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

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

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expand (-i(1+i)^n-1+i+(1+i)^n)/((1+i)^n)

expand

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

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

Solution steps

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

Apply the fraction rule:

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

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

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

Apply rule

\frac{a}{a} = 1

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

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

Cancel

\frac{i ( 1 + i ) ^{n}}{( 1 + i ) ^{n}} : i

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

\frac{( 2 ( x + 2 ) ( n ^{\frac{1}{4}} ) )}{( n + 1 ) ^{\frac{1}{4}}}

\frac{Z ^{2} \cdot K}{( Z - 0.95 ) \cdot ( Z + 0.95 )}

\frac{9 x + 6}{( x + 3 ) x ( x - 4 )}

(3 a y + C e^{6 x})^{3}

\frac{x y}{x ^{2}}

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