1. ## Find the error!

$\displaystyle x \in \mathbb{R}, e^{x} = e^{x \frac{2 \pi i}{2 \pi i}} = e^{2 \pi i \frac{x}{2 \pi i}} = (e^{2 \pi i})^{\frac{x}{2 \pi i}} = (cos(2 \pi) + i \text{ } sin(2 \pi))^{\frac{x}{2 \pi i}} = 1$

2. ## Re: Find the error!

I think the error is in the step $\displaystyle e^{x\frac{2i\pi}{2i\pi}}=e^{2i\pi\frac{x}{2i\pi}$ For similar reasons to why this is incorrect:
$\displaystyle e^{i\pi}=-1$
$\displaystyle e^{2i\pi}=1$
$\displaystyle ln(e^{2i\pi})=ln(1)$
$\displaystyle 2i\pi=0$

3. ## Re: Find the error!

Originally Posted by Lord Voldemort
$\displaystyle x \in \mathbb{R}, e^{x} = e^{x \frac{2 \pi i}{2 \pi i}} = e^{2 \pi i \frac{x}{2 \pi i}} = (e^{2 \pi i})^{\frac{x}{2 \pi i}} = (cos(2 \pi) + i \text{ } sin(2 \pi))^{\frac{x}{2 \pi i}} = 1$
When x is real valued we cannot say that $\displaystyle 1^x = 1$ for all x. For example, what happens when x = 1/2?

-Dan

4. ## Re: Find the error!

Originally Posted by Shakarri
I think the error is in the step $\displaystyle e^{x\frac{2i\pi}{2i\pi}}=e^{2i\pi\frac{x}{2i\pi}$ For similar reasons to why this is incorrect:
$\displaystyle e^{i\pi}=-1$
$\displaystyle e^{2i\pi}=1$
$\displaystyle ln(e^{2i\pi})=ln(1)$
$\displaystyle 2i\pi=0$
I don't understand though. Isn't that step just the associative property for multiplication?

Originally Posted by topsquark
When x is real valued we cannot say that $\displaystyle 1^x = 1$ for all x. For example, what happens when x = 1/2?

-Dan
I'm confused: 1^(1/2) = sqrt(1) = 1, right?

5. ## Re: Find the error!

Originally Posted by Lord Voldemort
I'm confused: 1^(1/2) = sqrt(1) = 1, right?
What about 1^(1/2) = -1? Mind you the exponent in your problem is going to be imaginary so this isn't a direct example...but my point is that $\displaystyle 1^x$ is, in general, multi-valued for $\displaystyle x \in \mathbb{C}$.

-Dan

6. ## Re: Find the error!

Greetings mathematicians,

On a much lighter note -
The error is here:

XD
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