[tex]5\left(\dfrac{1+i}{1-i}\right)^6+6\left(\dfrac{1-i}{1+i}\right)^5=\\\\5\left(\dfrac{(1+i)^2}{1+1}\right)^6+6\left(\dfrac{(1-i)^2}{1+1}\right)^5=\\\\5\left(\dfrac{(1+i)^2}{2}\right)^6+6\left(\dfrac{(1-i)^2}{2}\right)^5=\\\\5\cdot \dfrac{(1+i)^{12}}{64}+6\cdot \dfrac{(1-i)^{10}}{32}=\\\\5\cdot \dfrac{-64}{64}+6\cdot \dfrac{-32i}{32}=\\\\-5-6i[/tex]
