Smith chart as a nomogram for multiplication

Question

The Smith chart is a plot of the following Möbius transformation:

$$ \Gamma(z) = \frac{z-1}{z+1} $$

typically restricted to $\Re(z) \ge 0$ so that $|\Gamma(z)| \le 1$.

On page 36 of this slide deck, an interesting property of the Smith Chart is mentioned:

Let $a$ and $b$ be two positive real numbers. Draw a line on the Smith Chart that goes through $\Gamma(ai)$ and $\Gamma(-bi)$. Then the intersection of this line with the real axis is $\Gamma(ab)$.

Thus, the Smith chart can be used as a nomogram for multiplication, as shown in this picture for the case $2\cdot3=6$.

I have verified this by brute-force algebraic manipulations with SymPy. Is there a more elegant proof of this property?

Yes, I decided to use the traditional engineer's notation since the topic is closely related to electrical engineering — anonymous-ish, Mar 30 '25 at 09:45
Hi, welcome to Math SE. You've asked for a proof of where two lines meet. Their pre-images are circlines. I bet there's a nice proof that uses that. — J.G., Mar 30 '25 at 09:59
But, anon, you posted it to a math site, not an engineering site. When in Rome, do as the Romans do. — Gerry Myerson, Mar 30 '25 at 10:31
In electronics the $i$ usually stands for a variable current (from intensity of current ). So, for imaginary unit, a $j$ is used not to create confusion. — Davide Munari, Mar 30 '25 at 10:44
I changed it to i because in this context it will be probably less confusing. We engineers sometimes forget that the "j" notation feels weird to people from other fields :) — anonymous-ish, Mar 30 '25 at 10:52

Davide Munari · Accepted Answer · 2025-03-30T20:06:55.177

The Möbius transformation transforms lines into circles and viceversa.

Since we search a point on the line that passes through $\Gamma(ai)$ and $\Gamma(-bi)$, on the anti-transformation we search a point on a circle.

If we look at the anti-transformation formula $$z = \frac{1+\Gamma(z)}{1-\Gamma(z)}$$ and we take the limit

$$\lim_{\Gamma(z) \to \infty} \frac{1+\Gamma(z)}{1-\Gamma(z)} = -1$$

we realize that one of the point of the circle we are searching on is $z = -1$.

As we have 3 points of the circle: $${ai, -bi, -1}$$ we may start to search the center of the circle.

The imaginary part of the center is surely $\frac{a-b}{2}$

For the real part of the center we fall back to geometry and we search the intersection of 2 diameters. One diameter is the line $y = \frac{a-b}{2}$. For the other diameter we take the perpendicular to the chord of the points $(-1, 0)$ and $(0,a)$ We find this diameter has equation $$y = -\frac{1}{a} \left(x+\frac{1}{2}\right) + \frac{a}{2}$$

We cross the 2 diameters to find $$\frac{a-b}{2} = -\frac{1}{a} \left(x+\frac{1}{2}\right) + \frac{a}{2}$$ which reduces to $$2x+1 = ab$$.

Now, the point we are searching has only the real part, since it's the anti-transformation of a real point. So we have a circle that has 2 points with only the real part (or geometrically, they are on the $x$ axis). They are linked by the equation $$\frac{p_1+p_2}{2} = x$$ where $x$ is the center (the real part).

If $p_1 = -1$ we have $p_2 = 2x+1 $ or, with the formula for the center we found before $p_2 = ab$.

More information can be found by searching online for "circular inversion", "bilinear transform" and "Mobius transform".

PS. Thank to Euclid and his Intersecting Chords theorem we may even skip the explicit chords steps and immediately deduce that the point is at $ab$

https://en.wikipedia.org/wiki/Intersecting_chords_theorem

Awesome! In my attempts, I missed the fact that the preimage of the line contains -1. Once we know all three points on the circle we can just invoke the Intersecting chords theorem for an even shorter proof. — anonymous-ish, Mar 30 '25 at 19:36

Smith chart as a nomogram for multiplication

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