Given two points $z_1, z_2 \in \mathbb{C}$ such that $|z_1| < 1$ and $|z_2|<1$, there exists $K > 0$ such that for any points $z \neq 1$ in the closed triangle with vertices $z_1, z_2,$ and 1, $$\frac{|1-z|}{1-|z|} \leq K.$$ Find the smallest possible value of $K$ if $z_1 = \frac{1+i}{2}, z_2 = \frac{1-i}{2}.$

I let $z = x + iy$ in the closed triangle and use two dimensional calculus to find the absolute maximum on the closed triangle region. I find that the smallest possible $K$ is 1 in this case(Actually, it is at the point $z= \frac{1}{2} + i \frac{1}{2})$. I am not sure if this is true since the closed triangle region is not closed, so not compact. So I am not sure the method of finding absolute maximum in multivriate calculus can be applied here.

Can anyone recommend ?

  • $\begingroup$ It can't be $K = 1$. Since we always have $1-\lvert z\rvert \leqslant \lvert 1-z\rvert$, the quotient is bounded below by $1$. For a set $S$ where $K = 1$ is a possible choice, we must have equality $1-\lvert z\rvert = \lvert 1-z\rvert$ for all $z\in S$. That implies $S\subset [0,1]$. For $z = \frac{1}{2} + \frac{i}{2}$, we have $\lvert z\rvert = \lvert 1-z\rvert = \frac{1}{\sqrt{2}}$, and hence $$\frac{\lvert 1-z\rvert}{1-\lvert z\rvert} = \frac{1/\sqrt{2}}{1-1/\sqrt{2}} = \frac{1}{\sqrt{2}-1} = \sqrt{2}+1.$$ $\endgroup$ – Daniel Fischer Feb 12 '15 at 14:34
  • $\begingroup$ So you mean K should be $\sqrt{2}+1$ $\endgroup$ – Both Htob Feb 12 '15 at 16:32
  • $\begingroup$ At least. You need to check whether the quotient can attain a larger value somewhere on the triangle. $\endgroup$ – Daniel Fischer Feb 12 '15 at 16:38
  • $\begingroup$ Is the way to figure $K$ relying on the fact that $K$ depends on only $z_1, z_2$ and the triangle inequality $||a|-|b|| \leq |a-b|$ which implies that the smallest $K$ is when the equality holds. $\endgroup$ – Both Htob Feb 12 '15 at 16:57
  • $\begingroup$ If we call the triangle $\Delta$, then what you're looking for is $$\sup_{z\in \Delta\setminus \{1\}} \frac{\lvert 1-z\rvert}{1-\lvert z\rvert}.$$ You can use some geometric considerations to reduce the set of $z$ you need to consider. $\endgroup$ – Daniel Fischer Feb 12 '15 at 17:04

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