What is ((2-3i)^2-(2+3i)(3-2i))/(3i^{17)-1}-4/(5i^{-25)} ?

The solution to ((2-3i)^2-(2+3i)(3-2i))/(3i^{17)-1}-4/(5i^{-25)} is -17/5+6i

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simplify ((2-3i)^2-(2+3i)(3-2i))/(3i^{17)-1}-4/(5i^{-25)}

simplify

\frac{( 2 - 3 i ) ^{2} - ( 2 + 3 i ) ( 3 - 2 i )}{3 i ^{17} - 1} - \frac{4}{5 i ^{- 25}}

- \frac{17}{5} + 6 i

Solution steps

\frac{( 2 - 3 i ) ^{2} - ( 2 + 3 i ) ( 3 - 2 i )}{3 i ^{17} - 1} - \frac{4}{5 i ^{- 25}}

\frac{( 2 - 3 i ) ^{2} - ( 2 + 3 i ) ( 3 - 2 i )}{3 i ^{17} - 1} = \frac{17 ( - 1 + 2 i )}{5}

\frac{4}{5 i ^{- 25}} = \frac{4 i}{5}

= \frac{17 ( - 1 + 2 i )}{5} - \frac{4 i}{5}

Join

\frac{17 ( - 1 + 2 i )}{5} - \frac{4 i}{5} : \frac{17 ( - 1 + 2 i ) - 4 i}{5}

= \frac{17 ( - 1 + 2 i ) - 4 i}{5}

Expand

17 (- 1 + 2 i) - 4 i : - 17 + 30 i

= \frac{- 17 + 30 i}{5}

Apply the fraction rule:

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

= \frac{- 17}{5} + \frac{30}{5} i

Simplify

\frac{- 17}{5} + \frac{30}{5} i : - \frac{17}{5} + 6 i

= - \frac{17}{5} + 6 i

\frac{14}{15} - \frac{1}{6}

(2 x)^{0}

2 - \frac{3}{2}

∣ x - 7 ∣ > 1

x^{2} - 4 x y - 12 y^{2}

What is ((2-3i)^2-(2+3i)(3-2i))/(3i^{17)-1}-4/(5i^{-25)} ?
The solution to ((2-3i)^2-(2+3i)(3-2i))/(3i^{17)-1}-4/(5i^{-25)} is -17/5+6i