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

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

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

simplify

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

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

Solution steps

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

Apply exponent rule:

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

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

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

Apply the fraction rule:

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

= \frac{i}{1 - i}

Multiply by the conjugate

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

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

Simplify

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

= \frac{i - 1}{2}

Rewrite in standard complex form:

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

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

4 x (x - 9)

\frac{x y}{x}

\frac{lo g _{10} ( 9 )}{lo g _{10} ( 3 )}

\frac{x ^{3} + 9 x ^{2}}{2 x} \cdot \frac{x ^{2} - 14 x + 45}{x ^{2} - 81}

1 \cdot 1 0^{5}