Write the following numbers in a + bi form: (a) $ \frac{-2 + i}{3 + 4i} $ = $ \_\_\_ $ + $ \_\_\_ $i....

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Write the following numbers in a + bi form:
(a) $ \frac{-2 + i}{3 + 4i} $ = $ \_\_\_ $ + $ \_\_\_ $i.
(b) $ \frac{3}{4i} + \frac{2}{2i} $ = $ \_\_\_ $ + $ \_\_\_ $i.
(c) $ (i)^3 $ = $ \_\_\_ $ + $ \_\_\_ $i.

SnapAndSolve AI · Accepted Answer

To simplify $ \frac{-2 + i}{3 + 4i} $, multiply numerator and denominator by the conjugate of the denominator: $ 3 - 4i $.$$ \frac{(-2 + i)(3 - 4i)}{(3 + 4i)(3 - 4i)} = \frac{-6 + 8i + 3i - 4i^2}{9 + 16} = \frac{-6 + 11i + 4}{25} = \frac{-2 + 11i}{25} $$$$ = -\frac{2}{25} + \frac{11}{25}i $$ To simplify $ \frac{3}{4i} + \frac{2}{2i} $, rewrite each term:$$ \frac{3}{4i} = \frac{3}{4i} \times \frac{-i}{-i} = -\frac{3i}{4i^2} = \frac{3i}{4} $$ (since $ i^2 = -1 $)$$ \frac{2}{2i} = \frac{2}{2i} \times \frac{-i}{-i} = -\frac{2i}{2i^2} = i $$Combine the terms: $ \frac{3}{4} + i $ $ (i)^3 = i^2 \cdot i = (-1) \cdot i = -i $$$ = 0 - i $$

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Write the following numbers in a + bi form: (a) \( \frac{-2 + i}{3 + 4i} \) = \( \_\_\_ \) + \( \_\_\_ \)i....

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