What is ((2inpi+2)e^{-inpi})/(n^2) ?

The solution to ((2inpi+2)e^{-inpi})/(n^2) is (2(-1)^n}{n^2}+i\frac{2pi(-1)^n)/n

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simplify ((2inpi+2)e^{-inpi})/(n^2)

simplify

\frac{( 2 inπ + 2 ) e ^{- inπ}}{n ^{2}}

\frac{2 ( - 1 ) ^{n}}{n ^{2}} + i \frac{2 π ( - 1 ) ^{n}}{n}

Solution steps

\frac{( 2 inπ + 2 ) e ^{- inπ}}{n ^{2}}

(2 inπ + 2) e^{- inπ} = (2 iπn + 2) (cos (- πn) + i sin (- πn))

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

(2 inπ + 2) (cos (- nπ) + sin (- nπ) i) : 2 (- 1)^{n} + 2 πi (- 1)^{n} n

= \frac{2 ( - 1 ) ^{n} + 2 πi ( - 1 ) ^{n} n}{n ^{2}}

Rewrite

\frac{2 ( - 1 ) ^{n} + 2 π ( - 1 ) ^{n} ni}{n ^{2}}

in standard complex form:

\frac{2 ( - 1 ) ^{n}}{n ^{2}} + \frac{2 π ( - 1 ) ^{n}}{n} i

= \frac{2 ( - 1 ) ^{n}}{n ^{2}} + \frac{2 π ( - 1 ) ^{n}}{n} i

3 (1 - \frac{1}{e})

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

- 3 (4 - r = 9)

\frac{a s}{( s ^{2} + a ^{4} ) ^{2}}

(a c x^{2} - b y) (c x + d y)

What is ((2inpi+2)e^{-inpi})/(n^2) ?
The solution to ((2inpi+2)e^{-inpi})/(n^2) is (2(-1)^n}{n^2}+i\frac{2pi(-1)^n)/n