## log(-1)?

For the discussion of math. Duh.

Moderators: gmalivuk, Moderators General, Prelates

### log(-1)?

What is wrong with this reasoning?

log(-1) = log(1/-1) = log(1) - log(-1)
So
2 log(-1) = log(1) = 0
log(-1) = 0.
mollwollfumble

Posts: 26
Joined: Wed Sep 14, 2011 3:13 am UTC

### Re: log(-1)?

The problem with that reasoning is that it's only valid as the second part of the proposition, "If log(-1) is a real number, then...".

Similarly, we could have
sqrt(-1) = sqrt(1/-1) = sqrt(1)/sqrt(-1)
(sqrt(-1))^2 = sqrt(1) = 1
sqrt(-1) = +/- 1
In the future, there will be a global network of billions of adding machines.... One of the primary uses of this network will be to transport moving pictures of lesbian sex by pretending they are made out of numbers.
Spoiler:
gmss1 gmss2

gmalivuk
Archduke Vendredi of Skellington the Third, Esquire

Posts: 19283
Joined: Wed Feb 28, 2007 6:02 pm UTC
Location: Here, There, Everywhere (near Boston, anyway)

### Re: log(-1)?

The problem is 1) you have to check whether it actually works out the other way around e0 is not equal to -1, it is equal to 1 and 2) log(x) actually has multiple values, we just normally write the real valued one, but since log(1) also equals 2i*pi your reasoning should look like this:
2log(-1)=log(1)=n*2i*pi
log(-1)=n*i*pi, but since e2i*pi=1, then:
log(-1)=n*i*pi only for odd n
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
tomtom2357

Posts: 429
Joined: Tue Jul 27, 2010 8:48 am UTC