Answers

2014-04-17T15:28:47+05:30
Let square root be in form x+iy
therefore  (x+iy)(x+iy)=3+4i
x^2-y^2+2xyi=3+4i
take real and imag. parts separately
x^2-y^2=3,  2xy=4,      
x^4+y^4-2x^2y^2=9

x^4+y^4=9+8=17
( x^{2} +y^2)^2= x^{4} +y^4+2x^2y^2=17+8=25 \\ x^{2} +y^2=5
as
 x^{2} +y^2+ x^{2} -y^2=5+3
x^2=4,y^2=1
so x=2,-2 and y=1,-1
as xy>0
the roots are
2+i,-2-i


0