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## Homework Statement

Hello!

a)Prove that [tex](a+bi)^n[/tex] and [tex](a-bi)^n[/tex], n [tex]\in \mathbb{N}[/tex] are conjugate complex numbers;

b)Prove that quotient of any two numbers from the set of [tex]\sqrt[n]{1}[/tex] is again number from the set of [tex]\sqrt[n]{1}[/tex]

c)Prove that reciproca value of any number from the set of [tex]\sqrt[n]{1}[/tex], is again number from the set [tex]\sqrt[n]{1}[/tex]

## Homework Equations

[tex]z=r(cos\alpha+isin\alpha)[/tex]

[tex]\bar{z}=r(cos\alpha-isin\alpha)[/tex]

[tex]w_k=\sqrt[n]{r}(cos\frac{\alpha + 2k\pi}{n}+isin{\alpha + 2k\pi}{n}) ; k=0,1,2,...,n-1[/tex]

## The Attempt at a Solution

a) [tex](a+bi)=r(cos\alpha+isin\alpha)[/tex]

[tex] (a+bi)^n=r^n(cosn\alpha+isinn\alpha)[/tex]

[tex] (a-bi)=r(cos\alpha-sin\alpha)[/tex]

[tex] (a-bi)^n=r^n(cosn\alpha-sinn\alpha)[/tex]

Is this enough to prove that they are conjugate complex numbers?

b)[tex]E_k=\sqrt[n]{1}=cos\frac{2k\pi}{n}+isin{2k\pi}{n} , k=0,1,2,3,...,n-1[/tex]

Should I make like this?

[tex]E_n_-_1=\sqrt[n]{1}=cos\frac{2(n-1)\pi}{n}+isin\frac{2(n-1)\pi}{n}[/tex]

What should I do next?

c)If I know how to prove b) I will prove c)

In this case just [tex](\sqrt[n]{1})^-^1=\sqrt[n]{1}[/tex], right?