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Решите задачу:

$( \frac{3}{4}- \frac{1}{6}i)( \frac{6}{7}+ \frac{8}{3}i)= \frac{3}{4} \frac{6}{7}+ \frac{3}{4} \frac{8}{3}i- \frac{1}{6} \frac{6}{7}i- \frac{1}{6} \frac{8}{3} i^{2}= \frac{9}{14}+2i- \frac{1}{7}i- \frac{4}{9}*(-1)=$ $\frac{9}{14}+ \frac{4}{9}+(2-\frac{1}{7})i=\frac{9*9+4*14}{14*9}+1\frac{6}{7}i=\frac{137}{126}+1\frac{6}{7}i=1\frac{11}{126}+1\frac{6}{7}i$