$AB$ is a diameter of $D(0,1)$. Find $C\in \overline{D(0,1)}$ s.t. $|AC|\cdot|BC|$ is maximal

Question

Let $AB$ be a diameter of the circle $S(0,1)$. Find all the dots $C$ in $\overline{D(0,1)}$ such that $|AC|\cdot|BC|$ is maximal.

With very basic geometry calculus I have figured out that $C$ is the intersection of $\overline{D(0,1)}$ with the perpendicular bisector of $AB$. But this question is from complex analysis course so I try to find a way of solving this with complex analysis. I suspect that this is related somehow with Mobius transformations but I don't find how.

Answer 1

Since we can rotate the disk about the origin without affecting distances, we can assume, without loss of generality, that $A=(-1,0)$ and $B=(1,0)$.

Let $C$ be represented by $c\in\mathbb{C}$, where $|c|\le 1$. \begin{align*} &|AC|\,{\cdot}\,|BC|\\[4pt] =\;&|c+1|\,{\cdot}\,|c-1|\\[4pt] =\;&|c^2-1|\\[4pt] \le\;&|c|^2+1\\[4pt] \le\;&1^2+1\\[4pt] =\;&2\\[4pt] \end{align*} Claim:$\;$For $|c|\le 1$, the equation $|c^2-1|=2$ holds if and only if $c^2=-1$, or equivalently, $c=\pm i$.

Accepting the claim for now, it follows that the maximum value of $|AC|\,{\cdot}\,|BC|$ is $2$, which is realized if and only if $C=(0,1)$ or $C=(0,-1)$.

Rotating back, it follows that the maximum value of $|AC|\,{\cdot}\,|BC|$ is $2$, which is realized if and only if $C$ is an endpoint of the diameter perpendicular to $AB$.

Proof of the claim:

• One way is to argue geometrically . . .

  The expression $|c^2-1|$ is equal to the distance from $c^2$ to $1$.

  Given that $|c|\le 1$, we also have $|c^2|\le 1$, hence, since $-1$ is the only point in the closed standard unit disk whose distance from $1$ is equal to $2$, it follows that for $|c|\le 1$, we have $|c^2-1|=2$ if and only if $c^2=-1$, as claimed.$\\[4pt]$

• As an alternative, we can argue algebraically . . .

  Suppose $|c^2-1|=2$, for some $c\in\mathbb{C}$ with $|c|\le 1$.

  If $|c| < 1$, then $|c^2-1|\le |c|^2+1 < 1^2+1 = 2$, contradiction.

  Hence we must have $|c|=1$.

  Then $\bar{c}={\large{\frac{1}{c}}}$, hence \begin{align*} &|c^2-1|=2\\[4pt] \iff\;&|c^2-1|^2=4\\[4pt] \iff\;&(c^2-1)(\bar{c}^2-1)=4\\[4pt] \iff\;&1-c^2-\bar{c}^2+1=4\\[4pt] \iff\;&c^2+\bar{c}^2=-2\\[4pt] \iff\;&c^2+\frac{1}{c^2}=-2\\[4pt] \iff\;&c^4+1=-2c^2\\[4pt] \iff\;&c^4+2c^2+1=0\\[4pt] \iff\;&(c^2+1)^2=0\\[4pt] \iff\;&c^2+1=0\\[4pt] \iff\;&c^2=-1\\ \end{align*} as claimed.