Question

Помогите решить комплексные числа пожалуйста!!!!!!!)

Answer 1

$\frac{10i}{3+i}-\frac{3-4i}{1-2i}=\frac{10i(3-i)}{(3+i)(3-i)}-\frac{(3-4i)(1+2i)}{(1-2i)(1+2i)}=\\\\=\frac{30i-10i^2}{9-i^2}-\frac{3+6i-4i-8i^2}{1-4i^2}=\frac{30i+10}{9+1}-\frac{2i+3+8}{1+4}=\\\\=\frac{30i+10-2(2i+11)}{10}=\frac{26i-12}{10}=2,6i-1,2\\\\2)\; \; i^4=(i^2)^2=(-1)^2=1\\\\i^6=(i^2)^3=(-1)^3=-1\\\\i^{13}=i\cdot i^{12}=i\cdot (i^2)^6=i\cdot 1=i\\\\i^{22}=(i^2)^{11}=(-1)^{11}=-1\\\\i^{23}=i\cdot i^{22}=-i$

$\frac{1}{i^4}-\frac{4}{i^6}+\frac{1}{i^{13}}-\frac{5}{i^{22}}+\frac{1}{i^{23}}=\frac{1}{1}-\frac{4}{-1}+\frac{1}{i}-\frac{5}{-1}+\frac{1}{-i}=\\\\=1+4+\frac{1}{i}+5-\frac{1}{i}=1+4+5=10$