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

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

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

simplify

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

1 - i

Solution steps

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

Apply exponent rule:

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

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

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

Apply the fraction rule:

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

= \frac{1 + i}{i}

Multiply by the conjugate

\frac{- i}{- i}

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

Simplify

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

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

Rewrite in standard complex form:

1 - i

= 1 - i

2 x \times 2

- 5 \cdot 1

lo g_{16} (\frac{1}{8})

(a + b) \cdot (a + b)

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