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Was Euler wrong? 2*Pi=0?

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Date: 03/13/2002 at 17:06:43
From: Warren
Subject: Was Euler wrong? 2*Pi=0?

It is well known that e^(Pi*i) = -1, according to Euler's formula.
While I was surfing the Internet last week, however, I stumbled
across a website with an interesting proof that shows that 2*Pi = 0
by using Euler's famous equation. As far as I can tell, all of the
steps are mathematically sound. I've been puzzling over this problem
over the past few days and I can't seem to make much sense of it.
Here's the proof:

Let x = e^(Pi)

1.  x^i = -1
2.  (x^i)^i = (-1)^i
3.  x^(-1) = (-1)^i
4.  [x^(-1)]^i = [(-1)^i]^i
5.  x^(-i) = (-1)^(-1)
6.  x^(-i) = -1
7.  x^(-i) = x^i
8.  e^(-Pi*i) = e^(Pi*i)
9.  [e^(-Pi*i)]^i = [e^(Pi*i)]^i
10.  e^(Pi) = e^(-Pi)
11.  ln[e^(Pi)] = ln[e^(-Pi)]
12.  Pi = -Pi
13.  2*Pi = 0

The key step is #7, where step #1 is combined with step #6. I've even
checked this on my TI-83 calculator: when I enter e^(Pi*i) it returns
a -1, and, likewise, when I enter e^(-Pi*i) it returns a -1. If both
are equal to -1, this implies that e^(Pi*i) = e^(-Pi*i). Raise both
sides to the power of i and you end up with e^(-Pi) = e^(Pi), which
makes no sense whatsoever. One value is approximately 23.141 and the
other is about 0.043, yet they are equal? From this, you can do some
more mathematical manipulation and end up with 2*Pi = 0. If this were
true, then that would mean that the circumference of any circle is 0.
Obviously, this can't be true.

If you can help clarify this situation, or come up with a possible
answer as to why this proof is not mathematically sound, I'd be very
grateful. Thank you.
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Date: 03/13/2002 at 23:25:52
From: Doctor Peterson
Subject: Re: Was Euler wrong? 2*Pi=0?

Hi, Warren.

See if this explanation of a very similar "proof" helps:

Find the Flaw
http://mathforum.org/dr.math/problems/dan.08.02.01.html

It is very tricky; even though I wrote that answer, I had trouble with
this one. Your step 8 is fine; it still just says that -1 = -1. (In
fact, most of the previous steps could be left out.) But whereas in
"Find the Flaw" the problem lies in taking the logarithm, here step 10
as well as logs, can have multiple values. That is mentioned at the

Imaginary Exponents and Euler's Equation - Dr. Math FAQ
http://mathforum.org/dr.math/faq/faq.euler.equation.html

What you've done here is to show, not that -pi = pi, but that raising
any number, even -1, to an imaginary power can give multiple values,
and therefore is not allowed in a proof. And that's what false proofs
like this are really all about: teaching us to be careful when we do
the "obvious" in algebra!

Here are some pages from the Dr. Math archives that more directly
explain the idea that complex powers are multivalued:

e^(i*pi) = -1: pi = 0 ?
http://mathforum.org/dr.math/problems/koehler10.17.97.html

Exponentiation
http://mathforum.org/dr.math/problems/pomerance8.16.97.html

I'll add a little further discussion of my own.

We can write any complex number as r e^(it). Let's calculate this
number raised to a complex power:

(r e^(it))^(a + bi) = (r e^(it))^a * (r e^(it))^(bi)
= r^a e^(iat) r^bi e^(-bt)
= r^a e^(iat) e^(ln(r)bi) e^(-bt)
= r^a e^(-bt) e^[(at + b ln(r))i]
\_________/    \____________/
abs val          angle

But wait a minute: the angle t is not uniquely defined for a given
number. Any angle t + 2k pi could have been used, for any integer k.
Let's repeat using any such angle:

(r e^(i(t + 2k pi)))^(a + bi)
= r^a e^(-b(t+2k pi)) e^[(a(t+2k pi) + b ln(r))i]
= r^a e^(-bt) e^(-2kb pi) e^[(at + b ln(r))i] e^(2ka pi i)
= r^a e^(-bt) e^[(at + b ln(r))i]  e^(-2kb pi) e^(2ka pi i)
\_________/    \____________/    \_________/    \____/
abs val          angle          dilation     rotation
\___________________________/    \______________________/
principal value                 varies with k

This tells us that the absolute value of a complex power has
infinitely many values, whose spacing depends on b, while the angle
can take different values dependent on a. In fact, if a is an integer,
the angles will all be equivalent, but when it is not an integer, the
angle will spiral around while the absolute value changes. Weird,
isn't it? But in a way it's not that surprising; we see the same with
fractional real exponents, which are likewise multivalued (there are
two square roots and three cube roots, for example). Would you expect
imaginary numbers to be better behaved than fractions when you use
them as exponents?

In your case, you have a pure imaginary exponent and a real base:

(-1)^i = (e^((1 + 2k)pi i))^i = e^(-(1+2k)pi)

So you get infinitely many positive real numbers. Your "proof" just
assumes that two of them are equal, namely those for k=0 and -1.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
College Imaginary/Complex Numbers
College Number Theory
High School Imaginary/Complex Numbers
High School Number Theory

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