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I'm trying to solve the following problem:

"Show that any straight line in $\mathbb R^2$ can be represented via the complex equation $\overline a z+a \overline z+b=0$ ; $a\neq 0 \in \mathbb C,b \in \mathbb R$"

"Is the choice of $a,b$ unique for each line? is it unique with the constraint $|a|=1, b\leq 0$? Give a geometric interpetation for $a$."

What I did:

It may be ugly, but I wrote $z=x+iy$ which is our variable, and $a=\alpha+i\beta$ which is our parameter.

After inputting that in $\overline a z+a \overline z+b=0$. I got $2\alpha x+2\beta y+b=0$, this is indeed a representation of a line in $\mathbb R^2$.

My problem arises in the "is this representation unique." The answer is ofcourse no, since we can multiply by whatever number we wish, for example $\alpha x+\beta y+0.5b=0$ is the exact same line as $2\alpha x+2\beta y+b=0$.

But I don't know if this representation is unique under the constraints $|a|=1, b\leq 0$. This is where I could use some assistance.

Regarding geometric interpretation, $a \in \mathbb C$ is the slope of the straight line I would assume?

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  • $\begingroup$ Under the constraints, rewrite as: $x = -\beta y -\frac{b}{2}$ $\endgroup$ – Loreno Heer Nov 5 '14 at 22:01
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You've done fine. To be a line, $Ax + By + C = 0$ must have $A^2 + B^2 \ne 0$, so you can scale up your $a$ to get an $a$ with modulus one When you do, $b$ might be positive; if so, negate $a$.

In short: $a$ and $b$ are not unique. With the constraint, they're ALMOST unique. In the case where $b = 0$, you can negate $a$ and get the same line.

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