e^(ix) Paradox

Tue, 3 Aug 2004 17:19:00 EDT (-0400)

I suddenly started wondering about the term e^(2*pi*i)...

This prompted me to think:
One to the power of anything is still one, right? ie 1^x = 1.
(a^b)^c = a^(b*c), right?
e^(i*pi) = -1, right?
so e^(2*i*pi) = 1, because the square of negative one is one, right?
and e^(ix) can equal e^[2*pi*i*(x/(2*pi)] anytime, right?
This equals [e^(2*pi*i)]^[x/(2*pi)] = [1]^[x/(2*pi)] = 1, right? (??)
So how can e^(ix) ever be anything but 1?

Can you spot the fallacy? See my answer at bottom.
Here's some illuminating discussion from Usenet, thanks to the existence of search engines such as Google:
I "googled on" "exp(i always equals 1" and found via:
http://groups.google.ca/groups?q=exp(i+always+equals+1&hl=en&lr=&ie=UTF-8&selm=1993Nov13.163828.28823%40galois.mit.edu&rnum=6
the following:

Message 1 in thread
From: djvelleman@amherst.edu (djvelleman@amherst.edu)
Subject: Complex Exponential
Newsgroups: sci.math
Date: 1992-06-23 01:12:26 PST

In article <953@kepler1.rentec.com>, andrew@rentec.com (Andrew Mullhaupt) writes:
>
> You can take any of the three properties of the real exponential:
>
>  exp(a + b) = exp(a) * exp(b)
>
>  d/dx exp(x) = exp(x)
>
>  exp(x) = 1 + x + x^2/2 + x^3/6 + ...
>
> and see what function they give you in the complex plane which agrees on
> the reals with the real exponential. Any one of these determines the
> exponential function for_all_ complex values, and (miraculously) in the
> same way.

  This is true for the second and third property, but not for the first.  It's
clear that the third property completely determines the value of exp, and the
second is even stronger than necessary--simply requiring that the derivative
exists at all points in the complex plane is sufficient.  But there are
lots of functions on the complex plane which satisfy the first property, and
agree with the exponential function on the reals, but are not equal to the
exponential function.  For example:
            f(x + iy) = exp(x).
  (But of course the main point of the posting--that the definition of the
exponential function for complex numbers is not arbitrary--is correct.)
  Dan Velleman
  Dept of Math & CS
  Amherst College


Message 2 in thread
From: Michael Weiss (columbus@strident.think.com)
Subject: Re: Complex Exponential
Newsgroups: sci.math
Date: 1992-06-23 03:53:30 PST

   In article <953@kepler1.rentec.com>, andrew@rentec.com (Andrew Mullhaupt) writes:
   >
   > You can take any of the three properties of the real exponential:
   >
   >  exp(a + b) = exp(a) * exp(b)
   >
   >  d/dx exp(x) = exp(x)
   >
   >  exp(x) = 1 + x + x^2/2 + x^3/6 + ...
   >
   > and see what function they give you in the complex plane which agrees on
   > the reals with the real exponential. Any one of these determines the
   > exponential function for_all_ complex values, and (miraculously) in the
   > same way.

In article <1992Jun23.093638.1@amherst.edu> djvelleman@amherst.edu writes:

     This is true for the second and third property, but not for the first.  It's
   clear that the third property completely determines the value of exp, and the
   second is even stronger than necessary--simply requiring that the derivative
   exists at all points in the complex plane is sufficient.  But there are
   lots of functions on the complex plane which satisfy the first property, and
   agree with the exponential function on the reals, but are not equal to the
   exponential function.  For example:
        f(x + iy) = exp(x).
     (But of course the main point of the posting--that the definition of the
   exponential function for complex numbers is not arbitrary--is correct.)
     Dan Velleman
     Dept of Math & CS
     Amherst College


There are in fact 2^c solutions to the functional equation f(x+y)=f(x)f(y)
(where c=cardinality of the continuum).  It's a little easier to talk about
solutions to g(x+y)=g(x)+g(y) (just let f(x)=exp(g(x))).  Let H be a Hamel
basis for C, i.e. a basis for C as a vector space over Q.  Any
Q-linear function C-->C will do for g; hence we can pick the values of g on
each element of H at will.

We can pick an H which can be decomposed into a disjoint union H_R + H_C,
where H_R is a Hamel basis for R (over Q).  We can demand that g|H_R is the
identity, and hence g(x)=x on the reals, and still have 2^c distinct
solutions.

This all uses the axiom of choice, of course.


Message 3 in thread
From: Andrew Mullhaupt (andrew@rentec.com)
Subject: Re: Complex Exponential
Newsgroups: sci.math
Date: 1992-06-24 05:59:56 PST

In article <1992Jun23.093638.1@amherst.edu> djvelleman@amherst.edu writes:
>In article <953@kepler1.rentec.com>, andrew@rentec.com (Andrew Mullhaupt) writes:
>>
>>  exp(a + b) = exp(a) * exp(b)
>>
> But there are
>lots of functions on the complex plane which satisfy [this] property, and
>agree with the exponential function on the reals, but are not equal to the
>exponential function.  For example:
>            f(x + iy) = exp(x).

True. However if the function is restricted to having a derivative at
any point, then exp(x) is the only function. Thanks for pointing this out,
and I'm sorry if I've confused anyone needlessly.

Later,
Andrew Mullhaupt


Message 4 in thread
From: Wouter den Otter (otterw@ct.utwente.nl)
Subject: complex exponential
Newsgroups: sci.math
Date: 1993-11-12 11:09:44 PST

At the coffee-break we ran into some curious problem. What is
the solution to (exp(2*pi*i))^(pi*i)?

One way one could argue that exp(2*pi*i) equals 1, and therefore
the solution of 1^x equals 1.

On the other hand, (x^a)^b=x^(a*b), so the solution should be
exp(-2*pi^2).

My feeling prefers the second case, but I am not sure.

Anyone?

--
Groetjes,
  Wouter.
___________________________________________________________________________


Message 5 in thread
From: Thomas Andrews (thomaso@centerline.com)
Subject: Re: complex exponential
Newsgroups: sci.math
Date: 1993-11-12 11:49:16 PST

I tried email, but it bounced.

In article <otterw.753121111@utctu1.ct.utwente.nl>,
Wouter den Otter <otterw@ct.utwente.nl> wrote:
>At the coffee-break we ran into some curious problem. What is
>the solution to (exp(2*pi*i))^(pi*i)?
>
>One way one could argue that exp(2*pi*i) equals 1, and therefore
>the solution of 1^x equals 1.
>
>On the other hand, (x^a)^b=x^(a*b), so the solution should be
>exp(-2*pi^2).
>
>My feeling prefers the second case, but I am not sure.
>
>Anyone?
>

The problem is that while exp(y) has a simple power series, and thus
is well-defined for every number,

 x^y

is not well-defined for arbitrary x.  Why, you ask?  Because
x^y is tradition defined as:

 exp(y*log(x))

and log(x) (log base e) is multivalued as a complex function.
For example, log(1)=0 is acceptable, but so is log(1)=2*pi*i.

To make it single-valued and continuous, we have to refuse to define
it for some set of values, usually choses as the set non-positive
reals.

Then we can define the range as x+yi with -pi<y<+pi.  If we allow
y==+pi, the we can also define it for the negative numbers, too, but
the logorythm function would not be continuous at those negative values.

In any event, if we make it a single-valued function, we don't
get to use the formula:

 log(xy)=log(x)+log(y).

Instead, we get:

 log(xy)=log(x)+log(y)+k*2*pi*i

where k is some integer.

So lets look at your problem:

 1^(pi*i)=exp(0)^(pi*i)=exp(0*pi*i)=1

and

 1^(pi*i)=exp(2*pi*i)^(pi*i)=exp(2*pi*pi*i*i)=exp(-2*pi^2)

Here, we have taken two different logs of 1 (0 and 2*pi*i) and
thus gotten different values for 1^y.  Notice, you could have looked
at the simpler case:

 1^(1/2)=exp(0*1/2)=1

and

 1^(1/2)=exp(2*pi*i*1/2)=exp(pi*i)=-1.

In other words, the square root of 1 is +1 or -1.  It is easy to show
that the these are the only possible values for 1^(1/2),
the exponent becomes irrational, the problem becomes sicker.

Example: For irrational real values of y, the set of possible values for

 1^y

is a dense subset of the unit circle in the complex plain.

Things become even more problematic when the exponent has an imaginary
component.
--
Thomas Andrews
CenterLine Software


Message 6 in thread
From: Timothy Y. Chow (tycchow@cayley.mit.edu)
Subject: Re: complex exponential
Newsgroups: sci.math
Date: 1993-11-13 10:05:11 PST

In article <otterw.753121111@utctu1.ct.utwente.nl> otterw@ct.utwente.nl (Wouter den Otter) writes:
<At the coffee-break we ran into some curious problem. What is
<the solution to (exp(2*pi*i))^(pi*i)?
<
<One way one could argue that exp(2*pi*i) equals 1, and therefore
<the solution of 1^x equals 1.
<
<On the other hand, (x^a)^b=x^(a*b), so the solution should be
<exp(-2*pi^2).

Since I posted a very similar problem not long ago, without answering it,
maybe now is a good time to explain the situation.

The best way to define a^b where a and b are complex and a is nonzero is by
the formula a^b = exp(b*log(a)), where exp() is defined using the familiar
power series expansion exp(z) = 1 + x + x^2/2! + x^3/3! + ... and log() is
its MULTIPLE-VALUED inverse.  The multiple-valuedness of log() is the key
to all such puzzles about complex exponentials.

Why is log() multiple-valued?  An easy way to see this is by example: since
exp(2*pi*n*i) = 1 for all integers n, log(1) can have any of the values
{0, 2*pi*i, -2*pi*i, 4*pi*i, ... }.  Similarly, the log of any nonzero
complex number can take infinitely many distinct values.

To go back to your example, (exp(2*pi*i))^(pi*i) = 1^(pi*i) because, as you
correctly note, exp(2*pi*i) = 1.  However, from our definition, 1^(pi*i) is
exp(pi*i*log(1)), and log(1) is MULTIPLE-VALUED.  If we take log(1) = 0 then
we get 1, but if we take log(1) = 2*pi*i then we get exp(-2*pi^2).  Or in
other words, both 1 and exp(-2*pi^2) are possible values of 1^(pi*i).  This
is not so bizarre; after all, 1 and -1 are both possible values of 1^(1/2).

[Caveat: if a and b are real and a is positive, many books abuse notation
and by a^b mean the unique positive real value, so that 1^(1/2) = 1 and not
-1.  Also, e^z is often used to mean exp(z) rather than exp(z*log(e)).]

The rule (x^a)^b = x^(a*b) is not always valid, if we use the definition
of ^ that I have given above.
--
Tim Chow     tycchow@math.mit.edu
Where a calculator on the ENIAC is equipped with 18,000 vacuum tubes and weighs
30 tons, computers in the future may have only 1,000 vacuum tubes and weigh
only 1 1/2 tons.                               ---Popular Mechanics, March 1949
--------------------------------------------------------------------------------


Message 7 in thread
From: Larry Mintz (kabir@citenet.net)
Subject: COMPLEX EXPONENTIAL
Newsgroups: sci.math
Date: 1999/01/05
         a                               b
I know  z  =aexp(log z ) So what would  z    be in exponential
                                         a

[z^a is not equal to a*exp(log(z)). z^a would be equal to (exp(log(z)))^a,
or exp(a*log(z)). -Jim]

form ?  I am trying to prove the theorem, (I have done it

before, but lost the proof, that by writing b=u+iv

           | b |
show that  |z  |  is independent of  a.
           | a |

Larry-kabir@citenet.net

Here's my take on it now, as of Aug 8, 2004
I now think the problem is in the last two steps,
[e^(2*pi*i)]^[x/(2*pi)] = [1]^[x/(2*pi)] = 1
where we make e^(2*pi*i) = 1, thus losing information about what complex angle we had. 1^x could be expressed as e^[x*ln(1)] (to adopt "ln" for "natural logarithm", and ln(1) can have many values, because any of e^(2*n*pi*i) can equal 1, for n = ...-2,-1,0,1,2... (any integer). To choose which "n" we were working with, we just have to remember what he had, so there's no point in doing the step in which we converted the expression to 1. The final answer will be a different answer (a different complex number) depending upon what "n" we had and what value of x we have. -- see Message 6 in thread above.)

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Last updated Aug 8, 2004