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Was Euler wrong? 2*Pi=0?

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.

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   

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 
is already bad before you've done that. That's because complex powers, 
as well as logs, can have multiple values. That is mentioned at the 
bottom of this page:

   Imaginary Exponents and Euler's Equation - Dr. Math FAQ   

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 ?   


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   
Associated Topics:
College Imaginary/Complex Numbers
College Number Theory
High School Imaginary/Complex Numbers
High School Number Theory

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