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For $0<\alpha<1$, we define $\Omega_\alpha$ to be the union of the disc $D(0;\alpha)$ and the line segments from $z=1$ to points of $D(0;\alpha)$. Now, let $r=|z|$. Given $0<\alpha<1$, $\frac{|z-r|}{1-r}$ is bounded in $\Omega_\alpha$.

I thought this problem is very easy, but I cannot prove it.

Something I tried:

Let $z=x+iy$. Thus, $$|z-r|=|x-\sqrt{x^2+y^2}+yi|=2[x^2+y^2-x\sqrt{x^2+y^2}]$$ Thus,$$\frac{|z-r|}{1-r}=2[\frac{r^2-x}{1-r}+x]=x-(r+1)+\frac{1-x}{1-r}$$

I am very confused. Any help will be appreciated.

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up vote 3 down vote accepted

I shall argue about points $z\ne1$ in the closure of $\Omega_\alpha$.

If $r:=|z|\leq\alpha$ then $|z-r|\leq|z-1|\leq1+\alpha$ and therefore $${|z-r|\over 1-r}\leq{1+\alpha\over 1-\alpha}\ .$$ For the case $\alpha\leq r<1$ you have to draw a figure. If such an $r$ is given then for all corresponding $z$ we have $|z-r|\leq |z'-r|$ where $|z'|=r$ and $z'$ lies on a tangent from the point $1$ to the circle with radius $\alpha$. It follows that $$|z-r|\leq|z'-r|\leq|z'-1|=\sqrt{1-\alpha^2}-\sqrt{r^2-\alpha^2}$$ and therefore $${|z-r|\over 1-r}\leq{1+r \over\sqrt{1-\alpha^2}+\sqrt{r^2-\alpha^2}}\leq{2\over\sqrt{1-\alpha^2}}\ .$$

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Thank you very much!! Your explanation is very clear and useful! – Y. Fan Sep 24 '11 at 13:03

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