Why isn't $e^{2\pi xi}=1$ true for all $x$?

We know that $$e^{\pi i}+1=0$$and $$e^{\pi i}=-1$$

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

Because $1$ is the multiplicative identity,$$(e^{2\pi i})^x=1^x$$$$e^{2\pi xi} =1$$should also hold true.

But we also know that $$e^{xi}=\cos(x)+i\sin(x)$$and so$$e^{2\pi xi}=\cos(2\pi x)+i\sin(2\pi x)$$which does not equal 1 for all values of $x$.

Now I realize I probably didn't break math, so I must be making an invalid assumption. What is wrong with my reasoning?

Solution 1:

The notion that $(a^b)^c=a^{bc}$ has to be abandoned in complex analysis.

Or, you have to allow that $a^b$ is a multi-valued function and then you can actually say that (one of) the values of $1^x$ is $\cos(2\pi x)+i\sin(2\pi x)$. With multi-valued functions you can say "All of the values of $a^{bc}$ are values of $(a^b)^c$," but not visa versa.

Multi-valued exponentiation can be seen as an extension of the idea that there are two "square roots," and, while we usually take $\sqrt{x}$ to be the positive one, we might sometimes prefer to think of $\sqrt{x}$ as a multi-valued function. For example, if $\sqrt{x}$ is multivalued, then you can write the quadratic formula as:

$$\frac{-b+\sqrt{b^2-4ac}}{2a}$$

and no longer have that pesky $\pm$ symbol from the usual formula, being implicit in the multi-valued $\sqrt{x}$ function. But the obvious problem with multi-valued functions is that the above "looks like" it is describing a single root, when it is describing two roots.

The other problem with multivalued functions is, what would one mean by:

$$a^{b} + a^{2b}?$$Most of the time when you see something like this, you probably don't want to pick from all values of $a^{b}$ and all values of $a^{2b}$, but rather you want to pick the same "branch," which amounts to picking the same value for $\log a$ for each term, amongst the infinitely many possible values for $\log a$.

So, in short: Exponentiation in complex numbers is irritating and no fun.

Solution 2:

The property $(a^b)^c = a^{bc}$, true in the case ${\text{positive}^\text{real}}$ (with $a^b$ always positive) isn't true in the complex case. Example from the link:

$$(1-i)^{2i} \ne ((1-i)^2)^i.$$

Solution 3:

In complex numbers exponentiation rules are a bit different, in this case $$(e^{2 \pi i})^x\not\equiv e^{2 \pi i x}$$

Solution 4:

Basically, you don't have the equality $$\left(a^b\right)^c = a^{bc}$$ for complex numbers. The way I explain this to myself is that for real values, $a^b$ is defined as the limit of $a^{q_n}$ where $q_n$ is a rational number and $q_n\rightarrow b$ as $n\rightarrow\infty$. As $a^q$ is defined through roots for rational numbers, and roots become complicated for complex numbers, it makes sense that the original rule will become shaky.