Find the square root of x-isqrt(x^4+x^2+1)?

#x-isqrt(x^4+x^2+1)#

1 Answer
Jan 9, 2018

#sqrt(x-isqrt(x^4+x^2+1))=sqrt((x^2+x+1)/2)+isqrt((x^2-x+1)/2)#

Explanation:

Let #sqrt(x-isqrt(x^4+x^2+1))=a+ib#

then #x-isqrt(x^4+x^2+1)=(a^2-b^2)+2iab#

As real and imaginary components should be equal, we have

#a^2-b^2=x# (A) and #2ab=sqrt(x^4+x^2+1)#

or #(a^2-b^2)^2=x^2# and #4a^2b^2=x^4+x^2+1#

and as #(a^2+b^2)^2=(a^2-b^2)^2+4a^2b^2#

= #x^2+x^4+x^2+1#

= #x^4+2x^2+1=(x^2+1)^2#

Hence #a^2+b^2=x^2+1# (B)

Adding (A) and (B) we get #2a^2=x^2+x+1#

or #a=sqrt((x^2+x+1)/2)#

and subtracting (B) from (A) and processing similarly we get

#b=sqrt((x^2-x+1)/2)#

and hence

#sqrt(x-isqrt(x^4+x^2+1))=sqrt((x^2+x+1)/2)+isqrt((x^2-x+1)/2)#