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!

 A: 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.    
A: 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 $|1/z|$.
A: My first thought is to use Möbius transformations, which are "constructed" out of rotations, translations, and inversions.
A: 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 $1/z$.
