What is ((1+i)^n)/((1-i)^n)-2 ?

The solution to ((1+i)^n)/((1-i)^n)-2 is (-2(1-i)^n+(1+i)^n)/((1-i)^n)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify ((1+i)^n)/((1-i)^n)-2

simplify

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

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

Solution steps

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

Apply the fraction rule:

a + \frac{b}{c} = \frac{a c + b}{c}

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

8^{x} - \frac{1}{2 ^{y}} + 2

6 x - 2 y

\frac{( a ^{10} + b ^{10} )}{( a ^{- 1} 0 + b ^{- 1} 0 )}

\frac{( z - 3 i )}{( 5 i )}

e^{2} t + e^{- 2} t

What is ((1+i)^n)/((1-i)^n)-2 ?
The solution to ((1+i)^n)/((1-i)^n)-2 is (-2(1-i)^n+(1+i)^n)/((1-i)^n)