Question 62402
31.) (a+3i)+(2+bi)=i
2A+ABi+6i+3B(i)^2=i..........i^2=-1
(2A-3B)+i(AB+6)=i..REAL PARTS AND IMAGINARY PARTS ON BOTH SIDES OF = MUST BE SAME. HENCE
2A-3B=0....A=3B/2=1.5B
AB+6=1
1.5B*B=-5
B^2=-5/1.5=-10/3
B= i*SQRT(10/3)
A=i*1.5*SQRT(10/3)
 
33.) (5-the square root of -4)-(a+bi)=6 
USING SAME LOGIC AND METHOD AS ABOVE
5-4i-A-Bi=6
(5-A)+i(-4-B)=6
5-A=6
A=5-6=-1
-4-B=0
B=-4
I know how to do the problems that do not have an equal to (ie =i or =6) but I do not understand how to do these. Please help