If #(x+iy)^3# is an imaginary number and #|x+iy|=8#, find #x# and #y#?

1 Answer

Answer:

#x=+-4sqrt3 and y=+-4#

Explanation:

#(x+iy)^3#
#=x^3+3x^2iy+3x(iy)^2+(iy)^3#
#=x^3+3x^2iy-3xy^2-iy^3#
#(x^3-3xy^2)+i(3x^2y-y^3)#
As the value is purely imaginary hence
#(x^3-3xy^2)=0#
#x^2=3y^2#
It's also given
#x+iy=8#
#=>।x+iy।=8#
#sqrt(x^2+y^2)=8#

i.e. #x^2+y^2=64# or #3y^2+y^2=64#

or #4y^2=64# i.e. #y=+-4#

and #x^2=48# i.e. #x=+-4sqrt3#
From those equations we get the answer.