Mathematicians, I have a question

[Duckers]

why is ln(-1) = (pi)i ?

that's what my calculator says, can anyone prove it?

 

[Duckers]

ok...

 

[Soccer55]

If I remember correctly, you shouldn't even be able to take the log or ln of a negative number because the definition of log(x) and ln(x) is the power to which you raise the base in order to get x (ie: log 100 = 2 in base 10 because 10^2 = 100). Since there is no number y for which x^y = 0, it shouldn't even be possible to do on your calculator. I could be mistaken, but I believe this is correct. What calculator did you use anyway? I tried it on my TI-89 and I got a "Domain Error".

-Tom

 

[Duckers]

yeah, I am aware that e^((pi)i) should be equal to -1, because that's what my calculator says!!!

I used a Ti-83+, make sure that you set your calculator in a+bi (imaginary) mode.

I asked this because I never thought that there could be a relation between e and pi.

 

[RockEater]

ok, one of the really cool theorums of math (I think it came from Euler) is

e^(i(pi))=-1

I'm not gonna do the whole proof, but it comes from properties of complex numbers, that if complex number Z=r(coso+sino) then Z^n=r^n(cosno+isinno)
by somthing Euler says e^io=cos0+isino, use pi for o and get e^ipi=-1

pretty cool, huh? relates everything important, including 1, e, i, and pi

I just learned it last semester, so hopefully someone else can explain better why that works, I'm only a freshman

btw, take natural logarithm of both sides, you get what you found on your calculator

peace

 

[Duckers]

Thanks for explaining that to me.

Well, actually I didn't understand what you said but that's ok

I hate imaginary numbers