What is the error on this prove that says $-1=1$ So $e^{i\phi}=\cos(\phi)+i\sin(\phi)$ and let $\phi=2\pi$ so $e^{i2\pi}=\cos(2\pi)+i\sin(2\pi)=1$
Now $$e^{i2\pi}=1$$ and by taking the square root of both sides $$\sqrt{e^{i2\pi}}=\sqrt{1}$$ since $\sqrt{e^{i2\pi}}=e^{\frac{i2\pi}{2}}=e^{i\pi}$ and using Euler's identity $e^{i\pi}=-1$ so $$-1=1$$ I can't find an error in the proof can someone find it? 
 A: The dangerous thing you did was to take "the" square root.  
A: Hint you should use  Di Moiver formula for complex number where it is available and suitable only for the integer exponent and  not to use the role of exponentiation for complex number as it used in reals
look  we have  $e^{i2\pi}=(e^{i \pi})^2=(-1)^2=1$ in this direction the exponent is integer and equal $2$ then Di Moiver formula was applied here, but in the opposite direction to take square root which means the exponent is rational number not integer and equal $\frac 1 2$ then here you are used the extension of Di Moiver formula as shown below in  Note  in order to find the $2$- roots of your equation , we conclude that your problem is your equality between your roots .
Note: And you are just used  the extension version of de Moivre's formula to find the nth roots of your equation  such that they are $-1$ and $1$ .
A: (1) We have $1^2=1$ and $(-1)^2=1$. Both $1$ and $-1$ are square roots of $1$.
(2) If $e^{i2\theta}=2^{i2\pi}$, $\cos 2\theta=\cos 2\pi$ and $\sin2\theta=\sin2\pi$. But $2\theta$ is not necessarily $2\pi$. It can be any multuiple of $2\pi$. 
$$2\theta=2n\pi \quad \textrm{where}\quad n\in\mathbb{Z}$$
$$\theta=n\pi \quad \textrm{where}\quad n\in\mathbb{Z}$$
So $e^{i\theta}$ can be $e^{in\pi}$ for any $n\in\mathbb{Z}$. We ususally take $\theta=0$  or $\pi$ as the value of $e^{i\theta}$ is equal to either one of them for all other possible angles.
(3) So, the correct statement is $\{e^0,e^{i\pi}\}=\{-1,1\}$. This does not cause any problem.
A: Here a clarification without complex numbers.
Your surprising result is based on the following:
$$(-1)^2 = 1^2 \not \Rightarrow -1 = 1$$
