left( frac{1}{1 - 2i} + frac{3}{1 + i} right) | vedclass

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(D) First,simplify the first term: $\frac{1}{1 - 2i} = \frac{1 + 2i}{(1 - 2i)(1 + 2i)} = \frac{1 + 2i}{1 + 4} = \frac{1 + 2i}{5}$.Next,simplify the se

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$\frac{1}{4} + \frac{9}{4}i$

Vedclass · Answer

(D) First,simplify the first term: $\frac{1}{1 - 2i} = \frac{1 + 2i}{(1 - 2i)(1 + 2i)} = \frac{1 + 2i}{1 + 4} = \frac{1 + 2i}{5}$.Next,simplify the second term: $\frac{3}{1 + i} = \frac{3(1 - i)}{(1 + i)(1 - i)} = \frac{3 - 3i}{1 + 1} = \frac{3 - 3i}{2}$.Adding them: $\frac{1 + 2i}{5} + \frac{3 - 3i}{2} = \frac{2(1 + 2i) + 5(3 - 3i)}{10} = \frac{2 + 4i + 15 - 15i}{10} = \frac{17 - 11i}{10}$.Now,simplify the second bracket: $\frac{3 + 4i}{2 - 4i} = \frac{(3 + 4i)(2 + 4i)}{(2 - 4i)(2 + 4i)} = \frac{6 + 12i + 8i + 16i^2}{4 + 16} = \frac{6 + 20i - 16}{20} = \frac{-10 + 20i}{20} = \frac{-1 + 2i}{2}$.Finally,multiply the two results: $\left( \frac{17 - 11i}{10} \right) \left( \frac{-1 + 2i}{2} \right) = \frac{-17 + 34i + 11i - 22i^2}{20} = \frac{-17 + 45i + 22}{20} = \frac{5 + 45i}{20} = \frac{1}{4} + \frac{9}{4}i$.

$\left( \frac{1}{1 - 2i} + \frac{3}{1 + i} \right) \left( \frac{3 + 4i}{2 - 4i} \right) = $

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