What's e^(ipi)? (1 Viewer)

[KeypadSDM]

It's mad, clearly

 

[milton]

when i first saw

, that was pretty much my reaction

 

[Riviet]

I've kind of accepted it after deriving the result myself.

 

[Mumma]

I like this proof ...

z = cosx + isinx

dz/dx = -sinx + icosx
= i(cosx+isinx)
= iz

dz/dx = iz
dz/z = i dx
ln[z] + c = xi

x = 0, z = 1
thus c = 0
xi = lnz
therefore z = cosx + isinx = e^(xi)

cos[pi] + isin[pi] = e^(i Pi) = -1

 

[haque]

Lol that's from euler's notation where rcis@=re^i@- we can prove this result with the power series for sine and cosine as well
e^x=1+x+x^2/2! +.....(they in fact got us to prove a variation in the 2005 hsc)
sinx=x-x^3/3!+x^5/5! +...
cosx=1-x^2/2!+x^4/4!-...
e^i@=1 +i@ -@^2/2!...
=(1-@^2/2! +@^4/4!-...) +i(@-@^3/3!+@^5/5!-..)=cos@+isin@
Too bad learning extra stuff didin't help for hsc lol

 

[haque]

I like this proof ...

z = cosx + isinx

dz/dx = -sinx + icosx
= i(cosx+isinx)
= iz

dz/dx = iz
dz/z = i dx
ln[z] + c = xi

x = 0, z = 1
thus c = 0
xi = lnz
therefore z = cosx + isinx = e^(xi)

cos[pi] + isin[pi] = e^(i Pi) = -1

u can prove more stuff through differentiaton e.g the fact that the acceleration of a particle in circular motion is always directed towards centre etc.

 

[acmilan]

Complex numbers are so versatile. You can use partial fractions, derivatives and complex numbers to prove real properties like int 1/(1+x^2) = atanx + C!

 

[acmilan]

Still proud of your proof, ac? =p

it's not mine if it were i'd be famous

 

[haque]

it's not mine if it were i'd be famous

Fame isn't everythuing tho-u can still be successful without being famous(lol thats my target_"to be successful")

 

[KeypadSDM]

Fame isn't everythuing tho-u can still be successful without being famous(lol thats my target_"to be successful")

That's a copout and a half.

 

[haque]

Lol i'm not copping out but it takes extraordinary talent for that type of fame-which i don't have(guys like terrence tao and that do tho heh)

 

[SeDaTeD]

I like this proof ...

z = cosx + isinx

dz/dx = -sinx + icosx
= i(cosx+isinx)
= iz

dz/dx = iz
dz/z = i dx
ln[z] + c = xi

x = 0, z = 1
thus c = 0
xi = lnz
therefore z = cosx + isinx = e^(xi)

cos[pi] + isin[pi] = e^(i Pi) = -1

Though true, you haven't really defined complex integration. Over what path are you integrating?

I prefer exp, cos and sin defined by their power series.

 

[KeypadSDM]

But that still requires complex analysis. Both do, to be fair. The complex integration can be done using anti-differentiation, irrespective of a path integral (by the uniqueness theorem).

I think.

Start from:
ln[z] + c = xi
Then dz/dx = ix.
[Differentiation is independent of the complex nature of constants, just use the limits.]
Uniqueness implies that the soln to dz/dx = ix is:
ln[z] + c = xi

No complex analysis required. Roughly speaking.

To be fair, it's been at least 18 months since I did analysis, so I don't know if I'm right.

 

[SeDaTeD]

Wouldn't we have had to define the complex logarithm function first? If I remember correctly, it was defined as the inverse of the complex exp function, and then extended over the complex plane, excluding negative reals and zero. And to do that we would have had to define the complex exp function, so I see it as more of a verification than a proof.