If $(\cos \alpha + i \sin \alpha )^n = 1$ then $(\cos \alpha - i \sin \alpha )^n = 1$ Prove that if $(\cos \alpha + i \sin \alpha )^n = 1$ then $(\cos \alpha - i \sin \alpha )^n = 1$.
What should I use? De Moivre's formula? Exponential form? I tried, but It doesn't work.

Yet another approach: Since $(\cos\alpha+i\sin\alpha)^n=1,$ then $$(\cos\alpha-i\sin\alpha)^n=(\cos\alpha-i\sin\alpha)^n(\cos\alpha+i\sin\alpha)^n=\bigl((\cos\alpha-i\sin\alpha)(\cos\alpha+i\sin\alpha)\bigr)^n$$ Now, expand $(\cos\alpha-i\sin\alpha)(\cos\alpha+i\sin\alpha)$. What can we do then?
