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Kainui
So we know that the pattern goes that i^3=-i and i^7 is also -i. So if you set them equal to each other, i^3=i^7 and then take the log to both sides you get: 3ln(i)=7ln(i) And then you can simplify this down by dividing out the ln(i) to get 3=7. Obviously this isn't true, so what am I doing wrong?
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Give me as deep an answer as possible, I've taken linear algebra and differential equations and I'm planning on taking real analysis soon and want to understand all this stuff perfectly. I know how to take the natural log of i, and it is defined as an imaginary number, so I don't think that explains it well enough for me.
it looks like logs of imaginary numbers don't have those rules... http://en.wikipedia.org/wiki/Complex_logarithm
ln(i) is actually a multivalued function - so cannot just cancel out ln(i) from both sides of your identity since the identity would only be valid for DIFFERENT values of ln(i) on the left and right hand sides of that identity. You could form a similar case with real numbers - if x=1, then:\[x^3=x^7\]therefore:\[3\ln(x)=7\ln(x)\]divide both sides by ln(x) to get: 3 = 7 In this case the flaw is in dividing by zero.
good find @TuringTest - snippet from that article: A common source of errors in dealing with complex logarithms is to assume that identities satisfied by ln extend to complex numbers. It is true that \(e^{Log z} = z\) for all z ≠ 0 (this is what it means for Log z to be a logarithm of z), but the identity \(Log e^z = z\) fails for z outside the strip S. For this reason, one cannot always apply Log to both sides of an identity \(e^z = e^w\) to deduce z = w. Also, the identity Log(z1z2) = Log z1 + Log z2 can fail: the two sides can differ by an integer multiple of 2πi
Again the key point here is that ln(i) is a MULTI-VALUED function
Yes, it discusses two ways to fix the problem, one by restricting the domain, the other by some Reimann manifold craziness I won't pretend to understand. New to me :)
The Reimann manifold is way beyond my understanding as well :)
Totally awesome, thank you both.