Conditions for when a line in $\mathbb{C}$ is tangent to a point on a circle

Question

I am working on the following problem from Chapter 1, Section 5 of Conway's "Functions of One Complex Variable":

Let $C$ be the circle {z:|z-c|=r}, r>0; let $a=c+r\text{ cis }\alpha$ and put $$ L_\beta=\left\lbrace z:\text{Im}\left(\frac{z-a}{b}\right)=0\right\rbrace $$ where $b=\text{cis }\beta$. Find necessary and sufficient conditions in terms of $\beta$ that $L_\beta$ be tangent to $C$ at $a$.

Here, Conway uses the notation $\text{cis }\alpha$ to denote $(\cos\alpha+i\sin\alpha)$; also, he shows in the section mentioned that $L_\beta$ is simply the line in $\mathbb{C}$ containing $a$ in the direction of $b$.

I've been able to convince myself in pictures that the necessary and sufficient condition is that $\beta=\alpha\pm \pi/2$. One direction is:

Define $$ L_\alpha=\left\lbrace z:\text{Im}\left(\frac{z-c}{d}\right)=0\right\rbrace $$ where $d=\text{cis }\alpha$. Then $L_\alpha$ is the line containing $c$ in the direction of $a$. Assuming $\beta=\alpha\pm\pi/2$, define unit vectors $z_1=\text{cis }\alpha$ and $z_2=\text{cis }\beta$. If we were to add $a$ to each of these, then $z_1\in L_\alpha$ and $z_2\in L_\beta$. So, it suffices to show that $z_1$ and $z_2$ are perpendicular when considered as vectors in $\mathbb{R}^2$. Since $z_1=(\cos\alpha,\sin\alpha)$ and $z_2=(\cos\beta,\sin\beta)$, then $$z_1\cdot z_2=\cos\alpha\cos\beta+\sin\alpha\sin\beta=\cos\alpha-\beta=\cos(\pm \pi/2)=0.$$

My questions are:

1. Is this enough to show one direction, or would I have to show something more to show that $L_\beta$ is tangent to $C$ at $a$? Or maybe I've missed the mark completely?

2. How would I begin the other direction? Should I use the definition of "tangent to $C$ at $a$" that says $L_\beta$ contains only the point $a$ in the circle $C$? Or something else?

Answer 1

Here is a "computational" proof.

This issue is translation-invariant and enlargment (i.e., homothety)-invariant; you may thus assume $c=0$ and $r=1$.

Therefore,

• you can parametrize the circle as the set of $z$ such that $z=e^{i \theta}.$

• the "equation" of the straight line becomes

$$\Im(\dfrac{z-e^{i \alpha}}{e^{i \beta}})=0 \ \iff \ \Im((z-e^{i \alpha})e^{-i \beta})=0.$$

A straight line is tangent to a circle if it has only one common point with it. Thus, it remains to see under which condition the following equation in $\theta$ has a unique solution:

$$\Im((e^{i \theta}-e^{i \alpha})e^{-i \beta})=0 \ \iff \ \Im(e^{i (\theta-\beta)}-e^{i (\alpha-\beta)})=0$$ $$ \ \iff \ \sin(\theta-\beta)-sin(\alpha-\beta)=0.$$

Two angles have the same sine if they are either equal or supplementary (modulo $2 \pi$).

We can therefore infer two cases, delivering two solutions in $\theta$:

$$\begin{cases}\theta-\beta&=&\alpha-\beta & \iff & \theta&=&\alpha \\ \theta-\beta&=&\pi - (\alpha-\beta) & \iff & \theta & = &\pi - \alpha + 2\beta.\end{cases}$$

The solution is thus unique if and only if the two previous solutions are equal:

$\alpha=\pi - \alpha + 2\beta \ $ modulo $ \ 2 \pi\ \iff \ 2 \alpha=2\beta+\pi \ $ mod. $2\pi \iff \exists k \in \mathbb{Z}$ s.t. $2 \alpha=2\beta+k2\pi$

Dividing by 2, we find condition : $ \alpha=\beta+k\pi \ \iff$

$$\alpha=\beta+\dfrac{\pi}{2} \ \text{modulo} \ \pi.$$