[tex]a) \frac{(2+i)(3-5i)+(2-i)(3+5i)}{9+25}= \frac{6-10i+3i+5+6+10i-3i+5}{34}= \\ = \frac{22}{34}= \frac{11}{17} \\ b) \frac{(1+i \sqrt{3})^2+(1-i \sqrt{3})^2 }{4}= \frac{1+2i \sqrt{3}-3+1-2i \sqrt{3}-3 }{4} = \frac{-4}{4}=-1 \\ c) \frac{1+3i-2}{1-3i-2}+ \frac{1-3i-2}{1+3i-2}= \frac{-1+3i}{-1-3i}+ \frac{-1-3i}{-1+3i} = \\ = \frac{1-6i-9+1+6i-9}{10}= \frac{-16}{10}=- \frac{8}{5} [/tex]
