 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Are all complex numbers real? (Posted on 2006-08-24) Let's "prove" that every complex number z is real.

If z=0 it's obvious. For all other complex numbers z=r*e^(θi), where r is a real number, and i=√-1.

Now, z= r*e^(θi)= r*(e^(2πi))^(θ/2π). Now as we know that e^(2πi)=1 we can write z =r*(1)^(θ/2π) → z=r.

What's wrong with this?

 No Solution Yet Submitted by atheron Rating: 3.3333 (3 votes) Comments: ( Back to comment list | You must be logged in to post comments.) re(3): The Specific Problem (fleshed out a bit more) | Comment 9 of 14 | (In reply to re(2): The Specific Problem (fleshed out a bit more) by Richard)

Nothing much to worry here, we just have to get used to the fact that exponentiation a^b is not a unique operation. Even for real a and b it is so: a^b is only unique if b is an integer. If b = p/q with coprime p and q, then there are precisely q "versions" of a^b. And if b is irrational then there is an infinite (countable) number of choices. However, when we write things like Pi^Pi = 36.46... we always exclude the other (complex) choices as we are trained to consider the real-valued choice only.

I'd like to cite Fredericos posting her, because it nicely illustrates the problem for b=2 without even considering complex number:

-1 = (-1)^1 =* ((-1)^2)^(1/2) = 1^(1/2) =# 1

Ambiguity comes into play for the equations =* and =#, the other equal signs are fine. The equation =* chooses 1^(1/2) = -1, whereas equation =# chooses 1^(1/2) = 1. Due to the different choices, we end up with different results. No surprise here. Same for the original puzzle.

Finally, if we want to keep our good old (a^b)^c = a^(bc) without worrying about ambiguity, I am afraid we have to stick with real a, b, c and positive a and the convention that a power is always a positive, real number.
 Posted by JLo on 2006-08-28 09:59:09 Please log in:

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