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I Equations for functions in the complex domain

  1. Oct 3, 2016 #1
    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##?
     

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  2. jcsd
  3. Oct 3, 2016 #2

    andrewkirk

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    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##.
     
    Last edited: Oct 3, 2016
  4. Oct 3, 2016 #3
    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)##.
     
  5. Oct 3, 2016 #4

    andrewkirk

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    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.
     
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