Proving conjugates and de moivre's theorem (1 Viewer)

[x.Exhaust.x]

How would you prove de moivre's theorem?

i.e. (costheta+isintheta)^n=cosntheta+isinntheta

Is it through induction, by following the steps and the induction hypothesis? If so, would anyone happen to have a clear proof on it, labelling the induction hypothesis?

How you prove clearly:

a) z1 - z2 =
______
z1 - z2

b)
______ __ __
(z1/z2) / z1 / z2

Exam tomorrow. Epic fail

 

[lyounamu]

How would you prove de moivre's theorem?

i.e. (costheta+isintheta)^n=cosntheta+isinntheta

Is it through induction, by following the steps and the induction hypothesis? If so, would anyone happen to have a clear proof on it, labelling the induction hypothesis?

How you prove clearly:

a) z1 - z2 =
______
z1 - z2

b)
______ __ __
(z1/z2) / z1 / z2

Exam tomorrow. Epic fail

Have you got 4 unit fitzy? It explains quite simply.

It just proves that cis@ x cis% = cis(@+%)

And that's why we prove De Moivre's theorem.

Yeah, you can use induction of course which is actually a better method.

By the way, I don't understand your questions as in "what are the questions?"

 

[Trebla]

How would you prove de moivre's theorem?

i.e. (costheta+isintheta)^n=cosntheta+isinntheta

Is it through induction, by following the steps and the induction hypothesis? If so, would anyone happen to have a clear proof on it, labelling the induction hypothesis?

How you prove clearly:

a) z1 - z2 =
______
z1 - z2

b)
______ __ __
(z1/z2) / z1 / z2

Exam tomorrow. Epic fail

For all of them, refer to a textbook. These are standard syllabus proofs. De Moivre's Theorem can be proved by induction and the other two, simply let z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂ then prove.

 

[gurmies]

z1 - z2 =
______
z1 - z2


That isn't true..unless you have z1 conj and z2 conj...

 

[Harkaraj]

Ye true, there seems something wrong with that, it should be z1 conj - z2 conj = (z1 - z2) conj

 

[pyrodude1031]

1.
a+ib METHOD

cj(z-w) = cj(a+ib-c-id) = cj[(a-c)+i(b-d)] = (a-c)-i(b-d) = (a-ib)- (c-id) = cj(a+ib) - cj(c+id) = cj(z) - cj(w)

cis METHOD

cj {z-w} = cj {acisA-bcisB} =cj {acosA-bcosB + i(asinA-bsinB)} = acosA-bcosB - i(asinA-bsinB) = (acosA-iasinA) - (bcosB-ibsinB) = acis(-A) - bcis(-B) = cj(acisA) - cj(bcisB) = cjz - cjw

2.
cj(z/w) = cj {(a+ib)/(c+id)} = cj { [(ac+bd)+i(bc-ad)]/(c^2+d^2)} = [(ac+bd)-i(bc-ad)]/(c^2+d^2)
cjz/cjw = (a-ib)/(c-id) = [(ac+bd)-i(bc-ad)]/(c^2+d^2) = above

or

cj(z/w) = cj (acisA/bcisB) = cj { (a/b) cis(A-B) = (a/b)cis(B-A) = (a/b) cis(-A - (-B)) = (acis(-A)) / (bcis(-B)) = cj(z) / cj(w)

where I use z = acisA ,w = bcisB
= a+ib = c+id

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