simplify (-1+i/n)^3

Solution

simplify

(- 1 + \frac{i}{n})^{3}

- \frac{n ^{2} - 3}{n ^{2}} + \frac{3 n ^{2} - 1}{n ^{3}} i

Solution steps

(- 1 + \frac{i}{n})^{3}

Apply Perfect Cube Formula:

(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}

(- 1 + \frac{i}{n})^{3} = (- 1)^{3} + 3 (- 1)^{2} \frac{i}{n} + 3 (- 1) (\frac{i}{n})^{2} + (\frac{i}{n})^{3}

= (- 1)^{3} + 3 (- 1)^{2} \frac{i}{n} + 3 (- 1) (\frac{i}{n})^{2} + (\frac{i}{n})^{3}

(- 1)^{3} = - 1

3 (- 1)^{2} \frac{i}{n} = 3 \cdot \frac{i}{n}

3 (- 1) (\frac{i}{n})^{2} = 3 \cdot \frac{1}{n ^{2}}

(\frac{i}{n})^{3} = - \frac{i}{n ^{3}}

= - 1 + 3 \cdot \frac{i}{n} + 3 \cdot \frac{1}{n ^{2}} - \frac{i}{n ^{3}}

Simplify

- 1 + 3 \cdot \frac{i}{n} + 3 \cdot \frac{1}{n ^{2}} - \frac{i}{n ^{3}} : - \frac{n ^{2} - 3}{n ^{2}} + \frac{3 n ^{2} - 1}{n ^{3}} i

= - \frac{n ^{2} - 3}{n ^{2}} + \frac{3 n ^{2} - 1}{n ^{3}} i

(z + \frac{1}{4}) (z - \frac{1}{2})

(a + x)^{2}

(x cos (x) + sin (x))^{2}

\frac{4 e ^{- 3 t}}{( s - 3 ) ( s + 1 )}

(2 x + \frac{1}{x ^{2}}) (5 - x) (2 x^{5} + 1)^{3}

What is (-1+i/n)^3 ?
The solution to (-1+i/n)^3 is -(n^2-3)/(n^2)+(3n^2-1)/(n^3)i