Image of a family of circles under $w = 1/z$

Question

Given the family of circles $x^{2}+y^{2} = ax$, where $a \in \mathbb{R}$, I need to find the image under the transformation $w = 1/z$. I was given the hint to rewrite the equation first in terms of $z$, $\overline{z}$, and then plug in $z = 1/w$.

However, I am having difficulty doing this.

I completed the square in $x^{2}+y^{2}=ax$ to obtain $\left(x - \frac{a}{2} \right)^{2} + y^{2} = \left(\frac{a}{2} \right)^{2}$.

Then, given that $\displaystyle x = Re(z) = \frac{z+\overline{z}}{2}$ and $\displaystyle y = Im(z)= \frac{-(z-\overline{z})}{2i}$, I made those substitutaions and my equation became $\left( \frac{z+\overline{z}-a}{2}\right)^{2} - \left(\frac{\overline{z}-z}{2} \right)^{2} = \left(\frac{a}{2} \right)^{2}$.

Then, sbustituting in $z = \frac{1}{w}$, this became $\displaystyle \frac{\left(\frac{1}{w} + \frac{1}{\overline{w}} - a \right)^{2}}{4} - \frac{\left(\frac{1}{\overline{w}} - \frac{1}{w} \right)^{2}}{4} = \frac{a^{2}}{4}$.

Beyond this, my algebra gets very wonky.

Could someone please tell me what my final result should be? Knowing that would allow me to work backwards and then apply these methods to other problems (of which I have many to do!).

Thanks.

Answer 1

Write $z=x+iy$, so $x^2+y^2=z\bar{z}$, and $x=\frac{z+\bar{z}}{2}$. Thus the circles can be described by $$ z\bar{z}=a\frac{z+\bar{z}}{2} $$ Upon doing $z=1/w$, you get $$ \frac{1}{w\bar{w}}=\frac{a}{2}\frac{\bar{w}+w}{w\bar{w}} $$ that becomes $$ a\frac{\bar{w}+w}{2}=1 $$ Writing $w=X+iY$, you get $$ aX=1 $$

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More generally, if you have the circle $x^2+y^2+ax+by+c=0$, with $z=x+iy$ you get $$ z\bar{z}+a\frac{z+\bar{z}}{2}+b\frac{z-\bar{z}}{2i}+c=0 $$ Upon changing $z=1/w$, the equation becomes $$ \frac{1}{w\bar{w}}+a\frac{\bar{w}+w}{2w\bar{w}} +b\frac{\bar{w}-w}{2iw\bar{w}}+c=0 $$ and removing $w\bar{w}$ from the denominator, $$ 1+a\frac{w+\bar{w}}{2}-b\frac{w-\bar{w}}{2i}+cw\bar{w}=0 $$ or, with $w=X+iY$, $$ 1+aX-bY+c(X^2+Y^2)=0 $$ So the circle becomes a straight line if $c=0$ (that is, it passes through the origin), otherwise it is transformed into a circle.

Similarly, the line $ax+by+c=0$ becomes $$ a\frac{\bar{w}+w}{2w\bar{w}}+b\frac{\bar{w}-w}{2iw\bar{w}}+c=0 $$ and finally $$ aX-bY+c(X^2+Y^2)=0 $$ so it becomes a circle if $c\ne0$, or a straight line through the origin otherwise.

Answer 2

Hint: This transformation can be seen as successive transformations $$Z = \frac{z}{|z|^2} , \,\,\,w = \overline {Z}$$

since $z \cdot \overline{z} = |z|^2$.

Then $z = \frac{1}{w} = \frac{\overline w}{|w|^2}$ and $$x = \frac{u}{u^2 + v^2} \,\,\,\text{and} \,\,\,\,y = \frac{-v}{u^2+v^2}$$

where $w = u + iv$.

Edit: Now study $$A(x^2+y^2) + Bx + Cy + D = 0$$

complete the square and make the substitutions to see that is taken to

$$D(u^2 + v^2) + Bu - Cv + A = 0 $$

which is a circle or a line., depending on whether $D \neq 0$ or $D=0.$

See what happens when a cirle $A \neq 0$ is passing or not through the origin.

Answer 3

This can also be solved by inversive geometry. In (traditional circle) inversive geometry, we invert figures with respect to an inversion circle centered at (say) $O$ with radius $r$. Each point $P$ inverts to a point $P'$ such that $O, P, P'$ all lie on a line, with $P$ and $P'$ on the same side of $O$, and with $OP \cdot OP' = r^2$.

One useful property of this inversion is that circles invert to other circles, or to lines. (More about this in a moment.) The transformation $w = \frac{1}{z}$ is almost inversive with respect to the unit circle at the origin, except that the result is also reflected across the $x$-axis. However, since all the circles (including the unit circle at the origin) are symmetric with respect to the $x$-axis, we can ignore the reflection.

With that in mind, let us observe first that the family of circles all intersect the $x$-axis in two points: at $(0, 0)$ and at $(a, 0)$.

As mentioned above, all circles invert to either circles, or lines. They invert to circles if they do not pass through the inversion circle's center, but if they do pass through that center, as they do here, they invert to lines. Furthermore, if the circles intersect, as they also do here, that line passes through the two intersection points. The distance of that line from the center is the reciprocal of the furthest distance of that circle from the center. Since the circles have greatest distance $1, 2, 3, \ldots$, the distances of the lines from the center must be $1, \frac{1}{2}, \frac{1}{3}, \ldots$, as below:

So, in general, the circle

$$ x^2+y^2 = ax $$

inverts to

$$ x = \frac{1}{a} $$