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Popular >

simplify (1-i)^{-1}

Solution

simplify

(1 - i)^{- 1}

Solution

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

Solution steps

(1 - i)^{- 1}

Apply exponent rule:

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

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

= \frac{1}{1 - i}

Multiply by the conjugate

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

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

Simplify

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

= \frac{1 + i}{2}

Rewrite in standard complex form:

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

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

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Frequently Asked Questions (FAQ)

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