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Ilyes39
[tex] \frac{7-4i}{2+i} =\frac{(7-4i)(2-i)}{(2+i)(2-i)} = \frac{14-7i-8i+4i^2}{2^2 -i^2} \\\\ \frac{14-15i-4}{4+1} = \frac{10-15i}{5}= \frac{5(2-3i)}{5} \text{ because } i^2=-1.\\\\ =2-3i. [/tex]

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