What is (i^3+1)/(i^9-2) ?

The solution to (i^3+1)/(i^9-2) is -3/5+1/5 i

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simplify (i^3+1)/(i^9-2)

simplify

\frac{i ^{3} + 1}{i ^{9} - 2}

- \frac{3}{5} + \frac{1}{5} i

Solution steps

\frac{i ^{3} + 1}{i ^{9} - 2}

Simplify

i^{3} : - i

= \frac{- i + 1}{i ^{9} - 2}

Simplify

i^{9} : i

= \frac{- i + 1}{i - 2}

Rewrite

- i + 1

in standard complex form:

1 - i

Rewrite

i - 2

in standard complex form:

- 2 + i

= \frac{1 - i}{- 2 + i}

Multiply by the conjugate

\frac{- 2 - i}{- 2 - i}

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

Simplify

\frac{( 1 - i ) ( - 2 - i )}{( - 2 + i ) ( - 2 - i )} : \frac{- 3 + i}{5}

= \frac{- 3 + i}{5}

Rewrite in standard complex form:

- \frac{3}{5} + \frac{1}{5} i

= - \frac{3}{5} + \frac{1}{5} i

27 t^{3} - 1

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

\frac{5}{9} \cdot \frac{3}{5}

2 x^{2} - 11 x y - 21 y^{2}

3 a + 9 b + 7 a + 7