Undergrad Equations for functions in the complex domain

[TheCanadian]

When working in the complex domain ($z = x + iy$), how does one write the equation of a line?

I have attached a problem I was working on (and have the solution), but am curious as to why the definition of a line is given by $ax + by = c$. Are not $x$ and $y$ also variables that take on strictly real values? Should not this equation for this function (upon which an arbitrary point $z^* = x^* + iy^*$ will be reflected) be written: $y = -i(\frac {a}{b})x + \frac {c}{b}$ since we are discussing the $z$-plane where the imaginary axis corresponds to the value of $y$?

 

[andrewkirk]

A line is a set of points in the Complex plane that can be specified by three real parameters $a,b,c$ as follows:
$$l_{abc}\triangleq \{x+iy\ :\ x,y\in\mathbb R\wedge ax+by=c\}$$
The equation they have given is a slightly less explicit way of saying that.

Your approach uses $x$ and $y$ in different ways. Note that in yours $y$ is complex rather than real, so it would be better to replace $y$ by $z$. A limitation of your approach is that it cannot specify a vertical line, as that requires $b=0$.

 

[TheCanadian]

A line is a set of points in the Complex plane that can be specified by three real parameters $a,b,c$ as follows:
$$l_{ab}\triangleq \{x+iy\ :\ x,y\in\mathbb R\wedge ax+by=c\}$$
The equation they have given is a slightly less explicit way of saying that.

Your approach uses $x$ and $y$ in different ways. Note that in yours $y$ is complex rather than real, so it would be better to replace $y$ by $z$. A limitation of your approach is that it cannot specify a vertical line, as that requires $b=0$.

Thank you for the response; yes, I see the flaw in my approach as $y$ is imaginary. I guess I'm lacking intuition in this problem of what exactly it means for a point $z$ that is complex being reflected over a line: $ax + by = c$. Isn't this line purely real (e.g. plotted on such a graph)? It appears based on what you're saying that if we use the form $ax + by = c$, that this accounts for $y$ being the imaginary part despite not being explicitly imaginary based on this equation. It just feels like solving this problem amounts to solving it over a normal 2-dimensional Cartesian grid with both real axes. I guess the answer really shouldn't be different if $y$ was the imaginary axis or a real axis if $z$ is converted to just an ordered pair $(x,y)$.

 

[andrewkirk]

I guess the answer really shouldn't be different if $y$ was the imaginary axis or a real axis if $z$ is converted to just an ordered pair $(x,y)$.

Correct. Bear in mind that $ax+by=c$ is an equation, not a line. It is a(n understandable) informal, but common, bending of terminology to call it a 'line'. The line is a set of points $l_{abc}$, defined as in my post #2.

For your problem, you can consider the number plane as being $\mathbb R^2$ for the first part, where you work out the coordinates of the reflected points. None of that uses any properties of complex numbers. Then you need to convert from $\mathbb R^2$ to $\mathbb C$ for the last part, in order for the formula to make sense, as it contains multiplication and division of points in the plane.