What is (3((4i)/n)^2+8((4i)/n)-5)(4/n) ?

The solution to (3((4i)/n)^2+8((4i)/n)-5)(4/n) is (-192-20n^2)/(n^3)+i(128)/(n^2)

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simplify (3((4i)/n)^2+8((4i)/n)-5)(4/n)

simplify

(3 (\frac{4 i}{n})^{2} + 8 (\frac{4 i}{n}) - 5) (\frac{4}{n})

\frac{- 192 - 20 n ^{2}}{n ^{3}} + i \frac{128}{n ^{2}}

Solution steps

(3 (\frac{4 i}{n})^{2} + 8 (\frac{4 i}{n}) - 5) (\frac{4}{n})

Remove parentheses:

(a) = a

= (3 (\frac{4 i}{n})^{2} + 8 \cdot \frac{4 i}{n} - 5) \frac{4}{n}

3 (\frac{4 i}{n})^{2} = - \frac{48}{n ^{2}}

= \frac{4}{n} (8 \cdot \frac{4 i}{n} - \frac{48}{n ^{2}} - 5)

8 \cdot \frac{4 i}{n} = \frac{32 i}{n}

= \frac{4}{n} (\frac{32 i}{n} - \frac{48}{n ^{2}} - 5)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{4 ( - \frac{48}{n ^{2}} + \frac{32 i}{n} - 5 )}{n}

Join

- \frac{48}{n ^{2}} + \frac{32 i}{n} - 5 : \frac{- 48 + 32 in - 5 n ^{2}}{n ^{2}}

= \frac{4 \cdot \frac{- 5 n ^{2} + 32 in - 48}{n ^{2}}}{n}

Multiply

4 \cdot \frac{- 48 + 32 in - 5 n ^{2}}{n ^{2}} : \frac{4 ( - 5 n ^{2} + 32 in - 48 )}{n ^{2}}

= \frac{\frac{4 ( - 5 n ^{2} + 32 in - 48 )}{n ^{2}}}{n}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{( - 48 + 32 in - 5 n ^{2} ) \cdot 4}{n ^{2} n}

n^{2} n = n^{3}

= \frac{4 ( - 5 n ^{2} + 32 in - 48 )}{n ^{3}}

Rewrite

\frac{( - 48 + 32 in - 5 n ^{2} ) \cdot 4}{n ^{3}}

in standard complex form:

\frac{- 192 - 20 n ^{2}}{n ^{3}} + \frac{128}{n ^{2}} i

= \frac{- 192 - 20 n ^{2}}{n ^{3}} + \frac{128}{n ^{2}} i

(6 - 5 i)^{2}

4 - 3 x + 2 (5.5) = 9

8 x = 128

p^{2} q + r^{2} - pq r - p r

∣ 8 x - 15 ∣ - 4 > 29

What is (3((4i)/n)^2+8((4i)/n)-5)(4/n) ?
The solution to (3((4i)/n)^2+8((4i)/n)-5)(4/n) is (-192-20n^2)/(n^3)+i(128)/(n^2)