0

I have just started learning about euler's identity and I came across something that I couldn't make sense of, but it seemed to be very simple, so some clarification would be appreciated :)

If for any real $x$,

$$e^{i2\pi x} = ((e^{\pi i})^{2})^{x}=((-1)^{2})^{x}=1^{x}=1$$

Then that implies that for any real $x$:

$$1=e^{i2\pi x}=\cos{(2\pi x)} + i\sin{(2\pi x)}$$

Which is not true, for example:

$$\cos{(2\pi\frac{1}{4})}+i\sin{(2\pi\frac{1}{4})} = \cos{\frac{\pi}{2}}+i\sin{\frac{\pi}{2}} = 0 + i = i \ne 1 $$

Obviously I missed something, but I can't figure out what. Does anyone know?

Community
  • 1
Drew Banks
  • 1

2 Answers2

1

The first line is the issue. The problem is that you never defined exponentiation properly: What is $a^b$?. If $a$ is a positive real, there is no much mistery, is the usual exponentiation that we know. But things get weirder if $a$ is negative, If $b$ is an integer, everything is well defined, e.g. $(-1)^4=1$ makes total sense. But what happens when $a$ is negative and $b$ is a fraction? For instance, what if $a=-1$ and $b=0.5$, can we make sense of $(-1)^{0.5}$? you may think it as square root of $-1$, so let's say that $(-1)^{0.5}=i$.But then you think about it... wait what about saying $(-1)^{0.5}=-i$? Things get even more complicated if $b$ is an arbitrary real number, for instance what is $(-1)^\pi$?

As you can see, there are a lot of issues when the base is negative. So, technically, $a^b$ is not well defined for $a<0$. How do we define $a^b$ in general then? The best we can do is to define the following:

For $a\in\mathbb{C}\setminus(-\infty,0]$ and $b\in\mathbb{C}$. We define $a^b:=e^{b\operatorname{Log}(a)}$. Where $\operatorname{Log}(z)$ is the "principal branch" of the logarithm, which is defined only for $z\in\mathbb{C}\setminus(-\infty,0]$.

So, as a rule of thumb, to not make mistakes like this, always work with positive basis (preferrably $e$), since in this case all the properties of exponentials are still satisfied.

Julian Mejia
  • 4,438
0

On an intuition level: In your example x = 1/4. So in your first line you have the fourth root of one, which is not only equal to 1, but to i, -1, -i and 1.

Kero
  • 15