- This can be written i√6/(i√3*i√4)=-i√6/2√3=-i√(6/3)/2=-i√2/2. So a=0 and b=-√2/2 for a+ib.
- Another way is to calculate √(-6/((-3)(-4))=√(2/-4)=√-½=i√½=i√2/2. In this case b=√2/2 and a=0.
Which is correct? First note that b²=½ for both (1) and (2). But (1) is the correct answer because of the way the expression is given as separate square roots. Each one should be evaluated separately as shown in (1).
In (2), √2/√-4)=1/√-2=1/(i√2). 1/i=-i because if we multiply through by i we get 1=-i²=+1. Hence the negative sign on the final answer. Therefore a=0 and b=-√2/2 or -1/√2 before rationalisation.