# Ln(-1) is undefined

1. Mar 22, 2005

### T@P

ok dont 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 realise 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: ^

Last edited by a moderator: Mar 23, 2005
2. Mar 22, 2005

### 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.

3. Mar 22, 2005

### 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. $\varphi = n \pi$

4. Mar 23, 2005

### whozum

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

Last edited: Mar 23, 2005
5. Mar 23, 2005

### dextercioby

What is

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

Daniel.

6. Mar 23, 2005

### vincentchan

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

~vinnie

7. Mar 23, 2005

### whozum

I see whats 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 dont know where Im going with this..

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

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

8. Mar 23, 2005

### vincentchan

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

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

Last edited: Mar 23, 2005
9. Mar 23, 2005

### Integral

Staff Emeritus
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

Last edited: Mar 23, 2005
10. Mar 23, 2005

### 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 wierd because then i would appear to be real... only its not because $$-1^1.3$$ isnt quite real...
*edit* this is messed up, the point 3 doesnt 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

11. Mar 24, 2005

### Moo Of Doom

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

$$-1^{1.3}$$

12. Mar 24, 2005

### T@P

ah thanks.

13. Mar 24, 2005

### K.J.Healey

Yep, thats what I was about to say. Ln is periodic about your primary branch (arbitrary, isnt it?) just like exp is.
Ahh complex analysis, how i miss you...