Expressing as $z(t) = x(t) + iy(t)$ My exercise requires to express the equation of a line connecting two point, $z_1 = -3 +2i$ and $z_2 = 3 - 5i$, as  $z(t) = x(t) + iy(t)$.
We know that the equation for a line is $$ Re\left [ (m+i)z + b \right ] = 0$$ where $z = x+iy$
The slope is calculated as $$m = \frac{Im(z_1) - Im(z_2)}{Re(z_1) - Re(z_2)}$$
For $z_1 = -3 + 2i$ and $z_2 = 3 - 5i$, we calculate the slope to be: $m = \frac{7}{-6}$.
$$b =\frac{ [|z_2| - |z_1|]}{2(Re(z_2) - Re(z_1))}$$
$$ = \frac{ \sqrt{34 }- \sqrt{13} }{2(6)}$$
Then the line is $Re\left [ (-\frac{7}{6} + i )z + \frac{ \sqrt{34 }- \sqrt{13} }{2(6)} \right ] = 0$
However, I do not know how to get it into the required form. 

The form your exercise is requesting is often called the "parametric" version of the equation, just to throw the terminology out there. 
To find the slope, think in terms of the growth of the line. This is expressed by $z_2-z_1$, the vector between the two points, or $6-7i$, with the calculation done. 
All we need to do now to fully define the line segment is add the initial point/position vector, $z_1$ (not that different from with the reals, you may notice). This gives us: 
$z(t)=(-3+2i)+t(6-7i).$
