I am just learning about Euler's identity. When messing around with it, I am getting some unsettling identities, which, I believe, is probably due to my lack of application of certain rules.
Starting with:eπi=−1
Is this statement true:
πi=ln(−1)
→ln(−1∗n)=πi+ln(n)
Also this one clearly does not hold up:
ln(1)=0
ln(1)=ln(−1∗−1)=ln(−1)+ln(−1)=πi+πi
→2πi=0→2=0
Please advise me on what I am missing.
Answer
Like square root, log is treated as a "multi-valued function." It's just that when we restrict to the positive real line, we can make it single-valued. (Just as we do with square root in algebra 1.)
So the first few things you derived are right: one of the logarithms of −1 is πi. The others all differ from it by multiples of 2πi. So when you find that log(1)=0 and log(1)=2πi, that's fine: it just says that there are many values of the log. (Sorta like writing 112=1 and 112=−1, which each mean that the thing on the right, squared, gives you 1.)
To say this a bit more elegantly: there is no single log function defined on the whole complex plane (except the origin) that's continuous on the whole complex plane (except the origin), just as there's no continuous square root function defined on the whole complex plane.
By long tradition, we agree to not try to make log continuous on the negative real axis; we can then define a single consistent logarithm that's continuous everywhere except the negative real axis (and 0, where it's undefined). This single consistent log, for which log(1)=0, is called the "principal branch" of the logarithm. It's sometimes denoted Log, I believe.
There's another perfectly good logarithm-like function, which I'll call ^log: it's defined by
^log(x)=logx+2πi
It has all the properties of log except that ^log(1)=2πi instead of 0.
You can also define other log-like things, by adding 4πi, −6πi, etc. to log.
Taken together, we get something for which
log(ab)=log(a)+log(b)
where here "log" denotes any of the functions I mentioned above,
holds in a kind of limited sense: if you add log(a) and log(b), you'll get log(ab)...but only up to multiples of 2πi. That is to say:
log(ab)−log(a)−log(b)=2πi n
for some integer n.
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