So for my course on Complex Functions, I have the following problem:
Sketch the sets $\left\{z \in \mathbb{C}: \operatorname{Re}\,\left(\frac{z-a}{b}\right) > 0\right\} $ (also $<, =$) for the complex numbers $a = 4 + 7i$, $b=1+3i$.
I thought maybe writing the expression inside $\operatorname{Re}\,(...)$ out, with $z=x+iy$, but not sure if this is the right way to get started! A little help would be appreciated.
I would do \begin{align*} \operatorname{Re}\left[\frac{z-a}{b}\right]&=\operatorname{Re}\left[\frac{x+iy-4-7i}{1+3i}\right]\\ &=\operatorname{Re}\left[\frac{(x-4)+i(y-7)}{1+3i}\cdot\frac{1-3i}{1-3i}\right]\\ &=\operatorname{Re}\left[\frac{(x-4)+i(y-7)-3(x-4)i+3(y-7)}{1+9}\right]\\ &=\operatorname{Re}\left[\frac{(x+3y-25)+i(y-3x+5)}{10}\right]\\ &=\frac{x+3y-25}{10}. \end{align*} Does that get you started?