If one root of the equation $$i x^2-2(i+1) x+(2-i)=0$$ is $$(2-i)$$, then the other root is

If $$|z-2|=|z-1|$$, where $$z$$ is a complex number, then locus $$z$$ is a straight line

If $${\left( {{{1 + i} \over {1 - i}}} \right)^m} = 1$$, then m cannot be equal to

$$(\sin \theta-i \cos \theta)^3$$ is equal to