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$\frac{1-2 i}{2+i}+\frac{4-i}{3+2 i}=$
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Verified Answer
The correct answer is:
$\frac{10}{13}-\frac{24}{13} i$
$\begin{aligned} & \frac{1-2 i}{2+i}+\frac{4-i}{3+2 i}=\frac{(1-2 i)(3+2 i)+(4-i)(2+i)}{(2+i)(3+2 i)} \\ & =\frac{50-120 i}{65}=\frac{10}{13}-\frac{24}{13} i\end{aligned}$
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