Answer
$\dfrac{1+i}{1+2i}+\dfrac{1-i}{1-2i}=\dfrac{6}{5}$
Work Step by Step
$\dfrac{1+i}{1+2i}+\dfrac{1-i}{1-2i}$
Evaluate the sum of fractions:
$\dfrac{1+i}{1+2i}+\dfrac{1-i}{1-2i}=\dfrac{(1+i)(1-2i)+(1+2i)(1-i)}{(1+2i)(1-2i)}=...$
$...=\dfrac{1-2i+i-2i^{2}+1-i+2i-2i^{2}}{1^{2}-(2i)^{2}}=\dfrac{2-4i^{2}}{1-4i^{2}}=...$
Substitute $i^{2}$ by $-1$ and simplify:
$...=\dfrac{2-4(-1)}{1-4(-1)}=\dfrac{2+4}{1+4}=\dfrac{6}{5}$
