Question

Z1=5-6i
z2=3+2i
найти произведение и частное комплексных чисел
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Answer 1

$1) z_{1} *z_{2} = 27-8i\\2) \frac{z_{1}}{z_{2}} = \frac{2}{13} - \frac{28}{13} i\\3) \frac{z_{2}}{z_{1}} = \frac{2}{61}+ \frac{28}{61} i$

Пошаговое объяснение:

$1) z_{1} *z_{2} = (5-6i)*(3+2i) = 15+10i-18i-12i^{2} = 15+12-8i = 27-8i\\2) \frac{z_{1}}{z_{2}} = \frac{5-6i}{3+2i} = \frac{(5-6i)*(3-2i)}{(3+2i)*(3-2i)} = \frac{15-10i-18i+12i^{2}}{9+4} = \frac{2-28i}{13} = \frac{2}{13} - \frac{28}{13} i\\3) \frac{z_{2}}{z_{1}} = \frac{3+2i}{5-6i} = \frac{(3+2i)*(5+6i)}{(5-6i)*(5+6i)} = \frac{15+10i+18i+12i^{2}}{25+36} = \frac{2+28i}{61} = \frac{2}{61}+ \frac{28}{61} i$