i = negative number? (1 Viewer)

[pranks85]

Check this out:

4 ln (i) = ln[(i)^4] = ln (1)

Therefore, ln (i) = 0. Which is clearly not true.

Therefore, i<0, or logarithm laws don't apply.

However, log laws do apply to complex numbers, as the solution of negative logs can be found in the complex field.

Ergo., i<0.

 

[turtle_2468]

That's a really stupid argument (IMHO... well, obviously the person is trying to fudge it, so maybe not! *l* just joking anyway...). Because you're mixing up real and complex numbers. What does it mean anyway for i<0? it means that you're assuming i to be a real number. But then you're using a "well-known fact" at the end that you can find logs of complex numbers...
btw, if you used the same method for ln(-1), you would find that ln(-1)=0, which is a much clearer contradiction. It's just without the complex number bit so it's easier to see how it's dodgy...

 

[pranks85]

We're constantly mixing real and complex numbers. We apply indice laws, multiplication, division, addition & subtraction laws, factorisation laws, even calculus laws to complex numbers, the way we do to real numbers.

The fact that it is positive or negative does not neccessarily place it within the real field, as the fact that being in the real field does not automaticall classify it as one of the two - 0, though in the real field, is neither positive or negative. The term (-i) is itself testimony to the fact that we perceive i as having some sign, though as to which, we're uncertain.

The idea was meant to be that you can't use the same method for ln (-1) [for which the solution is i*pi, by the way), because it's negative, and therefore, you cannot evaluate it within the real field, because, as you have pointed out, it proves a contradiction. Similarly, when applying logarithm laws to ln (i), you end up with a contradiction, and a complex solution. Therefore, my point was that there is a case to be made for i<0. The fact that both processes are questionable due to the same line in the proof - where ln (-1), or ln (i) is stated, is the basis of this argument - I am not attempting to prove that ln (i) = 0. of applying such laws to both negative and complex numbers is the basis upon which my

For the record: e^ix = cis (x)

e^i(pi/2) = cos (pi/2) + isin *pi/2) = i
Therefore, ln (i) = i*pi/2

 

[turtle_2468]

Okay... one at a time. We apply the laws to the complex numbers similar to how we apply them to real numbers, because they form a field (with rules similar to those ppl are used to in reals etc). The fact that it is positive or negative in the conventional sense DOES mean that it's in the reals. You're mixing the logic up here in the second paragraph btw... there is a number neither positive nor negative, but it doesn't show that complex numbers can be positive or negative. Hmm, how can I explain this... suppose we have cats as complex numbers, dotted cats as reals, and white or black dots as positive or negative correspondingly. You say that there's a blue dotted cat, but that doesn't prove that all cats have dots... btw tell me if you didn't get that analogy, I'll relate it further to the complex numbers.
More in the next message...

 

[turtle_2468]

And, in any case, 0 is a complex as well as a real number! The reals are just a subset of the complex numbers, remember...

The REAL reason behind the dodginess of the argument is this... when you apply Euler's formula, cis x = e^ix, and then apply the logarithm, you get dodgy solutions. Lots of them. Why? It's akin to how you get dodgy solutions when solving for expressions of the type sin^-1(3/5)... consider the above formula with logs taken i.e. {ln(cis x) = ix} (in paretheses because I'll show it's dodgy). The point is, you can increase x by 2*pi, and the LHS will remain the same and the RHS will "increase" by 2*pi*i. Hence, the equation above produces an infinite number of values of RHS with a spacing of 2*pi between the various solns. So the equation should actually read:
ln(cis x) = ix PLUS AN INTEGRAL MULTIPLE OF 2*PI

so, if you look back at the original eqn, (the first one in the first post), the reason why it stuffs up is because we have assumed RHS = 0, but LHS = 2*i*pi. So what you should actually have shown after the first line is ln(i) = 0 + an integral multiple of i*pi/2. Hope that made sense... as for the positivity or negativity of i, see my post on the reciprocal of i thread.

Yeah. Hope that helped!

 

[pranks85]

Interesting analogy with the cats.....but yeah, I get it.

Thanks for the rebuttal

 

[SoFTuaRiaL]

Originally posted by pranks85
Check this out:

4 ln (i) = ln[(i)^4] = ln (1)

Therefore, ln (i) = 0. Which is clearly not true.

Therefore, i<0, or logarithm laws don't apply.

However, log laws do apply to complex numbers, as the solution of negative logs can be found in the complex field.

Ergo., i<0.

basically, the laws of inequalities dont apply to complex numbers ..... u can never say a complex number is greater or lesser than another ....... 0 too is a complex number (0+0i). so u cant compare (0+1i) with (0+0i) .........
cheers