What is e^{(2pii(n-1))/n} ?

The solution to e^{(2pii(n-1))/n} is cos((2pin-2pi)/n)+isin((2pin-2pi)/n)

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simplify e^{(2pii(n-1))/n}

simplify

e^{\frac{2 πi ( n - 1 )}{n}}

cos (\frac{2 πn - 2 π}{n}) + i sin (\frac{2 πn - 2 π}{n})

Solution steps

e^{\frac{2 πi ( n - 1 )}{n}}

Apply imaginary number rule:

e^{ia} = cos (a) + i sin (a)

= cos (\frac{2 πn - 2 π}{n}) + i sin (\frac{2 πn - 2 π}{n})

(\frac{( - 3 x - 2 ) ^{2} x}{6 x - 3}) - 1

(4 e^{6 t} - e^{8 t}) 6 e^{6 t}

\frac{x - y}{i + y} \cdot t

(x - 3) (sin (π x))

\frac{1}{( 1 - 6 t ) ^{4}}

What is e^{(2pii(n-1))/n} ?
The solution to e^{(2pii(n-1))/n} is cos((2pin-2pi)/n)+isin((2pin-2pi)/n)