[tex]i=\sqrt{-1};\ i^2=-1\\\\\sqrt{-3}=\sqrt{(-1)(3)}=\sqrt{-1}\cdot\sqrt3=i\sqrt3\\---------------------\\\\\dfrac{1+2\sqrt{-3}}{1+\sqrt{-3}}=\dfrac{1+2i\sqrt3}{1+i\sqrt3}=\dfrac{1+2i\sqrt3}{1+i\sqrt3}\cdot\dfrac{1-i\sqrt3}{1-i\sqrt3}=\dfrac{(1+2i\sqrt3)(1-i\sqrt3)}{(1+i\sqrt3)(1-i\sqrt3)}\\\\Use\ (a+b)(a-b)=a^2-b^2\\\\=\dfrac{(1)(1)+(1)(-i\sqrt3)+(2i\sqrt3)(1)+(2i\sqrt3)(-i\sqrt3)}{1^2-(i\sqrt3)^2}\\\\=\dfrac{1-i\sqrt3+2i\sqrt3-2i^2(\sqrt3)^2}{1-i^2(\sqrt3)^2}=\dfrac{1+i\sqrt3-2(-1)(3)}{1-(-1)(3)}[/tex]
[tex]=\dfrac{1+i\sqrt3+6}{1+3}=\dfrac{7+i\sqrt3}{4}=\dfrac{7}{4}+\dfrac{\sqrt3}{4}i\\\\Answer:\ \boxed{\dfrac{7}{4}+\dfrac{\sqrt3}{4}i}[/tex]
