The most interesting math problem I've seen in a long time

[Eeezee]

e^z = 0

Solve for z

Beer for the first person to solve it

(Hint: A solution exists)

(Hint 2: It's not negative infinity)

And just to blow your mind, e^(2*pi*i) = 1

 

[So]

-infinity

 

[OdiN]

42

 

[Eeezee]

Originally posted by: So
-infinity

Incorrect, that is only true when taking a limit or making an approximation.

 

[jonessoda]

ln 0. Easy.

 

[So]

Originally posted by: jonessoda
ln 0. Easy.

Ooooh! Game over man, game over.

 

[So]

you know, the funny thing is, if you had said 10^z I would have figured out the answer was log(0) immediately. Damn you e, you tricked me once again!

 

[Goosemaster]

Originally posted by: jonessoda
ln 0. Easy.

lol....

 

[ShadoWing]

um.....z=1?

 

[So]

Originally posted by: ShadoWing
um.....z=1?

e^1 = e.

 

[chuckywang]

sine and cosine are never 0 at the same time, so it's not possible.

 

[coomar]

isn't ln 0 equivalent to negative infinity

 

[dopcombo]

He's just playing all of us.

I challenge OP to give the answer now!

 

[coomar]

well i visualize y=e^x and it only approaches zero at negative infinitity

 

[jman19]

Originally posted by: coomar
isn't ln 0 equivalent to negative infinity

I do believe that it is defined as that... but more acurately, ln x approaches -inf as x approaches 0+... x=0 is an asymptote of the function ln x, so I don't think it's valid to say that the solution is ln(0)... though I most definitely may be wrong

 

[Eeezee]

Originally posted by: coomar
well i visualize y=e^x and it only approaches zero at negative infinitity

That's right, it approaches zero but never reaches it. That would make negative infinity a solution to the limit, but it is not a solution to z in e^z = 0.

 

[Eeezee]

Originally posted by: jman19

Originally posted by: coomar
isn't ln 0 equivalent to negative infinity

I do believe that it is defined as that... but more acurately, ln x approaches -inf as x approaches 0+... x=0 is an asymptote of the function ln x, so I don't think it's valid to say that the solution is ln(0)... though I most definitely may be wrong

You're completely right

 

[arcenite]

Originally posted by: Eeezee

Originally posted by: jman19

Originally posted by: coomar
isn't ln 0 equivalent to negative infinity

I do believe that it is defined as that... but more acurately, ln x approaches -inf as x approaches 0+... x=0 is an asymptote of the function ln x, so I don't think it's valid to say that the solution is ln(0)... though I most definitely may be wrong

You're completely right

I told him that. I want my beer. Yesterday.

 

[chuckywang]

Someone find a solution to ze^z = -1.

 

[SparkyJJO]

I hate e....

 

[Goosemaster]

Originally posted by: chuckywang
Someone find a solution to ze^z = -1.

(-ln(0))

edith wait..damnit

 

[TuxDave]

Oh yeah?? Find one of the solutions for i^i

 

[MrDudeMan]

actually the answer is e^(i*pi) = 0. the reason it is puzzling is because an imaginary number is interacting with a real number to get nothing.

edit: oops, wrong

 

[stan394]

Originally posted by: Bigsm00th
actually the answer is e^(i*pi) = 0

eh e^(i*pi) +1 = 0

 

[TuxDave]

Originally posted by: Bigsm00th
actually the answer is e^(i*pi) = 0

I think e^(i*pi) = -1