/z)=

z+(1/z)=cosθ+i sinθ+cosθ-i sinθ so: z+(1/z)=2 cosθ

if z=cosθ+i sinθ, what is z-(1/z)=

z-(1/z)=cosθ+i sinθ-(cosθ-i sinθ) z-(1/z)=2 i sinθ

You can use the modulus- argument form of a complex number to express it in the exponential form which is what?

z= re^(iθ)

if z=cosθ+i sinθ, what is zⁿ+(1/zⁿ) =

zⁿ+(1/zⁿ) =cos nθ+i sin nθ+ cos nθ-i sin nθ =2 cos nθ

if z=cosθ+i sinθ, what is zⁿ-(1/zⁿ) =

zⁿ-(1/zⁿ) =cos nθ +i sin nθ-(cos nθ- i sin nθ) zⁿ-(1/zⁿ) =2 i sin nθ

for w, z∈C, what is ∑w(z^r) = (With n-1 on top and 0 on the bottom)

∑w(z^r) = w +wz+wz²+...+wz^(n-1) so ∑w(z^r) =w((z^n)-1)/(z-1)

for w, z∈C, what is ∑w(z^r) = (With ∞ on top and 0 on the bottom)

∑w(z^r) =w+wz+wz²+... ∑w(z^r) =w/(1-z), for |z|<1