Drawing a region in the complex plane Let $a,b \in \mathbb{C}$, $b \neq 0 $. and let 

$$ G_0 = \{ z \in \mathbb{C} : Im ( \frac{ z -a }{b} ) = 0 \} $$

Question is to draw this set. My attempt: Let $z = (x,y) , a = ( \alpha, \beta) , b = ( \alpha', \beta') $.  After doing computation I find that the $y$ coordinate of the complex number $w = \frac{ z -a }{b} $ is 

$$ \frac{-(x - \alpha) \beta' + (y- \beta) \alpha'}{(\alpha')^2 + (\beta')^2} = Im(w)$$

In particular, $Im(w) = 0 $ implies 

$$ (x-\alpha)\beta' = (y-\beta) \alpha' $$

iff

$$ y = \beta' x + ( \beta - \frac{ \alpha \beta'}{\alpha'} )$$

So, we have a line with slope $\beta'$ and $yintercept$ $\beta - \frac{ \alpha \beta'}{\alpha'} $. This is a correct drawing? thanks
Hint. A complex number has imaginary part equal to zero f and only if it is a real number.
Hint. A complex number has imaginary part equal to zero f and only if it is a real number.