I am going to "prove" that $\cos\left(\frac{2\pi}{n}\right) = 1$ for all $n \geq 1.$ While this is definitely not true, I have a chain of arguments that seemingly shows exactly that. Please help me understand where the error creeped in.

Here we go. We start with recalling a number of identities involving the complex exponential function. First, recall that $$e^{ix} = \cos{x} + i\sin{x}.$$ By trigonometry, it follows that $$e^{-ix} = \cos(-x) + i\sin(-x)= \cos{x} - i\sin{x}.$$ By these two identities, we have $$\cos{x} = \frac{e^{ix} + e^{-ix}}{2}.\tag{1}$$

Next, recall Euler's identity: $$e^{i\pi} = -1.$$

Squaring both sides, we obtain $$e^{2i\pi} = 1.\tag{2}$$

And now to the meat of the "argument". Letting $x = \frac{2\pi}{n}$ in (1), we get \begin{align} \large \cos\left(\frac{2\pi}{n}\right) &= \large \frac{e^{i\frac{2\pi}{n}} + e^{-i\frac{2\pi}{n}}}{2} \\ &= \large \frac{(e^{2 \pi i})^\frac1n + (e^{2 \pi i})^{-\frac1n}}{2} \\ & = \large \frac{1^\frac1n + 1^{-\frac1n}}{2} = \large \frac{1 + 1}{2} = 1. \end{align}

My guess is that I messed up the factoring of the exponents in step 2 of the above, but it's not clear to me why what I did is not allowed.

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    $\begingroup$ Complex numbers in general do not satisfy $a^{bc}=(a^b)^c$. Can you see where you use this? $\endgroup$
    – Jolien
    Feb 6 '18 at 7:36
  • 1
    $\begingroup$ There are $n$ distinct $n$th roots of $1$. Heck, there's two square roots of $1$. When you write $1^{\frac{1}{2}}$, do you mean $1$ or $-1$? Either one is a square root of $1$. Taking the $n$th root is a multivalued function. So your error is going from $\frac{1^{\frac{1}{n}}+1^{-\frac{1}{n}}}{2}$ to $\frac{1+1}{2}$. $\endgroup$
    – Bob Jones
    Feb 6 '18 at 7:38

The error lies in the equality$$e^{i\frac{2\pi}n}=\left(e^{2\pi i}\right)^{\frac1n}.$$The LHS is a concrete complex number, whereas the RHS could be any $n$th-root of $e^{2\pi i}$.

In general, the equality$$e^{ab}=(e^a)^b$$is meaningless.


Nope, the problem is that taking $n$th roots of a complex number (even if it is $1$) is more complicated than for real numbers - there are $n$ different solutions and generally none is preferred over the others.

Actually in the reals you have the same problem, to the smaller extent. Look at this "proof": $-1=\sqrt{(-1)^2}=\sqrt 1=1$. This is wrong because $\sqrt{\cdot}$ has "picked" the wrong root - the positive one rather than the one you started with. With reals, at least, you have a rule which root will be picked, so you have formulas such as $\sqrt{x^2}=|x|$.

Similarly, even if you had a rule which $n$th root to pick for a complex number (which, for complex numbers, you don't), it may still not give you the same value as the one you started with.


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