Question 459891

9+3b+(3a-b)i=6+6i   (separate real and imaginary values also separate variable on left side and constants on right side.)

=> 3b + (3a-b)i = 6 + 6i -9 

=> 3b + (3a-b)i = -3 + 6i 

now compare real and imaginary values..

i.e 3b = -3  

 and 3a-b = 6

from first equation b = -1 now, put this value on second 

3a+1 = 6 

=> a = 5/3 

so, a= 5/3, b = -1 

if any doubt, you are welcome to contact me..