If $(\cos \alpha + i \sin \alpha )^n = 1$ then $(\cos \alpha - i \sin \alpha )^n = 1$

Question

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.

Answer 1

If $\overline{z}$ denotes the complex conjugate of $z$, notice that $\overline{z^n} = \overline{z}^n$ for any integer. So

$$(a+ib)^n=1\implies\left(\overline{a+ib}\right)^n=\overline{(a+ib)^n}=\overline{1}=1.$$

Answer 2

$(\cos \alpha + i \sin \alpha )^n =e^{in\alpha}= 1 \implies e^{-in\alpha}=(\cos \alpha - i \sin \alpha )^n=1$

Answer 3

Another way of thinking about the problem is to let $z = \cos \alpha + i \sin \alpha $. Then, using the fact that $z^n = 1$, evaluate $\frac1{z^n}$

You should get that $\frac1z=\cos \alpha - i \sin \alpha $

Answer 4

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?