Solving Ln(-1) with Euler's Equation

[T@P]

ok don't kill me, hear me out before you say its undefined:

$$e ^ { \pi i}= -1$$

so taking the natural log of both sides yields:

$$ln(-1) = \pi i$$

$$\frac {ln(-1)} \pi = i$$

and plugging it into the well known equation,

$$e^{\varphi i} = cos(\varphi) + i sin(\varphi)$$

you substitute your i into the left side,

$$e^{\frac {ln (-1)} { \pi * \varphi}} = cos(\varphi) + i sin(\varphi)$$

$$-1^{\frac {\varphi} { \pi}} = cos(\varphi) + i sin(\varphi)$$

now the left side is arguably real, so if you solve for i, you have i in terms of a real number. some of the time. i realize that you phi/pi can be 1/2 which would get you back to i, but can anyone explain this to me?

thanks (excuse the shoddy latex)

ok for some reason instead of getting something to the power of i get a funny ) sign. please assume it means to the power of, like this sign: ^

 

[whozum]

Thats really hard to understand but remember that your final equation traces out a path in the complex plane, not in the real plane. it will simplify to exactly what you started with.

 

[vincentchan]

which part you don't understand??

$$(-1)^{\varphi/\pi}$$
the above function is real if and only if the exponent is integer ... i.e. $\varphi = n \pi$

 

[whozum]

you substitute your i into the left side,

$$e^(ln(-1 / \pi * \varphi) = cos(\varphi) + i sin(\varphi)$$

$$-1^(\varphi/\pi) = cos(\varphi) + i sin(\varphi)$$

$$e^{ln(-1)/\pi*\varphi)}$$ is not the same as $$e^{ln(-\varphi)/\pi)}$$

Perhaps that is your mistake?

 

[dextercioby]

What is

$$e^{\frac{\ln (-1)}{\pi}\phi} =...?$$

Daniel.

 

[vincentchan]

$$e^{ln(-1)/\pi*\varphi)}$$ is not the same as $$e^{ln(-\varphi)/\pi)}$$

Perhaps that is your mistake?

I truly believe he want to say $$(-1)^{\varphi/\pi}$$, he just didn't type it right, notice what he said earlier:

i realize that you phi/pi can be 1/2 which would get you back to i,

~vinnie

 

[whozum]

I see what's goin on here. You confused an exponential rule.

You have $$e^{ln(-1)*\varphi/\pi}$$ and you simplified this to $$-1^{\varphi/\pi}$$ because $$e^{ln(x)} = x$$. The problem with that is
$$e^{ln(-1)*\varphi/\pi} = e^{ln(-1)}*e^{\varphi/\pi}$$
which simplifies to
$$-e^{\varphi/\pi}$$.

Then you have
$$-e^{\varphi/\pi} = cos(\varphi) + i sin (\varphi)$$

$$-e^{\varphi} * e^{1/\pi} = cos(\varphi) + i sin (\varphi)$$

$$-e^{\varphi} * 1.375 = cos(\varphi) + i sin (\varphi)$$

I really don't know where I am going with this..

$$-e^{\varphi} * 1.375 - cos(\varphi) = i sin (\varphi)$$

$$i = \frac{-e^{\varphi} * 1.375 - cos(\varphi)}{sin(\varphi)}$$

 

[vincentchan]

$$e^{ln(-1)*\varphi/\pi} = e^{ln(-1)}*e^{\varphi/\pi}$$

Your algebra is not quite correct...
$$e^{ln(-1)\times\varphi/\pi}=(e^{ln(-1)})^{\varphi/\pi}=(-1)^{\varphi/\pi}$$

EDIT:
$$e^{a+b} = e^ae^b$$
$$e^{ab} = (e^a)^b$$

 

[Integral]

Your error is in your very first step. Since $ln( e^{ \pi i})$ has a complex argument you must use the complex valued Ln function, this is multiple valued. You have thrown out the baby with bath.

ln( $e^z$)=ln |$e^z$| + i arg ($e^z$)

so

ln(-1) = (2n+1) $\pi$ i

 

[T@P]

ouchy this is too complicated. but yes my latex is pretty bad and my main point tho was that you can express i in terms of -1 to the power something real which seems weird because then i would appear to be real... only its not because $$-1^1.3$$ isn't quite real...
*edit* this is messed up, the point 3 doesn't get put into superscript but i hope you see what i mean

also just some random trivia but the 'function' $$(-1)^x$$ is funny because the derivative is just $$(-1)^x * \pi * i$$

almost trignometric, which it is because of the whole $$e^\varphi i$$ thing...

i was bored during physics, what can i say

 

[Moo Of Doom]

*edit* this is messed up, the point 3 doesn't get put into superscript but i hope you see what i mean

Short LaTeX lesson: Use {} to keep things together. You write -1^1.3, while you should write -1^{1.3}. See:

$$-1^1.3$$

versus

$$-1^{1.3}$$

 

[T@P]

ah thanks.

 

[K.J.Healey]

Your error is in your very first step. Since $ln( e^{ \pi i})$ has a complex argument you must use the complex valued Ln function, this is multiple valued. You have thrown out the baby with bath.

ln( $e^z$)=ln |$e^z$| + i arg ($e^z$)

so

ln(-1) = (2n+1) $\pi$ i

Yep, that's what I was about to say. Ln is periodic about your primary branch (arbitrary, isn't it?) just like exp is.
Ahh complex analysis, how i miss you...