complex numbers

yh sorry the actual question is {(bi)/(1+ai)} - {(3b+4i)/(3a+b)}=0 solve for a and b both real?

i managed to do that but once i reach this point i got confused. (3abi)+(b^2i)-(4i)+(4a^2i)=(3b)-(ab^2)

i multiplied the equation on the left with 1-ai and then multiplied out the rest & got the following equations after equating real and imaginary parts:

$4a^2 + 4 = 3ab + b^2$ imaginary

$3a^2 b + ab^2 = 3a^2 b + 3b$ real

Edit: Now you can solve the two simultaneous equations to solve for a and b. If you're still stuck let me know.

(3abi)+(b^2i)-(4i)+(4a^2i)=(3b)-(ab^2)

TBH, I can't see a simple way to go from here.

Why not?

You get a= 3/b sub it into the first equation you just get a quadratic equation which can factorise.

$\frac{(bi)}{(1+ai)}-\frac{(3b+4i)}{(3a+b)}=0$ solve for a and b both are real numbers?