If #(4+i)/(2-3i) = x+iy# for real numbers #x, y#, then what are #x# and #y# ?
1 Answer
Mar 17, 2017
Explanation:
#x+iy = (4+i)/(2-3i)#
#color(white)(x+iy) = ((4+i)(2+3i))/((2-3i)(2+3i))#
#color(white)(x+iy) = (8+12i+2i+3i^2)/(2^2-(3i)^2)#
#color(white)(x+iy) = (5+14i)/(4+9)#
#color(white)(x+iy) = (5+14i)/13#
#color(white)(x+iy) = 5/13+14/13i#
So