# How do you solve 4+ 2i - ( a + 4i ) = 9- 2i?

Jun 20, 2018

Group the real part and imaginary part and find what you need to.

#### Explanation:

Okay, let me explain. What you have here is a complex number. Right now we have to put it in form $a + b i$. In that form we call $a$ the real part and $b$ the imaginary part. This because $b$ is multiplied by $i$, which is nowhere on the real line. This makes things much easier. You will often see the notation of $I m \left(z\right)$ as a function that gives you back the imaginary part and $R e \left(z\right)$ as the function that give you the real part (ex. $I m \left(4 + 3 i\right) = 3$, $R e \left(4 + 3 i\right) = 4$). I'm going yo use that notation, so you can start gettin used to it.

Let's make $4 + 2 i - \left(a + 4 i\right)$ into $a + b i$ form.
$4 + 2 i - \left(a + 4 i\right) = 4 + 2 i - a - 4 i = 4 - a + 2 i - 4 i = \left(4 - a\right) + \left(2 - 4\right) i = \left(4 - a\right) - 2 i$

So we know that $\left(4 - a\right) - 2 i = 9 - 2 i$. We can use the $R e$ function here now, to solve for $a$.

$R e \left(\left(4 - a\right) - 2 i\right) = R e \left(9 - 2 i\right)$
$4 - a = 9$

And so, we finally claim our victory:
$4 = 9 + a$
$a = - 5$