Question

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Вариант 12!!! 3 задание

Answer 1

$1)\; \; (2-8i)\cdot (-2-5i)-(3i+11)=-4-10i+16i+\underbrace {40i^2}_{-40}-3i-11=\\\\=-44+3i\\\\\\2)\; \; \frac{(-3+4i)\cdot (7+i)}{5+9i}=\frac{-21-3i+28i+4i^2}{5+9i}=\frac{-25+25i}{5+9i}=\frac{25(-1+i)(5-9i)}{(5+9i)(5-9i)}\\\\\\=\frac{25(-6+9i+5i-9i^2)}{25-81i^2}=\frac{25(3+14i)}{25+81}=\frac{25(3+14i)}{106}=\frac{75}{106}+\frac{350}{106}\, i=\frac{75}{106}+\frac{175}{53}\, i$