solve the equations below for Real numbers x and y;
^{(a).} (3+4i)^{2}-2(x-yi)=x+yi
(3+4i)^{2}-2(x-yi)=x+yi →[(3)^{2}+2*3*4i+(4i)^{2}]-2x+2yi=x+yi→9+24i-16-2x+2yi=x+yi→-2x-7+(2y+24)i=x+yi→-2x-7=x and 2y+24=y→x= -7/3 and y=-24
I cannot understand your second line. What is (b).(1+i ?