What is e^{i(2-n)pi}-e^{-i(2-n)pi} ?

The solution to e^{i(2-n)pi}-e^{-i(2-n)pi} is 0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify e^{i(2-n)pi}-e^{-i(2-n)pi}

simplify

e^{i (2 - n) π} - e^{- i (2 - n) π}

0

Solution steps

e^{i (2 - n) π} - e^{- i (2 - n) π}

e^{i (2 - n) π} = (- 1)^{n}

e^{- i (2 - n) π} = (- 1)^{n}

= (- 1)^{n} - (- 1)^{n}

Add similar elements:

(- 1)^{n} - (- 1)^{n} = 0

= 0

\frac{2 ( a + h ) - 5}{( a + h ) - 4}

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

(2 x y - sec^{2} (x)) d x

\frac{4}{( s - 3 ) ( s ^{2} + 9 )}

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

What is e^{i(2-n)pi}-e^{-i(2-n)pi} ?
The solution to e^{i(2-n)pi}-e^{-i(2-n)pi} is 0