What is (-i+2)^{-1} ?

The solution to (-i+2)^{-1} is 2/5+1/5 i

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simplify (-i+2)^{-1}

simplify

(- i + 2)^{- 1}

\frac{2}{5} + \frac{1}{5} i

Solution steps

(- i + 2)^{- 1}

Apply exponent rule:

a^{- 1} = \frac{1}{a}

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

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

Rewrite

- i + 2

in standard complex form:

2 - i

= \frac{1}{2 - i}

Multiply by the conjugate

\frac{2 + i}{2 + i}

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

Simplify

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

= \frac{2 + i}{5}

Rewrite in standard complex form:

\frac{2}{5} + \frac{1}{5} i

= \frac{2}{5} + \frac{1}{5} i

x (3 x + 2 (x - (2 x + 1)))

(\frac{16}{25})^{- \frac{3}{2}}

4 (5 y + 2 z) + 8 z - 4 (5 z - 2 y)

u^{3} - 64

x^{3} y - 36 x y