Last time we looked at some false proofs, which are often used to help students understand what does and does not constitute a valid proof, and in particular, to remind them to be careful in algebraic proofs, looking for issues like division by zero and taking square roots. This time, we’ll look at two similar (but sort of opposite) questions that use exponential and logarithmic functions of complex numbers to “prove” that unequal numbers are equal.

## Complex logarithms

First, here is a question from 2001 that seems to show that 1 = -1:

Find the Flaw I have carefully read the Dr. Math FAQ and still don't understand where the following proof goes wrong. e^(i*x) = cos(x) + sin(x)*i e^(2*pi*i) = 1 e^((2*pi)/i) = 1 e^(2*pi*i) = e^((2*pi)/i) ln(e^(2*pi*i)) = ln(e^((2*pi)/i)) 2*pi*i = (2*pi)/i i = 1/i i^2 = 1 i^2 = -1 Perhaps I should attempt to reclaim all points that I've ever lost on sign errors.

It does look as if signs might be meaningless … unless, of course, this is a false proof. But where does it go wrong?

Here we have a chain of reasoning starting from Euler’s Equation, which we have discussed in our FAQ here:

Imaginary Exponents and Euler's Equation

Dan has in effect taken \(x\) to be \(2\pi\) and \(-2\pi\), both of which yield a value of 1 from the formula, and derived from that a conclusion that 1 = -1. I think the following formulation of his work might be a little clearer:

e^(i*x) = cos(x) + sin(x)*i In particular, e^(2*pi*i) = cos(2*pi) + sin(2*pi)*i = 1 + 0i e^(-2*pi*i) = cos(-2*pi) + sin(-2*pi)*i = 1 + 0i Since these both equal 1, e^(2*pi*i) = e^(-2*pi*i) Since 1/i = -i, this can be written as e^(2*pi*i) = e^(2*pi/i) Taking logarithms, ln(e^(2*pi*i)) = ln(e^(2*pi/i) But then, 2*pi*i = 2*pi/i Dividing by 2*pi, i = 1/i Cross-multiplying, i^2 = 1 But this means -1 = 1

Last time I showed a way to locate the error, by checking whether each statement is true. That’s a little tricky here, but if you do it, you’ll find that taking the natural logarithm seems to be the problem.

Here is my initial reply:

The problem is that the log is a multivalued "function" when you work with complex numbers. Think about e^(ix); actually e^(ix + 2kpi) has the same value (for any integer k), since the sine and cosine are periodic. So there are many exponents that give the same value. Try graphing the complex numbers e^(2 pi i) and e^(-2 pi i) in polar form to see what I mean.

I referred to a couple pages that touched on this multi-valued nature of the log; I’ll have some more recent and detailed links below. If you look carefully at my version of the “proof” above, you’ll see that my last sentence is in fact present from the start: He *began* with two different powers of *e* that are equal. The key idea is that different exponents can result in the same power, so the exponential function of complex numbers is *not one-to-one*, and its inverse function, the natural logarithm, is therefore *not a (single-valued) function* in this context.

## What makes a proof false

Dan replied, showing that he was fully aware of that:

I'm afraid I wasn't very clear, or must be confused. I think that I understand the multivalued nature of e^(i*pi) and was trying to use it to prove that 1 = -1. Why can't I?

I’m not sure whether he meant what he said literally! I answered,

Hi, Dan. Well, you can't prove it because it's not true... Maybe what you mean is that you don't understand why the fact that e^(ix) has the same value for different x makes your proof invalid. Let's build up some background to explain what a valid proof requires.

The point I’ll be working toward is why we must only apply single-valued functions in a proof (related to the fact that we must only take reversible steps in solving an equation, unless we check for extraneous solutions at the end).

What I was trying to say is that, because ln(z) is multivalued (has more than one value for a given z, and therefore is not really a function as we usually think of it), you can't use it in a proof. When you prove something algebraically, what you are really doing istransforming an equation step by step, through a series ofequivalentequations,from a known fact to an unknown one. In order for each equation to be equivalent to the one before, each transformation must be valid and reversible. You are probably familiar with this from solving algebra problems, which is similar to proofs. If I have the equation 3x - 1 = 5 then I can add 1 to both sides, because any value of x that satisfies the equation also satisfies 3x = 6 and any value that satisfies the latter also satisfies the former. They are equivalent equations.

An example of a step that does *not* yield an equivalent equation is squaring, which can result in an equation that has *additional solutions* (as when we square \(x = 3\) and get \(x^2 = 9\)), which has two solutions: squaring is *not one-to-one*. Another is the square root we saw last time, which gives only *one of two* solutions.

The same is true in proofs. For example, if I start with 1 = 1 and then say that 1 - 1 = 0 I am subtracting one from each side. If the first equation is true, so is the second, since equal numbers remain equal when you subtract one. Also, if the second equation is true, so is the first one, because my transformation is reversible: I can add 1 to both sides of the second equation, and the first will be true.

Note that, typically, solving an equation and proving an algebraic fact are opposites: in the former, we are trying to *find when* the original equation is true, while in the latter we are trying to *show* that the final equation is *always* true.

In false proofs, generally there is one step that does not produce an equivalent equation. (Or, in proofs that depend only on implication rather than equivalence, the step attempts to reverse an irreversible transformation.) For example,dividing by zero is not valid. If I start with the fact that 1 * 0 = 2 * 0 and divide both sides by zero, I get 1 = 2 The first equation is true, but the second is not. When you divide by zero, you aretrying to reverse multiplication by zero. If a = b, it will be true that 0a = 0b, but not the other way around! You can't say that two numbers whose product with zero is zero are the same number, because multiplying by zero converts ALL numbers to zero, and therefore dividing by zero would convert zero to anything.

That is the basis of the division-by-zero “proofs” we saw last time: Multiplication by 0 is not one-to-one, so attempting to undo it (by dividing, or by canceling a common factor) results in different values.

As another example,taking the square root(or rather, removing a square) does not produce an equivalent equation. If I know that (1)^2 = (-1)^2 I can't say that therefore 1 = -1 because there are in fact TWO numbers whose square is 1. If a = b, then a^2 = b^2, but not vice versa, since squaring takes two numbers to the same value. Again, we are trying to undo an action (squaring this time, multiplication by zero last time) that is not reversible, so we are losing possibilities in our equation. If we really took the square root (that is, the positive square root) of each side, we would get |1| = |-1| which is valid.

This is the basis of the square root “proofs” from last time: We claimed to be taking the square root, but were really choosing a different root on each side, by removing the square without considering the double-valued nature of the square root in that sense.

In your attempted proof, you are taking the log of both sides, saying that e^(2*pi*i) = e^((2*pi)/i) implies ln(e^(2*pi*i)) = ln(e^((2*pi)/i)) But just as with the square root, since the log has multiple values (is not actually a function) for complex numbers, this is not necessarily true.If you take different values of the log, they will not be equal.In order to make this step valid, you would have to take the _principal value_ of the log, just as we took the positive value of the square root in order to correct our square root example.

The principal value of the log can be defined by choosing a particular range for the imaginary part.

To put it differently, if you avoid the "ln" notation and simply claim that e^(2*pi*i) = e^((2*pi)/i) implies 2*pi*i = (2*pi)/i thenyou are assuming that if the power is the same, the exponent must be the same, that is, that exponentiation is reversible, as it is for real numbers. Since this is not true (e^z can be the same for different z), it is not valid to drop "e^" from both sides. The resulting equation is not equivalent. Does that make the problem with your "proof" clearer?

That is, assuming that the logarithm is single-valued is the same as assuming that the exponential function is one-to-one; and the start of the “proof” made it clear that that is *not* true.

Here are a few pages from the Ask Dr. Math archives that discuss the “many-to-one” nature of the complex logarithm, and the “one-to-many” nature of its inverse function, the logarithm:

Functions of Imaginary Numbers Taking the Natural Log of e^(ki) Inconsistency in Complex Logarithms

## Complex exponents

Now let’s look at an even more twisted situation, from 2002:

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.

Warren has done the initial checks that I would do; like the last example, this is tricky to check by eye, as values like \((-1)^i\) are not familiar to most of us. It will turn out that even checking on a calculator is tricky.

I answered, first referring to the page we looked at above, and continuing:

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 is already bad before you've done that. That's becausecomplex 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 http://mathforum.org/dr.math/faq/faq.euler.equation.html What you've done here is to show, not that -pi = pi, but thatraising 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!

I referred to two older answers that dealt with the multiple-valued nature of the complex exponential:

e^(i*pi) = -1: pi = 0 ? Exponentiation

I then continued:

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

Here I wrote the base in polar (exponential) form, using its *modulus* (absolute value, or length) and *argument* (angle, usually written as *θ* rather than *t*); and I wrote the exponent in rectangular form, because that is the easiest way to deal with each.

But wait a minute:the angle t is not uniquely definedfor 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

Here I used the fact that the angle is not unique, but any coterminal angle could have been used; so we have an infinite set of values for the exponential expression.

This tells us that theabsolute valueof a complex power has infinitely many values, whose spacing depends on b, while theanglecan 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 multi-valued (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.

What is most surprising here, I think, is that such odd things can happen even when the base is real!