# Math Help - equate complex numbers

1. ## equate complex numbers

Is it possible to solve for a and b in this equation?

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

Just trying to help my son with his homework.

Thanks for any information

2. Yes it is, as $a+bi=c+di$ only if $a=c$ and $b=d$

So the two equations you have to solve simultaneously are:

$a+3b=6$ and $3a-b=6$

3. Thanks I did get that far but was wondering where to go from there or is it something just done in my head.

4. If you rearrange the first equation you can see that $a=6-3b$, so now you can plug this value of $a$ into the second equation, yielding $3(6-3b)-b=6$

Solving this gives $b=\frac{6}{5}$

Now if you plug this value of $b$ into either of the first two equations you can find the value of $a$

5. For more info on how to solve simultaneous equations you should have a look here Practical Algebra Lessons

6. WOW!

I actually think I understand it.

Thanks so much