Why $\min_{a\in \Bbb C}\max \{|1-ax|,|1-ay|\}$ with given $x,y\in \Bbb C$ is attained at $a=\frac{2}{x+y}$ Consider complex numbers. Given $x,y \in \Bbb C$, how can we see the following

$\min_{a\in \Bbb C}\max \{|1-ax|,|1-ay|\}$

is attained when $a=\frac{2}{x+y}$? The absolute value is the modulus of complex numbers. 
Many thanks!

The previous post is not good (too many changes in the question), and there was no answer, so I open a new one. 
 A: The estimate for the smallest possible value for the $\max$ is a bit off. 
OK, first of all we know that $x$ and $y$ are real. That's not a big omission. 
We have 
$$x(1-a y) - y(1-a x) = x-y$$
Conversely, if $x$, $y$ are both non-zero, any solution of 
$$x u - y v = x-y$$ is of form $u = 1-a y$, $v = 1- a x$ for some $a$. 
From the equality above for $u$, $v$ we get
$$(|x| + |y|)\max (|u|,|v|) \ge |xu| +|yv|\ge |xu-yv|=|x-y||$$ so
$$\max(|u|, |v|) \ge \frac{|x-y|}{|x|+ |y|}$$
This is the best that can be said, that is: this value can be achieved. 
However, for $a= \frac{2}{x+y}$ we get both $(1-ax) = \frac{y-x}{x+y}$ and $1-a y = \frac{x-y}{x+y}$ so both absolute values are $\frac{|x-y|}{|x+y|}$. That is not always the smallest possible value for the $\max$. Example: Take $x$, $y$ real of opposite sign. For $a=0$ we get both values $=1$, the smallest possible value. Note that $\frac{|x-y|}{|x+y|}$ is $>1$ in this case. 
Conclusion: the paper is a bit off, not too much. It's OK if $x$, $y$ are real of same sign ( or complex of same argument). 
