Question

(aman - RJ) Sendo i=$\sqrt{-1}$ , o resultado$\frac{1+2i}{1-3i}$ + $\frac{i}{1+3i}$ é igual a:

Answer 1

$\frac{1+2 \sqrt{-1} }{1-3 \sqrt{-1} } = \frac{1+3 \sqrt{-1} }{1+3 \sqrt{-1}} = \frac{1+3 \sqrt{-1}+2 \sqrt{-1} +6}{1+3 \sqrt{-1}-3 \sqrt{-1}+9} = \boxed{\frac{7+5 \sqrt{-1} }{10} }$

$\frac{ \sqrt{-1} }{1+3 \sqrt{-1}} = \frac{1-3 \sqrt{-1}}{1-3 \sqrt{-1}} = \boxed{\frac{ \sqrt{-1}-3 }{-2} }$

$\boxed{\frac{7+5 \sqrt{-1} }{10}+ \frac{ \sqrt{-1}\ -3 }{-2}}$