Question

If 1,${\alpha }_1$,${\alpha }_2$,${\alpha }_3$,............,${\alpha }_{n-1}$ are the roots of the equation x = $(1{)}^{\frac{1}{n}}$. Find the sum of the roots.

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Answer 1

x = $(1{)}^{\frac{1}{n}}$ = $(cos\theta +isin\theta {)}^{\frac{1}{n}}$

$\left[cos\right(2k\pi +0)+isin(2k\pi +0){]}^{\frac{1}{n}}$

${\alpha }^k$ =$cos\frac{2k\pi }{n}$ + i$sin\frac{2k\pi }{n}$

where k = 0,1,2,3,...............(n-1)

So,$n^{th}$ roots of the unit are $a^k$ (k=0,1,2,3,.........(n-1))

sum of the roots

1 + ${\alpha }_1$ + ${\alpha }_2$ + ${\alpha }_3$,............,${\alpha }^{n-1}$

= $\frac{1(1-a^{n-1})}{(1-a)}$ = $\frac{1-[cos2\pi +isin2\pi ]}{(1-a)}$ $$\left\{\begin{array}{c}Whenk=n\\ a^n=cos2\pi +isin2\pi \end{array}\right\}$$

= $\frac{1-1}{1-\alpha }$ = 0

Alternative

Since $x^n$ - 1 = 0

Sum of the root of the equation = -$\frac{coefficientof{x}^{n-1}}{coefficientof{x}^n}$

= $\frac{-0}{1}$ = 0.