What is e^{(pi(i-1))/3} ?

The solution to e^{(pi(i-1))/3} is 1/(2e^{pi/3)}+i(sqrt(3))/(2e^{pi/3)}

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simplify e^{(pi(i-1))/3}

simplify

e^{\frac{π ( i - 1 )}{3}}

\frac{1}{2 e ^{\frac{π}{3}}} + i \frac{3}{2 e ^{\frac{π}{3}}}

Solution steps

e^{\frac{π ( i - 1 )}{3}}

Apply imaginary number rule:

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

= e^{- \frac{π}{3}} (cos (\frac{π}{3}) + i sin (\frac{π}{3}))

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

e^{- \frac{π}{3}} = \frac{1}{e ^{\frac{π}{3}}}

= \frac{1}{e ^{\frac{π}{3}}} (cos (\frac{π}{3}) + i sin (\frac{π}{3}))

cos (\frac{π}{3}) = \frac{1}{2}

sin (\frac{π}{3}) i = \frac{3 i}{2}

= \frac{1}{e ^{\frac{π}{3}}} (\frac{3 i}{2} + \frac{1}{2})

Combine the fractions

\frac{1}{2} + \frac{3 i}{2} : \frac{1 + 3 i}{2}

= (\frac{1 + 3 i}{2}) \frac{1}{e ^{\frac{π}{3}}}

Remove parentheses:

(a) = a

= \frac{1}{e ^{\frac{π}{3}}} \cdot \frac{1 + 3 i}{2}

Multiply fractions:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{1 \cdot ( 1 + 3 i )}{e ^{\frac{π}{3}} \cdot 2}

1 \cdot (1 + 3 i) = 1 + 3 i

= \frac{1 + 3 i}{2 e ^{\frac{π}{3}}}

Rewrite

\frac{1 + 3 i}{e ^{\frac{π}{3}} \cdot 2}

in standard complex form:

\frac{1}{2 e ^{\frac{π}{3}}} + \frac{3}{2 e ^{\frac{π}{3}}} i

= \frac{1}{2 e ^{\frac{π}{3}}} + \frac{3}{2 e ^{\frac{π}{3}}} i

(\frac{x}{2 + 1})^{5}

\frac{( x - 3 ) ( x - 2 )}{( x + 4 ) ( 2 x - 1 )}

\frac{( 5 - t ) ^{2}}{( t + 6 )}

\frac{3 ( 1 + 3 )}{2}

(e^{- 1600 t} - e^{- 400 t})^{2}

What is e^{(pi(i-1))/3} ?
The solution to e^{(pi(i-1))/3} is 1/(2e^{pi/3)}+i(sqrt(3))/(2e^{pi/3)}