# Construct $\frac{1}{z}$ from $z \in \mathbb{C}$

Suppose $z \in \mathbb{C}$. How can we construct $\frac{1}{z}$ with tools without calculating? My teacher suggested something with a parallel line, but I couldn't figure it out.

A unit distance is not given!

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You can invert $z$ with respect to the unit circle and after take the conjugate. The geometric inverse is $w$ on the ray from 0 to $z$ such that $|w||z|=1$ (that is, $\vec w$ and $\vec z$ are parallel vectors). – Sigur Oct 23 '12 at 13:53
If $z=\rho e^{i\theta}$, then $\frac{1}{z}=\frac{1}{\rho}e^{-i\theta}$. Hence $1/z$ lies on the straight line that spans an angle $-\theta$ with the $x$-axis. Now the question is: how can we select a point on this line whose distance from the origin is $1/\rho$? – Siminore Oct 23 '12 at 13:55
The geometric inversion can be done using two orthogonal circles to the unit circle, containing $z$. The second intersection point is $w$. – Sigur Oct 23 '12 at 13:57
How do you construct orthogonal circles, @Sigur? – Thomas Andrews Oct 23 '12 at 14:26
@ThomasAndrews, I guess that the answers below are enough for you. – Sigur Oct 23 '12 at 20:51

Draw the reflection of the ray through 0 and $z$ with respect to the horizontal axis. Use the compass to find the point $A$ of absolute value $|z|$ on the axis and the point $B$ of absolute value 1 on the new ray.

Intersect the parallel line to $AB$ with the new ray to get $P=1/z$.

This works because the angle of $P$ is clearly the negative of the angle of $z$ and by similar triangles the absolute value of $P$ is $|1/z|$.

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We assume that in addition to the point $z$, we also are given the real axis, and a unit distance. So we can draw the unit circle.

If $z\ne 0$, then $\dfrac{1}{z}$ has norm the reciprocal of the norm of $z$. Its argument is the negative of the argument of $z$. So if we find the number $w$ on the ray through the origin and $z$ such that $w$ has norm the reciprocal of the norm of $z$, and reflect in the real axis, we will have constructed $\dfrac{1}{z}$. Reflection is easy using the Euclidean tools. (Thanks to Phira for pointing out an earlier error.)

This brings us to a classical problem of Euclidean geometry: Given a length $a\gt 0$, draw a line segment of length $\dfrac{1}{a}$.

Draw a pair of axes, which for convenience we call the $u$ and $v$ axes, meeting at some point $O$. (The axes do not need to be perpendicular.)

On the $u$-axis, draw a point $A$ such that $OA=a$, and a point $U$ such that $OU=1$. On the $v$-axis, draw a point $V$ such that $OV=1$. Join $AV$. Through $U$, draw a line parallel to $AV$. Suppose that this line meets the $v$-axis at $B$.

Because $\triangle OAV$ and $\triangle OUB$ are similar, we have $\dfrac{OB}{OU}=\dfrac{OV}{OA}$, and therefore $OB=\dfrac{1}{OA}=\dfrac{1}{a}$.

All the constructions described above can be done with compass and straightedge.

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@Phira: Thanks! I have made the correction. – André Nicolas Oct 23 '12 at 14:58

My first thought is to use Möbius transformations, which are "constructed" out of rotations, translations, and inversions.

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If you prefer a solution with inversion:

For a point $z$ outside the unit circle, intersect the circle with diameter $0z$ with the unit cercle. Intersect the line of the two intersection points with the line $0z$. The intersection point gives $1/\bar z$. Now, reflect with respect to the horizontal axis to find $1/z$.

You can check that this works, because you have lots of similar (rectangular) triangles.

For a point inside, draw the line $0z$, draw the line through $z$ orthogonal to it, intersect it with the unit circle to get the points $A$ and $B$. Draw the line through $A$ that is orthogonal to the radius $0A$, intersect it with $0z$ to get $1/\bar z$. Reflect it with respect to the horizontal axis to find $1/z$.

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