I know that $ax+by+c=0$, with $a^2+b^2>0$ is represented in the cartesian plane like a line. Now if $ax+by+c=0$ and the complex number is $z=x+iy$, then
$x=\frac{1}{2} (z+\bar{z})$ and $y=\frac{1}{2i} (z-\bar{z})$
My book show me the next transformation steps, to convert a cartesian equation line to complex:
$ax+by+c=0$
$a[\frac{1}{2} (z+\bar{z})]+b[\frac{1}{2i} (z-\bar{z})]+c=0$
$\frac{1}{2} (a-ib)z+\frac{1}{2} (a+bi)\bar{z}+c=0$
$\bar{A}z+A\bar{z}+2c=0$
$2Re(\bar{A}z)+2c=0$
$Re(\bar{A}z)+c=0$
Where $A=a+ib$ and $c\in\mathbb{R}$. If $c=0$ then $Re(\bar{A}z)=0$ In other words $z$ is perpendicular to $A$… (But I dont undestand some steps like number 3 or 4 or the conclusion) "What are the correct steps to convert the equation of the straight line to the complex plane??"
Convert in terms of complex numbers, the line whose equation is $2x-y=3$. I convert it, taking the coefficients $a=2,b=-1 and c=-3$. In consequence $A=2-i$ and $c=-3$ Then the line equation is $Re((3+i)z)-3=0$ ("Is this true??")
Best Answer
The procedure described in the books looks unnecessarily complicated. The most natural thing (to me) is to use the formula $$ \operatorname{Re}((a+ib)\overline{(u+iv)}) = au+bv = (a,b)\cdot (u,v) $$ which expresses the dot product of real vectors in terms of complex numbers. Here the conjugation can go over either factor: $$ \operatorname{Re}(\overline{(a+ib)}{(u+iv)}) = \operatorname{Re}((a+ib)\overline{(u+iv)}) = (a,b)\cdot (u,v) $$
In your problem, the expression $2x-y$ should be recognized as $(2,-1)\cdot (x,y)$. By the above, this is equal to $\operatorname{Re}(\overline{(2-i)}{(x+iy)})$. Hence the answer: $$\operatorname{Re}((2+i)z) = 3$$
Which is what you have at the end, except you've made a typo replacing $2$ by $3$.