What is ((3-4i)/5)^{-1} ?

The solution to ((3-4i)/5)^{-1} is 3/5+4/5 i

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simplify ((3-4i)/5)^{-1}

simplify

(\frac{3 - 4 i}{5})^{- 1}

\frac{3}{5} + \frac{4}{5} i

Solution steps

(\frac{3 - 4 i}{5})^{- 1}

Apply exponent rule:

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

(\frac{3 - 4 i}{5})^{- 1} = \frac{1}{\frac{3 - 4 i}{5}}

= \frac{1}{\frac{3 - 4 i}{5}}

Apply the fraction rule:

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

= \frac{5}{3 - 4 i}

Multiply by the conjugate

\frac{3 + 4 i}{3 + 4 i}

= \frac{5 ( 3 + 4 i )}{( 3 - 4 i ) ( 3 + 4 i )}

Simplify

\frac{5 ( 3 + 4 i )}{( 3 - 4 i ) ( 3 + 4 i )} : \frac{15 + 20 i}{25}

= \frac{15 + 20 i}{25}

Apply the fraction rule:

\frac{a + bi}{c} = \frac{a}{c} + \frac{b}{c} i

= \frac{15}{25} + \frac{20}{25} i

Simplify

\frac{15}{25} + \frac{20}{25} i : \frac{3}{5} + \frac{4}{5} i

= \frac{3}{5} + \frac{4}{5} i

(\frac{1}{2})^{3} + 2 (\frac{1}{2})

9 u^{2} + 24 u + 16

12 < 4 x < 24

w^{3} - 125

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