Question

Выполните действия с комплексными числами:

Answer 1

1)

$\frac {(i-1)(1+2i)}{3+i}=\frac{(i+2i^2-1-2i)(3-i)}{(3+i)(3-i)}=\frac {(-3-i)(3-i)}{(3+i)(3-i)}=-1$

2)

$\frac{(5-i)(1+5i)}{-5-12i}=-\frac{(5+25i-i-5i^2)(5-12i)}{(5+12i)(5-12i)}=-\frac {(10+24i)(5-12i)}{(5+12i)(5-12i)}=-\frac {2(5+12i)(5-12i)}{(5+12i)(5-12i)}=-2$

3)

$\frac{5}{3-5i}+\frac {5}{3+5i}=\frac {15+25i}{9-25i^2}+\frac {15-25i}{9-25i^2}=\frac{15+25i}{9+25}+\frac {15-25i}{9+25}=\frac{15+25i+15-25i}{34}=\frac {30}{34}=\frac {15}{17}$

4)

$\frac {1}{2+i}+\frac {1}{2-i}=\frac{2-i}{4-i^2}+\frac {2+i}{4-i^2}=\frac{2-i}{4+1}+\frac {2+i}{4+1}=\frac {2-i+2+i}{5}=\frac {4} {5}$

5)

$\frac {3-i}{3+i}+\frac {3+i} {3-i}=\frac {(3-i)^2}{9-i^2}+\frac {(3+i)^2}{9-i^2}=\frac{9-6i+i^2+9+6i+i^2}{9+1}=\frac {16}{10}=\frac {8}{5}=1,6$

6)

$\frac {4-5i}{4+i}+\frac {4+5i}{4-i}=\frac {(4-5i)(4-i)}{16-i^2}+\frac {(4+5i)(4+i)}{16-i^2}=\frac{16-4i-20i+5i^2+16+4i+20i+5i^2}{16+1}=\frac {22}{17}$