$e^{2\sqrt{2} \pi i}=(e^{2 \pi i})^{\sqrt{2}}=1$?

Question

This is probably stupid. But this true?$$e^{2\sqrt{2} \pi i}=(e^{2 \pi i})^{\sqrt{2}}=1$$ I feel like this is wrong but I cannot see how. Any help is appreciated. Thank you

Answer 1

For non-integer $\alpha$ and $\beta\notin\Bbb R$, the identity $e^{\beta\alpha}=(e^{\beta})^{\alpha}$ does not hold: actually, $(e^\beta)^\alpha$ needs some technicality to even make sense, while $e^{\alpha\beta}$ does not.

In your case, by Euler's identity $$e^{2i\pi\sqrt2}=\cos\left(2\pi\sqrt2\right)+i\sin\left(2\pi\sqrt2\right)\notin\Bbb R$$

Answer 2

Using the same logic, $i=(-1)^{\frac{1}{2}}=((-1)^2)^{\frac{1}{4}}=1^{\frac{1}{4}}=1$.

Sadly, that is not how it works in the realm of complex numbers.

We would have $\displaystyle e^{2\sqrt{2}πi}=+\cos(2\sqrt2π)+i\sin(2\sqrt{2}π).$

You know that $\cos(nπ)$ is $1$ and $\sin(nπ)$ is $0$ for multiples of $π$!

But, INTEGER multiples. So $2\sqrt{2}$ doesn't count.

Answer 3

The "obvious" identity $(a^b)^c=a^{bc}$ does not hold in complete generality.

It does hold for all complex numbers $a$ if $b$ and $c$ are restricted to be integers (with a minor caveat for $a=0$ if $b$ and $c$ and negative integers).

It does hold for all positive real numbers $a$ if $b$ and $c$ are restricted to be real numbers, provided you define the exponential notation so that $x^y$ is a positive real number when $x$ is a positive real and $y$ is real, e.g., $x^y=e^{y\ln x}=\sum_{n=0}^\infty{(y\ln x)^n\over n!}$ with $\ln x=\int_1^x{dt\over t}$.

There are other circumstances under which it holds, but in general it does not. In the problem at hand, $a=e$ is a positive real, and $c=\sqrt2$ is real, but $b=2\pi i$ is not, so you cannot expect the identity to hold.

Answer 4

$a^{mn}=(a^m)^n$ is true only if $a >0$ and $a \in \mathbb R$.

Here, in your question, $a=e^{2\sqrt 2 \pi i}$ isn't purely real, so you can't apply this. Because if you go to apply this property for imaginary numbers also then weird things such as following will occur -$$i=e^{i\pi/2}=e^{2\pi I /4}=(e^{2i\pi})^{1/4}=1$$

Answer 5

the existence of complex $n^{\text{th}}$ roots of unity should be enough to dispel the notion that $1^a$ can be regarded as a single-valued function of $a$ in the complex domain.

if $$ x=1^{\sqrt{2}} $$ then taking logarithms, since the logarithm has an infinite number of branches, $$ \log x\in \{2^{\frac32}n\pi i\}_{n \in \mathbb{Z}} $$ this is an infinite set of points equally spaced along the imaginary axis, including the origin. under the exponential map we have: $$ x \in \{e^{2^{\frac32}n\pi i}\}_{n \in \mathbb{Z}} $$ which represents a dense set of points on the unit circle, one of which is 1.