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Why the equation of an arbitrary straight line in complex plane is $zz_o + \bar z \bar z_0 = D$ where D $\in R$

I understand that a vertical straight line can be defined by the equation $z+\bar z= D$ because suppose $z =x+yi$ then $\bar z = x-yi$ Thus, $z+\bar z = x+yi+x-yi=2x$ which is an arbitrary vertical straight line in w-plane.

But why $zz_o + \bar z \bar z_0 = D$ is an arbitrary straight line in complex plane?

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4 Answers 4

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You know that a vertical straight line can be defined as $z+\bar z= D$, so if you rotate it's points with angle $\theta$ you get $(e^{i\theta}z)+ \overline{(e^{i\theta}z)}= D$ or $e^{i\theta}z + e^{-i\theta}\overline{z}= D$ and with arbitrary real $r\neq0$, $$re^{i\theta}z + re^{-i\theta}\overline{z}= rD$$ gives us $$zz_0 +\bar{z}\bar{z_0}= D_0$$ where $z_0=re^{i\theta}$ and $D_0=rD$.

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  • $\begingroup$ Why we can rotate its points by multiply $e^{i \theta}$ $\endgroup$
    – Parting
    Commented Apr 30, 2017 at 4:50
  • $\begingroup$ To find arbitrary sloped line! $\endgroup$
    – Nosrati
    Commented Apr 30, 2017 at 4:52
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HINT:

$$z+\bar z=2\text{Re}(z)\implies zz_0+\bar z\bar z_0=2\text{Re}(zz_0)$$

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    $\begingroup$ But we still only get an arbitrary vertical straight line not a straight line right? $\endgroup$
    – Parting
    Commented Apr 30, 2017 at 4:08
  • $\begingroup$ We get any line we want. Note $\text{Re}(zz_0)=xx_0-yy_0$ $\endgroup$
    – Mark Viola
    Commented Apr 30, 2017 at 4:09
  • $\begingroup$ I still don't get it. Once we get a real part a complex number. The equation becomes w=1,2,3,4 or any real numbers , but there are all vertical lines $\endgroup$
    – Parting
    Commented Apr 30, 2017 at 4:16
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    $\begingroup$ $xx_0-yy_0=D\implies y=\frac{x_0}{y_0}x-\frac{D}{y_0}$ for $y_0\ne 0$. That is the equation of a straight line. $\endgroup$
    – Mark Viola
    Commented Apr 30, 2017 at 4:18
  • $\begingroup$ is x the coordinate in real part and y the coordinate in imaginary part? $\endgroup$
    – Parting
    Commented Apr 30, 2017 at 4:38
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Hint: given any two points $z_1, z_2 \in \mathbb{C}\,$, then $z$ is collinear with $z_1, z_2$ iff there exists $\lambda \in \mathbb{R}$ such that $z-z_1 = \lambda(z-z_2)$. Eliminate $\lambda$ between the following, then define $z_0, D$ appropriately:

$$ \begin{cases} \begin{align} z-z_1 &= \lambda(z-z_2) \\ \bar z- \bar z_1 &= \lambda(\bar z- \bar z_2) \end{align} \end{cases} $$

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Hack the equation.

Substitute:

$$ \begin{cases} z = x + iy \\ z_0 = x_0 + iy_0 \end{cases} $$

Do some algebraic manipulations and you'll obtain the equation of a line.

Maybe what confuses you is that neither $z_0$ nor $D$ have a clear geometric interpretation in terms of intercepts, distance to the origin, etc...

Hope this helps!

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    $\begingroup$ I do the manipulation and I get $2xx_0-2yy_0 =R$ and is it a straight line about x and y? and is x the coordinate in real part and y the coordinate in imaginary part? $\endgroup$
    – Parting
    Commented Apr 30, 2017 at 4:36
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    $\begingroup$ That's the equation of a line in $\mathbb{R}^2$ which is isomorphic with $\mathbb{C}$ and $z = x + iy$ $\endgroup$
    – user410935
    Commented Apr 30, 2017 at 4:40

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