Proof of $ \exp(2πi)^{5/6}=\exp(5πi/3)=\exp(-iπ/3)$

Question

I have tried many times to convince my self that the following claim $ \exp(2πi)^{5/6}=\exp(5πi/3)=\exp(-iπ/3)$ being true but I can't .

Assume $ \exp(2πi)=\exp(10πi)$ then we have $ \exp(2πi)^5/6=\exp(5πi/3)=\exp(−iπ/3)$

and $ \exp(10πi)^{5/6}=\exp(50πi/6)=\exp(iπ/3)$ which it is a contradiction .

Now my question here : Does there exist a such proof for which De Moivre formula applied for rational number ?

Answer 1

Just because $e^{2\pi i} =e^{10\pi i} $ does not mean that $e^{2a\pi i} =e^{10a\pi i} $ for all $a$. The rule $$(a^b)^c=a^{bc} $$ does not hold in general for complex numbers.

Answer 2

I think The OP meant the answer of this question He may want to ask why this :$$\exp(i 10 \pi)^{5/6} = \exp(i \pi 50/6) = \exp(i \pi 8) . \exp (i \pi /3) = \exp (i \pi /3)$$ is true , But this is true for unit complex roots which it has six root as shown in all answers there , In particular this one, that is the reason why He asked for existence of such de formula which is not valide for Rational number and in the same time the part of this question is the same of his answer here

Answer 3

In both cases, you are raising a sixth root of $1$ to the fifth power. This gives you a sixth root of $1$ again. Since there are $6$ complex sixth roots of $1$, the symbol $\exp(2\pi i)^{5/6}$ is perhaps best understood as multi-valued, indicating any of the sixth roots of $1$, or perhaps it should be understood as indicating the set of all six of them, $\{e^{k\pi i/3}|k=0,1,2,3,4,5\}$. If we interpret $\exp(10\pi i)^{5/6}$, in the same manner, then the second interpretation gives no problem. In the first interpretation, $\exp(2\pi i)^{5/6}=\exp(10\pi i)^{5/6}$ simply means that there is some complex number which is a possible value of both expressions.

In any case, as others have said, rules of exponents are different for complex numbers.

Answer 4

Complex powers are multivalued functions and since sadly modern mathematics kind of frowns upon those, there is a lot of confusion around. Personally I think of them as (generally infinite) sets until we are in an analytic context where branches that give unique values can be fixed.

So here $A=${$\exp(2πi)^{5/6}$} ={$\exp (\frac{5}{6}\log \exp 2\pi i)$}={$\exp (\frac{5}{6}(2\pi i k)), k \in \mathbb Z$}. In particular $A$ contains $1, e^{\frac{5\pi i}{3}}, e^{\frac{10\pi i}{3}}= e^{\frac{4\pi i}{3}},e^{\frac{15\pi i}{3}}=-1, e^{\frac{20\pi i}{3}}=e^{\frac{4\pi i}{3}},e^{\frac{25\pi i}{3}}=e^{\frac{\pi i}{3}}$ as it has $6$ elements by periodicity

Similarly $B=${$\exp(10πi)^{5/6}$} ={$\exp (\frac{5}{6}\log \exp 10\pi i)$}={$\exp (\frac{5}{6}(10\pi i k)), k \in \mathbb Z$} and one can convince themselves that $B$ has also $6$ elements which are unsurpsingly the same as those in $A$

So as from the equality of sets {$-1,1$}={$-1,1$} one cannot conclude that $1=-1$, similarly from the equality sets $A=B$ one cannot conclude that distinct elements (eg $e^{\frac{5\pi i}{3}}$ in $A$ and $e^{\frac{\pi i}{3}}$ in $B$ )in one set are equal so there is no contradiction.