How does $e^\left(2i\pi\right) = 1$ result in $e^\left(i\pi\right) +1 = 0$?

Question

I can't see how $$e^\left(2i\pi\right) = 1$$ will result in: $$e^\left(i\pi\right) +1 = 0$$ thanks

Answer 1

It doesn't. However, $e^{i\pi}=-1$ does imply both $e^{i\pi}+1=0$ and $e^{2i\pi}=1$.

The more general result is that (multiplying by) $e^{i\theta}$ represents a counterclockwise rotation of the complex plane by an angle of $\theta$ about the origin. If $\theta=\pi$, then you rotate $180^\circ$, which is the same as multiplying by $-1$.

Answer 2

Algebra wll not do what you want. But algebra does this much:

Note if $z = e^{i\pi}$, then $z^2 = e^{2\pi i}$. Now if $e^{2\pi i}= 1$, that is, $z^2 = 1$, we get $z^2-1 = 0$ or $(z-1)(z+1) = 0$ so that either $z=1$ or $z=-1$.

The complex number $1$ has two square roots.