Circle inversion in the complex plane: properties and worked problems

What is circle inversion?

Inversion in the unit circle maps a nonzero complex number $z$ to

$$w=\frac1{\bar z}.$$

Here $\bar z$ is the complex conjugate of $z$. The origin is excluded because division by zero is undefined.

The geometric meaning

Write $z=r(\cos\theta+i\sin\theta)$, with $r>0$. Then

$$w=\frac1r(\cos\theta+i\sin\theta).$$

Thus $z$ and $w$ have the same argument, and their distances from the origin have product one. If they represent points $P$ and $Q$, then $Q$ lies on the ray $OP$ and

$$OP\cdot OQ=1.$$

Points on the unit circle stay fixed; points inside it move outside, and vice versa. Applying the transformation twice returns the original point.

Why it helps with locus problems

A locus problem asks what figure is traced by a transformed point. Knowing the general behavior of inversion lets you predict whether to expect a line or a circle before calculating.

The related map $w=1/z$ is inversion followed by reflection in the real axis. It has the same circle-and-line classification, but the resulting positions can differ, so do not confuse the two transformations.

Two important properties

A circle through the origin, with the origin removed, maps to a line not through the origin.

A circle not through the origin maps to another circle not through the origin.

The reverse statements also hold because inversion is its own inverse. A line through the origin maps to itself, excluding the origin.

A circle through the origin becomes a line

Let a circle have center $p\ne0$ and pass through the origin. Its equation is

$$|z-p|=|p|.$$

Substituting $z=1/\bar w$ and using $w\ne0$ gives

$$\left|\frac1{\bar w}-p\right|=|p|\quad\Longleftrightarrow\quad\left|w-\frac1{\bar p}\right|=|w|.$$

The last equation says that $w$ is equally distant from $0$ and $1/\bar p$. Its locus is the perpendicular bisector of the segment joining those points: a line not passing through the origin.

Conversely, every point on that line gives a valid nonzero $z=1/\bar w$ on the original circle. This reverse check establishes the whole locus.

A circle not through the origin becomes another circle

Let the original circle be $|z-c|=r$, with $r>0$ and $r\ne|c|$. Substitution gives

$$\left|\frac1{\bar w}-c\right|=r\quad\Longleftrightarrow\quad|\bar c w-1|=r|w|.$$

Both sides are nonnegative, so squaring preserves equivalence:

$$(|c|^2-r^2)|w|^2-\bar c w-c\bar w+1=0.$$

Complete the square:

$$\left|w-\frac{c}{|c|^2-r^2}\right|^2=\frac{r^2}{(|c|^2-r^2)^2}.$$

The image is therefore a circle with center and radius

$$\boxed{\frac{c}{|c|^2-r^2},\qquad\frac{r}{\bigl||c|^2-r^2\bigr|}.}$$

In the following diagrams, the red circle maps to the blue circle. The black circle is the unit circle.

When the original circle does not enclose the origin:

When the original circle encloses the origin:

Problem 1: a circle that maps to itself

Problem. In the complex plane with origin $O$, let $Q$ lie on the ray $OP$ and satisfy $OP\cdot OQ=1$.

1. If $P$ and $Q$ represent $z$ and $w$, express $w$ in terms of $z$. You may use $\bar z$.
2. As $P$ moves on the circle with center $2$ and radius $\sqrt3$, determine the locus of $Q$.

Part 1: translate the geometry into algebra

Because $Q$ lies on the ray $OP$, write $w=kz$ with $k>0$. The distance condition gives

$$|z||kz|=1,\qquad k=\frac1{|z|^2}.$$

Thus

$$\boxed{w=\frac{z}{|z|^2}=\frac1{\bar z},\quad z\ne0.}$$

If $P=O$, no point $Q$ satisfies the distance condition, which explains the exclusion.

Part 2: substitute the inverse transformation

The original circle is

$$|z-2|^2=3.$$

It does not pass through the origin, so its image is another circle. Substitute $z=1/\bar w$:

$$\left|\frac1{\bar w}-2\right|^2=3.$$

Multiplying by $|w|^2$ and expanding yields

$$|w|^2-2w-2\bar w+1=0,$$

or

$$\boxed{|w-2|^2=3.}$$

This circle does not contain $w=0$, and every point on it gives a valid original point. It is the complete locus.

The image is the same circle. In the general formula, $|c|^2-r^2=4-3=1$, so both center and radius remain unchanged.

Problem 2: move the inversion circle

Problem. Points $P$ and $Q$ are inverse with respect to a circle of center $O$ and radius $r$ if they lie on the same ray from $O$ and satisfy $OP\cdot OQ=r^2$.

For a positive real number $\alpha$, let $C_\alpha$ be the circle of center $\alpha$ and radius $\alpha$. Write the inverse of $z$ with respect to $C_\alpha$ as $w=f_\alpha(z)$.

1. Find $f_\alpha(z)$ in terms of $z$, $\bar z$, and $\alpha$.
2. Write $f_\alpha(i)=x+yi$ with real $x,y$, and express $x,y$ in terms of $\alpha$.
3. Determine the curve traced by $f_\alpha(i)$ as $\alpha$ varies over all positive real numbers.

Part 1: translate the center to the origin

Subtracting $\alpha$ from both points gives the same geometry as ordinary inversion, with radius $\alpha$:

$$w-\alpha=\frac{\alpha^2}{\overline{z-\alpha}}=\frac{\alpha^2}{\bar z-\alpha}.$$

Therefore

$$\boxed{f_\alpha(z)=\frac{\alpha\bar z}{\bar z-\alpha},\quad z\ne\alpha.}$$

Part 2: invert the point i

Substitute $z=i$, so $\bar z=-i$:

$$f_\alpha(i)=\frac{-\alpha i}{-i-\alpha}=\frac{\alpha}{\alpha^2+1}+\frac{\alpha^2}{\alpha^2+1}i.$$

Thus

$$\boxed{x=\frac{\alpha}{\alpha^2+1},\qquad y=\frac{\alpha^2}{\alpha^2+1}.}$$

Part 3: eliminate the parameter and check the range

Since $\alpha>0$, we have $x>0$ and $\alpha=y/x$. Substitution gives

$$x^2+\left(y-\frac12\right)^2=\frac14.$$

This identifies a circle, but we must still determine which part is traced. We know $x>0$, so the points lie on its right-hand semicircle.

Conversely, take a point on this circle with $x>0$. The circle equation implies $x^2=y(1-y)$ and $0<y<1$. Set $\alpha=y/x>0$. Then

$$\alpha^2+1=\frac{y^2+x^2}{x^2}=\frac{y}{x^2},$$

which recovers both parameter formulas. Hence every point on the open right-hand semicircle is reached:

$$\boxed{x^2+\left(y-\frac12\right)^2=\frac14,\quad x>0.}$$

The endpoints $0$ and $i$ are approached as $\alpha\to0^+$ and $\alpha\to\infty$, respectively, but neither is attained.

The original animations show how the inversion circle moves and how the image of $i$ traces the semicircle:

Problem 3: the reciprocal transformation

The following problem is from the University of Tokyo's 2017 science-track mathematics examination, Problem 3. Remember that $w=1/z$ includes a reflection compared with unit-circle inversion.

Problem. For $z\ne0$, set $w=1/z$.

1. Let $\alpha$ be a nonzero complex number, and let $L$ be the perpendicular bisector of the segment joining $0$ and $\alpha$. As $z$ moves along $L$, the locus of $w$ is a circle with one point removed. Find its center and radius.
2. Let $\beta$ be the cube root of $1$ with positive imaginary part. Determine and sketch the locus of $w$ as $z$ moves along the segment joining $\beta$ and $\beta^2$.

Part 1: write the line as an equality of distances

The condition is $|z|=|z-\alpha|$, or

$$z\bar\alpha+\bar z\alpha=|\alpha|^2.$$

Substituting $z=1/w$ and multiplying by $|w|^2$ yields

$$\bar\alpha\bar w+\alpha w=|\alpha|^2|w|^2.$$

Completing the square gives

$$\left|w-\frac1\alpha\right|^2=\frac1{|\alpha|^2}.$$

Hence the center is $1/\alpha$ and the radius is $1/|\alpha|$. The point $w=0$ is excluded, since $1/z$ is never zero for finite $z$. Every other point gives a valid $z=1/w$ on $L$.

Part 2: determine the correct arc

We have $\beta=-1/2+(\sqrt3/2)i$, and the segment can be parameterized by

$$z=-\frac12+it,\qquad-\frac{\sqrt3}2\le t\le\frac{\sqrt3}2.$$

Writing $w=u+iv$ gives

$$u=\frac{-1/2}{1/4+t^2},\qquad v=\frac{-t}{1/4+t^2}.$$

It follows that

$$(u+1)^2+v^2=1,\qquad-2\le u\le-\frac12.$$

The map is continuous and one-to-one on the segment. As $t$ runs from $-\sqrt3/2$ through $0$ to $\sqrt3/2$, $w$ runs from $\beta$ through $-2$ to $\beta^2$. Thus the locus is the arc

$$\boxed{|w+1|=1,\qquad\operatorname{Re}w\le-\frac12,}$$

including both endpoints. This is the longer arc on the left side of the circle with center $-1$ and radius $1$.

A useful habit for locus problems

First predict the type of figure using the inversion properties. Then substitute the inverse transformation, simplify the equation, and check excluded points and parameter ranges. An equation for a whole circle is not enough when the original point moves only along a segment or a restricted part of a curve.

Practice another kind of structural reasoning with integer problems or probability recurrence relations.