Evaluate `sqrt(-i)`.

1.	Evaluate `sqrt(-i)`.
Answer» Let `sqrt(-i)=(x+iy)." "...(i)` On squaring both sides of (i), we get `(-i)=(x-iy)^(2) rArr -i = (x^(2)-y^(2))-i(2xy)." "...(ii)` On comparing real parts and imaginary parts on both sides of (ii), we get `(x^(2)-y^(2))=0 and 2xy=1` `therefore" "(x^(2)+y^(2))=sqrt((x^(2)-y^(2))^(2)+4x^(2)y^(2))=sqrt(0^(2)+ 1^(2))=sqrt(0+1)=sqrt(1)=1` Thus, `(x^(2)-y^(2))=0" "...(iii) and (x^(2)+y^(2))=1" "...(iv)`. On solving (iii) and (iv), we get `x^(2)=(1)/(2) and y^(2)=(1)/(2)`. `therefore" "x = +-(1)/(sqrt(2))and y = +-(1)/(sqrt(2))`. Since `xy gt 0`, so x and y are of the same sign. `therefore" "(x=(1)/(sqrt(2)),y=(1)/(sqrt(2)))or(x=(-1)/(sqrt(2)),y=(-1)/(sqrt(2)))`. Hence, `sqrt(-i)=((1)/(sqrt(2))-i.(1)/(sqrt(2)))or((-1)/(sqrt(2))+i.(1)/(sqrt(2)))`.

Discussion