What is (32-32i)^{-6} ?

The solution to (32-32i)^{-6} is -1/(32^6*8)i

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simplify (32-32i)^{-6}

simplify

(32 - 32 i)^{- 6}

- \frac{1}{3 2 ^{6} \cdot 8} i

Solution steps

(32 - 32 i)^{- 6}

Apply exponent rule:

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

(32 - 32 i)^{- 6} = \frac{1}{( 32 - 32 i ) ^{6}}

= \frac{1}{( 32 - 32 i ) ^{6}}

Expand

(32 - 32 i)^{6} : 8 \cdot 3 2^{6} i

= \frac{1}{8 \cdot 3 2 ^{6} i}

Multiply by the conjugate

\frac{- i}{- i}

= \frac{1 \cdot ( - i )}{8 \cdot 3 2 ^{6} i ( - i )}

Simplify

\frac{1 \cdot ( - i )}{8 \cdot 3 2 ^{6} i ( - i )} : \frac{- i}{3 2 ^{6} \cdot 8}

= \frac{- i}{3 2 ^{6} \cdot 8}

Rewrite in standard complex form:

- \frac{1}{3 2 ^{6} \cdot 8} i

= - \frac{1}{3 2 ^{6} \cdot 8} i

(z - \frac{1}{z})^{5}

\frac{2}{5 - 6.01}

lo g_{- 1} (2)

20 x^{9}

6 \cdot 6^{- 2}