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

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

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

simplify

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

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

Solution steps

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

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

i^{- 1} = - i

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

Apply rule:

a - (- b) = a + b

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

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

Join

\frac{1 - i}{2} + 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

49 a^{2} b^{6}

6 x^{2} - 2 x + 2 x^{2} - 9 x - 10

18 x^{2} + 3 x - 15

- 6 \cdot (- 7)

\frac{3}{12} \div \frac{7}{10}