SOLUTION: 31.) (a+3i)+(2+bi)=i 33.) (5-the square root of -4)-(a+bi)=6 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.

Click here to see ALL problems on Complex Numbers

Question 62402: 31.) (a+3i)+(2+bi)=i
33.) (5-the square root of -4)-(a+bi)=6
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
Found 2 solutions by stanbon, venugopalramana:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
31.) (a+3i)+(2+bi)=i
Combine the Real number terms and the Complex number terms to get:
(a+2)+(3+b)i=0+1i
Equate the Real number of the left with the Real number on the right.
Do the same for the Complex number parts.
a+2=0 implies a=-2
3+b=1 implies b=-2
---------------------
33.) (5-the square root of -4)-(a+bi)=6
(5-2i)-(a+bi)= 6 + 0i
(5-a)+(-2-b)i=6+0i
5-a=6 implies that a=-1
-2-b=0 implies that b=-2
---------
Cheers,
Stan H.

Answer by venugopalramana(3286) (Show Source):
You can put this solution on YOUR website!
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