frac{1 - 2i}{2 + i} + frac{4 - i}{3 + 2i} = | vedclass

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(D) To solve $\frac{1 - 2i}{2 + i} + \frac{4 - i}{3 + 2i}$,we first rationalize each term or find a common denominator.Step $1$: Simplify the first te

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$\frac{10}{13} - \frac{24}{13}i$

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(D) To solve $\frac{1 - 2i}{2 + i} + \frac{4 - i}{3 + 2i}$,we first rationalize each term or find a common denominator.Step $1$: Simplify the first term by multiplying the numerator and denominator by the conjugate of the denominator $(2 - i)$:$\frac{1 - 2i}{2 + i} 	imes \frac{2 - i}{2 - i} = \frac{2 - i - 4i + 2i^2}{4 - i^2} = \frac{2 - 5i - 2}{4 + 1} = \frac{-5i}{5} = -i$.Step $2$: Simplify the second term by multiplying the numerator and denominator by the conjugate of the denominator $(3 - 2i)$:$\frac{4 - i}{3 + 2i} 	imes \frac{3 - 2i}{3 - 2i} = \frac{12 - 8i - 3i + 2i^2}{9 - 4i^2} = \frac{12 - 11i - 2}{9 + 4} = \frac{10 - 11i}{13}$.Step $3$: Add the two results:$-i + \frac{10 - 11i}{13} = \frac{-13i + 10 - 11i}{13} = \frac{10 - 24i}{13} = \frac{10}{13} - \frac{24}{13}i$.

$\frac{1 - 2i}{2 + i} + \frac{4 - i}{3 + 2i} = $

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