Square root of x2+1x2−4i(x−1x)−6 where xεR is equal to :
x2+1x2−4i(x−1x)−6
=x2+1x2+4ix−4i1x−4−2 (Using 1i=−i)
=x2+1x2+4i2+2×x×2i−2×1x×2i−2×x×1x ----------(Using i2=1)
=(x)2+(−1x)2+(2i)2+2×x(−1x)+2×(−1x)×2i+2×x×2i
=(x−1x+2i)2 (Using (a+b+c)2=a2+b2+c2+2ab+2bc+2ac))
Therefore, square root of x2+1x2−4i(x−1x)−6 is ±(x−1x+2i).
Hence, Option (A) is correct.