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Are all complex numbers real? (Posted on 2006-08-24) Difficulty: 3 of 5
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)

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Solution re: The Specific Problem (fleshed out a bit more) | Comment 6 of 14 |
(In reply to The Specific Problem by Richard)


Richard, you state that there is no identity such as

exp(ab)=(exp(a))^b.                (0)

This is correct, but it is not the whole truth.



You are correct, because the left side of your equation is something well-defined, namely the exp-function, which can be defined as

exp(a+bi):=e^a*(cos(b)+isin(b)).        (1)

where e^a is the "normal" real exponentiation. The right side of the equation is however not a priori well-defined, namely the exponentiation of two complex numbers (coming back to my previous post here).



It is not the whole truth, because one can still define exponentiation of complex numbers, only that it is not a unique operation. That should not surprise us, we know for example that the square root of a number is not uniquely defined either, we just have a convention to always interpret sqrt(x) as the positive square root of the real number x. Furthermore I claim that equation (0) does still hold.

How exactly is exponentiation of complex numbers defined? Easy, we define

a^b:=exp(b*ln(a))                (2)

where ln is the natural logarithm. The next step is to define what the natural logarithm is for complex numbers. Naturally we define that ln(y) is the solution for x of the equation

exp(x)=y.                    (3)

But wait, there is more than one solution! In fact, if x is a solution, then x+2πik is also a solution for every integer k, as can be seen from equation (1). So when we talk about the logarithm ln(y), we are really talking about ALL the numbers of the sort ln(y)+2πik, where ln(y) is ONE of the many solutions to (3). Consequently, when we write a^b, we really mean all the numbers exp(b*(ln(a)+2πik)), i.e. equation (2) must be written as

a^b:=exp(b*(ln(a)+2πik))            (4)

(It's a bit like writing sqrt(4):=+/-2.) BTW, when a and b are positive real numbers, we usually choose k=0 to obtain our "usual" power, but even for real numbers there is a multitude of powers. In particular, 1^b is not necessarily equal to 1; 1 is only ONE possibility! Again, no surprise because we are already familiar with the fact that 1^(1/2)=sqrt(1)=+/-1. So when would identity (0) hold? One could say it holds when it can be made true for ONE of the numbers a^b as defined in (4). So let's see if it does:

exp(ab)=(exp(a))^b
<=>
exp(ab)=exp(b(a+2πik)) for one integer k
<=>
ab=b(a+2πik)-2πin for some integers k and n
<=>
n=kb

Now if we set k=n=0 we can make (0) true, hence the identity does hold. In some way at least...



How can we use all this to figure out what went wrong in the puzzle's proof? Let's take a look at the right side of the equation (Writing "t" for "theta") using what we know about complex exponentiation:

RS=r*(e^(2πi))^(t/2π)=r*exp(t/2π*(2πi+2πik)=r*exp(ti+tik)=r*exp(ti)*exp(tik)

Now when we choose k=0, we get exactly what's on the left side of the equation, which is z. Atheron has however cleverly chosen k=-1, in which case we end up with the real number r.

Edited on August 26, 2006, 7:14 pm

Edited on August 26, 2006, 7:16 pm
  Posted by JLo on 2006-08-26 19:11:51

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