Is the equation $e^{2πi}=e^{4πi}$ true?

Question

Consider an integer: 1. We can represent this integer by $e^{2πi}$. Also it can be represented as $e^{4πi}$. Now, we have an equation as below:

$$e^{2πi}=e^{4πi}$$

If we take "$i$" power of each terms, we get:

$$(e^{2πi})^{i}=(e^{4πi})^{i}$$ $$e^{2πi^2}=e^{4πi^2}$$ $$e^{-2π}=e^{-4π}$$ Now both side of equation are positive real numbers and if we calculate the values we get: $$0.00186744 = 0.000003487$$ So where is the error here? Or was our initial assumption $e^{2πi}=e^{4πi}$ wrong?

I have an idea, and I want to get confirmation from experienced friends here.

Consider the unit circle. Here, the numbers corresponding to $0$, $2π$, $4π$, etc. radians will be positive real numbers. For example, $25$ is can be written as $25.e^{2πi}$ or $25.e^{4πi}$ or $25.e^{6πi}$ or etc. In fact, when we are given an integer n, we do not know which angle it belongs to. If we try to take the square root of all of them here ($\frac{1}{2}$ power of each), we get $5.e^{πi}$ or $5.e^{2πi}$ or $5.e^{3πi}$ respectively. Now, if you pay attention, half of these numbers correspond to $-5$ and the other half correspond to $+5$. In fact, when we are given $25$ and asked to take the square root, we have to answer $\pm 5$ because we do not know which radian $25$ belongs to. Maybe it could be $2π$, maybe it could be $4π$, or others... So there are a lot of integer $25$. So when we take the $i$ power of these, we get a lot of different numbers. I think this is where this contradiction begins. Can you help me to understand this concept?

Answer 1

tl; dr: To the question in the title, Yes (with a standard notational convention).

To the resulting false statement in the question body, "laws" of exponents, particularly $(b^{z})^{w} = b^{zw}$, are not generally true.

Your explanation about ambiguities of angle is indeed the root of the problem (as it were).

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There is a complex exponential function $\exp$ that agrees with the real exponential on the real axis. For complex $z$, when one writes $e^{z}$ without qualification, it conventionally means $\exp(z)$.

If $x$ and $y$ are real and $z = x + iy$, it turns out that $$ \exp(z) = e^{x}\cos y + ie^{x}\sin y. $$ Consequently, $\exp$ is periodic with period $2\pi i$. Particularly, $\exp(2\pi i) = 1 = \exp(4\pi i)$; or, in the conventional notation above, $e^{2\pi i} = e^{4\pi i}$.

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If we heedlessly manipulate expressions using real rules of exponentiation, we easily go awry. Formally, for every base $b$, we "define" $$ b^{z} = \exp(z\log b). $$ "Define" needs scare quotes here because the right-hand side requires careful attention to the meaning of "$\log$": There exists no single-valued function $\log$ satisfying $\log(\exp z) = z$ for all complex $z$. The best we can do is define a multi-valued (or set-valued) "total logarithm" (any two of whose values differ by a whole multiple of $2\pi i$) that satisfies $z \in \log(\exp z)$ for all complex $z$.

When working with complex exponentials, then, we have a dilemma:

• If we allow $\log$ to be multi-valued, then $b^{z}$ is multi-valued even if $b = e$: by literal interpretation of the definition, \begin{align*} e^{z} &= \exp(z\log e) \\ &= \exp\bigl(z(1 + 2\pi ki)\bigr) \\ &= \exp z \cdot \exp(2\pi kiz)\quad\text{for all integer $k$.} \end{align*}
• If we fix a single-valued choice of $\log$, then "laws" of exponents, particularly $(b^{z})^{w} = b^{zw}$ for complex $z$ and $w$, are not generally true.

With all this as preamble: There are two ways to resolve the proposed paradoxical argument of the question.

• If $\log$ is multi-valued, then $e^{-2\pi}$ and $e^{-4\pi}$ are both values of $1^{i}$. (Note carefully: Even the identity $1^{z} = 1$ is not true!)
• If we fix a (single-valued) choice of $\log$, then $\exp(z) = \exp(w)$ does not imply $z = w$; regardless of our choice we have $\log(\exp 2\pi i) \neq 2\pi i$ or $\log(\exp 4\pi i) \neq 4\pi i$ (or both). Consequently "taking $i$th powers of both sides" of $e^{2\pi i} = e^{4\pi i}$ by multiplying exponents is not mathematically correct.

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A sharp "paradox" that uses real laws of exponents in complex exponentiation was pointed out to me in the late 1980s by a classmate, David Day, who to my knowledge did not learn of it elsewhere: For all complex $z$, "we have" $$ e^{z} = e^{2\pi i(z/2\pi i)} = (e^{2\pi i})^{z/2\pi i} = 1^{z/2\pi i} = 1. $$