Let #z=(1+9i)/(-2+9i)#
Then usual Method to simplify this is to multiply both #Nr.# and #Dr.# by the conjugate of the complex no. of the #Dr.#, i.e., to say,
#z=(1+9i)/(-2+9i) xx (-2-9i)/(-2-9i)#
#=(-2-9i-18i-81i^2)/{-2^2-(9i)^2}#
#=(-2-27i-81(-1))/(4-81i^2)#
#=(79-27i)/85#
#=79/85-27/85i#.
But, we can, as well, solve it as follows :-
Suppose that, #z=x+iy=(1+9i)/(-2+9i)#
#rArr (x+iy)(-2+9i)=1+9i#
#rArr -2x+9ix-2iy+9i^2y=1+9i#
#rArr -2x+i(9x-2y)+9(-1)y=1+9i#
#rArr (-2x-9y)+i(9x-2y)=1+9i#
Comparing the Real and Imaginary Parts of both sides, we get,
#(1) : 2x+9y=-1, and, (2) : 9x-2y=9#
#:. 2xx(1)+9xx(2) rArr (4x+18y)+(81x-18y)=-2+81#
#rArr 85x=79# #rArr x=79/85#.
Using this with #(2)#, we have,
#2y=9(x-1)=9(79/85-1)=9(-6/85)#
#rArr y=-(9*6)/(85*2)=-27/85#, giving,
#z=x+iy=79/85-27/85i#, as before!
Enjoy Maths!
