Question

30 POINTS What is the standard form of (1+2√-3)/(1+√-3) then put it as a fraction

Answer 1

`i=√(-1);\ i^2=-1\\\\√(-3)=√((-1)(3))=√(-1)\cdot\sqrt3=i\sqrt3\\---------------------\\\\(1+2√(-3))/(1+√(-3))=(1+2i\sqrt3)/(1+i\sqrt3)=(1+2i\sqrt3)/(1+i\sqrt3)\cdot(1-i\sqrt3)/(1-i\sqrt3)=((1+2i\sqrt3)(1-i\sqrt3))/((1+i\sqrt3)(1-i\sqrt3))\\\\Use\ (a+b)(a-b)=a^2-b^2\\\\=((1)(1)+(1)(-i\sqrt3)+(2i\sqrt3)(1)+(2i\sqrt3)(-i\sqrt3))/(1^2-(i\sqrt3)^2)\\\\=(1-i\sqrt3+2i\sqrt3-2i^2(\sqrt3)^2)/(1-i^2(\sqrt3)^2)=(1+i\sqrt3-2(-1)(3))/(1-(-1)(3))`

`=(1+i\sqrt3+6)/(1+3)=(7+i\sqrt3)/(4)=(7)/(4)+(\sqrt3)/(4)i\\\\Answer:\ \boxed{(7)/(4)+(\sqrt3)/(4)i}`