# Solving Simultaneous Equations with Complex Numbers

I've just started studying Computer Science at university and have been thrown into the deep end with mathematics. I haven't done maths since 2013 so it is like learning it all over again.

We have began with complex numbers which I have never done before and is leaving me confused.

In one of the questions we were given, we are asked the following:

Solve for $a$ and $b$

$(a - 3bi) + (b - 2ai) = 4 + 6i$

($i^2=-1$)

Could someone give me a break down on how to solve this as I haven't been able to move past it.

• is this $$a+3bi+b-2ai=4+6i$$? – Dr. Sonnhard Graubner Sep 16 '17 at 13:47
• Equate real and imaginary parts to get $a+b = 4, -3b-2a = 6$. Now you can solve for $a,b$ – user348749 Sep 16 '17 at 13:47
• Hint. Just equate the real and imaginary parts. And learn to use mathjax if you want to ask questions on this site. math.meta.stackexchange.com/questions/5020/… – Ethan Bolker Sep 16 '17 at 13:48

So when you add two complex numbers, you need to keep track of the real parts and the imaginary parts separately. In your case, $a + b = 4$ and $-2a -3b = -6$. So $a =6$ and $b=-2$.
$z=a-3bi$ , $d=b-2ai$ and $x=4+6i$
What $\text{Re}(z)$ represents is the reel part of $z$ and what $\text{Im}(z)$ represents is the imaginary part
$$\text{Re}(z)=a, \quad \text{Im}(z)=-3b$$ $$\text{Re}(d)=b, \quad \text{Im}(d)=-2a$$ $$\text{Re}(x)=4, \quad \text{Im}(x)=6$$ sonra we get $$a+b=4$$ and $$-3b-2a=6$$ write $b$ in terms of $a$ and...