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Here is a complex analysis homework problem I can't quite figure out:

Prove that every line or circle in $\mathbb{C}$ is the solution set of an equation of the form $a|z|^2+\bar{w}z+w\bar{z}+b=0$, where $a,b\in\mathbb{R}$ and $w,z\in\mathbb{C}$. Conversely, show that every equation of this form has a line, circle, point, or the empty set as its solution set.

So far, I've tried to rewrite the equation of a line in $\mathbb{R}^2$ as $y=mx+b$ in $\mathbb{C}$, where $m$ is real and $x,b$ are complex. I know that a circle in the complex plane is given by $|z-a|=r$, where $a$ is the center and $r$ is the radius.

I also noticed that $\bar{w}z+w\bar{z}=2\text{Re}(\bar{w}z)$. I'm just not sure how all these pieces fit together in answering the question. Any help would be greatly appreciated.

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    $\begingroup$ Write the equation of the circle as $\lvert z-a\rvert^2 = r^2$. And write the equation of the line as $rx + sy = t$. $\endgroup$ Commented Sep 7, 2014 at 21:03
  • $\begingroup$ @DanielFischer Okay. Still not sure where to go from here. After examining the equation $a|z|^2+\bar{w}z+w\bar{z}+b=0$, I noticed that everything reduces to real numbers. How could the equation of a line or circle, which involves complex numbers, be the solution set of a real-valued equation? $\endgroup$
    – Ray B.
    Commented Sep 7, 2014 at 21:28
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    $\begingroup$ Generally, every complex equation corresponds to two real equations. Here we have the situation that one of the two real equations is trivial ($0 = 0$), since the complex equation is invariant under conjugation. You can switch between the real and the complex form of the equation by substituting $z = x+iy$ (and $w = u+iv$), resp. $x = \frac{1}{2}(z+\overline{z}),\; y = \frac{1}{2i}(z-\overline{z})$. If you take the equation $\lvert z-m\rvert^2 = r^2$ of a circle with centre $m$ and radius $r$, do you see how to transform that into an equation of the given form? $\endgroup$ Commented Sep 7, 2014 at 21:41

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You need to use the property of complex conjugation to express the circle equation

\begin{align} |z - z_0|^2 &= r^2 \Leftrightarrow \overline{(z - z_0)}(z-z_0) = r^2 \Leftrightarrow (\overline{z} - \overline{z_0})(z - z_0) = r^2 \Leftrightarrow \\ & z\overline{z} - \overline{z}z_0 - z\overline{z_0} + |z_0|^2 = r^2 \end{align}

Now make the substitution $z_0 = -\frac{\overline{\alpha}}{A}$ where $A \in \mathbb{R} \setminus \{0\}$

$$ z\overline{z} + \overline{z}\frac{\overline{\alpha}}{A} + z\frac{\alpha}{A} + \left|\frac{\overline{\alpha}}{A}\right|^2 - r^2 = 0 $$

Mutliplying the equation by $A$

$$ Az\overline{z} + \overline{z\alpha} + z\alpha + A\left(\left|\frac{\alpha}{A}\right|^2 - r^2\right) = 0 $$

And setting $B = A\left(\left|\frac{\alpha}{A}\right|^2 - r^2\right)$ yields the general form of the equation for a circle in the complex plane. This equation also describes lines which can be viewed as circles with infinite radius.

$$ Az\overline{z} + \overline{z\alpha} + z\alpha + B = 0 $$

When $A = 0$ it represents a line,

$$ \overline{z\alpha} + z\alpha + B = 0 $$

You can convice yourself that this equation describes a line by setting $z = x + iy$ and $\alpha = p + iq$.

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  • $\begingroup$ This makes a lot of sense, thanks! $\endgroup$
    – Ray B.
    Commented Sep 7, 2014 at 22:09
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A line or circle in the plane has an equation of the form $$ D(x^2+y^2)+Ax+By+C=0. $$ It's a line if $D=0$, but $A$ and $B$ are not both zero; it's a circle if $D\ne0$ and $A^2+B^2-4CD^2>0$. So the equation above represents either a line or a circle if $A^2+B^2-4CD^2>0$. The equation represents a single point if $A^2+B^2-4CD^2=0$ but $A$ and $B$ are not both zero (so $D\ne0$). It represents the empty set when $A^2+B^2-4CD^2<0$. (I'll assume that not all coefficients are zero.)

Set $z=x+iy$; then $$ x=\frac{z+\bar{z}}{2},\quad y=\frac{z-\bar{z}}{2i}=-i\frac{z-\bar{z}}{2},\quad x^2+y^2=z\bar{z}. $$ Substituting in the equation above we get $$ 2Dz\bar{z}+A(z+\bar{z}-iB(z-\bar{z})+2C=0 $$ or $$ 2Dz\bar{z}+(A-iB)z+(A+iB)\bar{z}+2C=0. $$ Setting $2D=a$, $A+iB=w$ and $2C=b$ we get the requested form.

The converse is just doing the converse substitution: set $D=a/2$, $C=b/2$, $A=(w+\bar{w})/2$ and $B=(w-\bar{w})/(2i)$. Also $A^2+B^2=w\bar{w}$.

The condition for a line/circle reads $$ w\bar{w}-4\frac{b}{2}\frac{a^2}{4}>0 $$ or $$ 2w\bar{w}-a^2b>0. $$ Single point when $2w\bar{w}=a^2b\ne0$, empty set when $2w\bar{w}-a^2b<0$.

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The equations for circles and lines over $\mathbb{R}$ are quadratic equations where the quadratic term is of the form $ax^2+ay^2$. From your observations about the reality of the linear term the claim follows.

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