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.$

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

Mahan1

Member

omegadot

Active Member

omegadot

Active Member

Mahan1

Member

omegadot

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)