# Thread: general form of lines and circles in complex plane

1. ## general form of lines and circles in complex plane

Prove the equation of a line and a circle in complex plane has a general form of :

$\alpha \ z\overline{z}+\beta \ z + \overline{\beta\ z}+ \gamma = 0$

where $\alpha,\gamma \in \mathbb{R}, \beta \in \mathbb{C}$

Hence, or otherwise, prove

If $z,z_{1},z_{2}$ are complex numbers

which satisfies below condition:

$|z-z_{1}| =k |z-z_{2}|, k \neq 0,1$

then they are locus of circles in complex plane.

2. ## Re: general form of lines and circles in complex plane

Originally Posted by Mahan1
Prove the equation of a line and a circle in complex plane has a general form of :

$\alpha \ z\overline{z}+\beta \ z + \beta \ \overline{z}+ \gamma = 0$

where $\alpha,\gamma \in \mathbb{R}, \beta \in \mathbb{C}$

Hence, or otherwise, prove

If $z,z_{1},z_{2}$ are complex numbers

which satisfies below condition:

$|z-z_{1}| =k |z-z_{2}|, k \neq 0,1$

then they are locus of circles in complex plane.
$\noindent The general equation of a circle in the complex plane should be$

$\alpha z \bar{z} + \beta z + \bar{\beta} \bar{z} + \gamma = 0.$

$\noindent \textbf{Proof for first part}$

$\noindent As should be well-known, the complex numbers z satisfied by the set of points given by$

$\{z \in \mathbb{C} : |z - z_0| = r, z_0 \in \mathbb{C}, r \in \mathbb{R}\}$

$\noindent Corresponds to a circle with centre z_0 and radius r.$

$\noindent Squaring the equation |z - z_0| = r gives |z - z_0|^2 = r^2.$

$\noindent From the result z \bar{z} = |z|^2 the squared equation above can be rewritten as$

\begin{align*}(z - z_0)\overline{(z - z_0)} &= r^2\\\Rightarrow (z - z_0)(\bar{z} - \bar{z}_0) &= r^2\\z \bar{z} - z \bar{z}_0 - \bar{z} z_0 + z_0 \bar{z}_0 &= r^2.\end{align*}

$\noindent Multiplying both sides of this equation by \alpha \in \mathbb{R} and rearranging gives$

$\alpha z \bar{z} - (\alpha \bar{z}_0) z - (\alpha z_0) \bar{z} + \alpha z_0 \bar{z}_0 - \alpha r^2 = 0. \quad (*)$

$\noindent Now as \alpha is real we have \alpha \bar{z}_0 = \overline{\alpha z_0}. Also, since z_0 \bar{z}_0 = |z_0|^2 \in \mathbb{R}, let \gamma = \alpha z_0 \bar{z}_0 - \alpha r^2 \in \mathbb{R} and let \alpha z_0 = -\bar{\beta} where \beta \in \mathbb{C}. Thus \overline{\alpha z_0} = -\beta.$

$\noindent So Eq. (*) becomes \alpha z \overline{z} + \beta z + \overline{\beta} \,\overline{z} + \gamma = 0, as required to prove.$

3. ## Re: general form of lines and circles in complex plane

Originally Posted by Mahan1
Prove the equation of a line and a circle in complex plane has a general form of :

$\alpha \ z\overline{z}+\beta \ z + \beta \ \overline{z}+ \gamma = 0$

where $\alpha,\gamma \in \mathbb{R}, \beta \in \mathbb{C}$

Hence, or otherwise, prove

If $z,z_{1},z_{2}$ are complex numbers

which satisfies below condition:

$|z-z_{1}| =k |z-z_{2}|, k \neq 0,1$

then they are locus of circles in complex plane.
$\noindent \textbf{Proof for the second part}$

$\noindent Squaring the equation |z - z_1| = k|z - z_2| and making use of the result w \bar{w} = |w|^2 where w \in \mathbb{C} we have$

\begin{align*}|z - z_1|^2 &= k^2|z - z_2|^2\\(z - z_1)\overline{(z - z_1)} &= k^2(z - z_2)\overline{(z - z_2)}\\(z - z_1)(\bar{z} - \bar{z}_1) &= k^2(z - z_2)(\bar{z} - \bar{z}_2)\\z\bar{z} - z\bar{z}_1 \bar{z}z_1 + z_1 \bar{z}_1 &= k^2(z \bar{z} - z \bar{z}_2 - \bar{z} z_2 + z_2 \bar{z}_2)\end{align*}

$\noindent After rearranging and collecting like terms we have$

$(1 - k^2) z \bar{z} + z(k^2 \bar{z}_2 - \bar{z}_1) + \bar{z} (k^2 z_2 - z_1) + (z_1 \bar{z}_1 - k^2 z_2 \bar{z}_2) = 0. \quad (*)$

$\noindent Since k \in \mathbb{R}, let \alpha = 1 - k^2 \in \mathbb{R}. Also, let \beta = k^2 z_2 - z_1 \in \mathbb{C}, then \bar{\beta} = \overline{k^2 z_2 - z_1} = k^2 \bar{z}_2 - \bar{z}_1. Finally, since z_1 \bar{z}_1 = |z_1|^2 \in \mathbb{R} and z_2 \bar{z}_2 =|z_2|^2 \in \mathbb{R}, let \gamma = z_1 \bar{z}_1 - k^2 z_2 \bar{z}_2 \in \mathbb{R}.$

$\noindent So Eq. (*) becomes \alpha z\bar{z} + \beta z + \bar{\beta} \, \bar{z} + \gamma = 0, which in the first part we proved is the general form for the equation of a circle in an Argand diagram. Hence proven.$

4. ## Re: general form of lines and circles in complex plane

Originally Posted by omegadot
$\noindent The general equation of a circle in the complex plane should be$

$\alpha z \bar{z} + \beta z + \bar{\beta} \bar{z} + \gamma = 0.$

$\noindent \textbf{Proof for first part}$

$\noindent As should be well-known, the complex numbers z satisfied by the set of points given by$

$\{z \in \mathbb{C} : |z - z_0| = r, z_0 \in \mathbb{C}, r \in \mathbb{R}}$

$\noindent Corresponds to a circle with centre z_0 and radius r.$

$\noindent Squaring the equation |z - z_0| = r gives |z - z_0|^2 = r^2.$

$\noindent From the result z \bar{z} = |z|^2 the squared equation above can be rewritten as$

\begin{align*}(z - z_0)\overline{(z - z_0)} &= r^2\\\Rightarrow (z - z_0)(\bar{z} - \bar{z}_0) &= r^2\\z \bar{z} - z \bar{z}_0 - \bar{z} z_0 + z_0 \bar{z}_0 &= r^2.\end{align*}

$\noindent Multiplying both sides of this equation by \alpha \in \mathbb{R} and rearranging gives$

$\alpha z \bar{z} - (\alpha \bar{z}_0) z - (\alpha z_0) \bar{z} + \alpha z_0 \bar{z}_0 - \alpha r^2 = 0. \quad (*)$

$\noindent Now as \alpha is real we have \alpha \bar{z}_0 = \overline{\alpha z_0}. Also, since z_0 \bar{z}_0 = |z_0|^2 \in \mathbb{R}, let \gamma = \alpha z_0 \bar{z}_0 - \alpha r^2 \in \mathbb{R} and let \alpha z_0 = -\bar{\beta} where \beta \in \mathbb{C}. Thus \overline{\alpha z_0} = -\beta.$

$\noindent So Eq. (*) becomes \alpha z \overline{z} + \beta z + \overline{\beta} \,\overline{z} + \gamma = 0, as required to prove.$
That proves the above equation is a general form of a circle, but the question says that it also is the form of a line in the complex plane.
To just add on to your proof:

1.prove the equation below represents a line in complex plane:

$\alpha z\overline{z} + \beta \ z +\overline{\beta}\ \overline{z} + \gamma = 0 \ \ \ \ (1)$

the equation of the line in general is :

$y = mx+b$

in complex plane, that translates into:

$\frac{z-\overline{z}}{2i} = m(\frac{z+\overline{z}}{2}) + b$

where $m \in \mathbb{R}$

$z-\overline{z} = m\ i \( z+\overline{z}) + 2b \ i$

$(m \ i-1)z +(m\ i+1) \overline{z} + 2b \ i= 0$

multiply by i:

$(-m-i)z +(-m+i) \overline{z} - 2b= 0$

for $\alpha = 0, \beta = -m-i, \gamma = -2b$

we get

$\ \beta \ z + \overline{\beta \ z} + \gamma = 0$

5. ## Re: general form of lines and circles in complex plane

Originally Posted by Mahan1
That proves the above equation is a general form of a circle, but the question says that it also is the form of a line in the complex plane.
To just add on to your proof:

1.prove the equation below represents a line in complex plane:

$\alpha z\overline{z} + \beta \ z +\overline{\beta}\ \overline{z} + \gamma = 0 \ \ \ \ (1)$

the equation of the line in general is :

$y = mx+b$

in complex plane, that translates into:

$\frac{z-\overline{z}}{2i} = m(\frac{z+\overline{z}}{2}) + b$

where $m \in \mathbb{R}$

$z-\overline{z} = m\ i \( z+\overline{z}) + 2b \ i$

$(m \ i-1)z +(m\ i+1) \overline{z} + 2b \ i= 0$

multiply by i:

$(-m-i)z +(-m+i) \overline{z} - 2b= 0$

for $\alpha = 0, \beta = -m-i, \gamma = -2b$

we get

$\ \beta \ z + \overline{\beta \ z} + \gamma = 0$
$\noindent Correct. Or from the second part, when k = 1 it is well known that the equation |z - z_1| = |z - z_2| gives the equation of a line for the perpendicular bisector joining the points z_1 and z_2. Since in our proof above we set \alpha = 1 - k^2 this is equal to zero when k = 1 meaning the first equation reduces to \beta z + \bat{\beta} \bar{z} + \gamma = 0 as you found. Note k = 1 is the only value for k which results in a line as k = 0 gives a point, k < 0 makes no sense, and k \neq 1 where k is positive was shown to give circles.$

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