What is ((-1-i)/(1-i))^{-1} ?

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

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simplify ((-1-i)/(1-i))^{-1}

simplify

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

i

Solution steps

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

Apply exponent rule:

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

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

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

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

Multiply by the conjugate

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

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

Simplify

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

= \frac{2 i}{2}

Cancel the common factor:

2

= i

(5 tan (5 x))^{2}

(3 y) (y^{2})

3 \frac{1}{4} + 2 \frac{3}{5}

\frac{9}{2} + 3

- ∣ - 5 ∣