general form of lines and circles in complex plane (1 Viewer)

Mahan1 · Feb 19, 2017

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.

omegadot · Feb 24, 2017

Mahan1 said:
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$, as required to prove.$

omegadot · Feb 24, 2017

Mahan1 said:
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.$

Mahan1 · Feb 26, 2017

omegadot said:
$\noindent The general equation of a circle in the complex plane should be$

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

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

$\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

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

omegadot · Feb 28, 2017

Mahan1 said:
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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general form of lines and circles in complex plane (1 Viewer)

Mahan1

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