# $1/(t-z_n)$ converges uniformly to $1/(t-z)$

Suppose that $z_0$ is a a complex number that does not lie on the real interval $[0,1]$. Next, suppose $z_n$ is a sequence of complex numbers in $C\backslash[0,1]$ that converge to $z_0$.

I am trying to show that

$$\frac{1}{t-z_n}$$

converges uniformly to

$$\frac{1}{t-z_0}$$

for all real numbers $t$ such that $0\le t\le 1$.

Really stuck. Any suggestions?

-
What does $C-[0,1]$ mean? – JohnD Jan 22 '13 at 1:49
Perhaps you prefer $C \setminus [0,1]$ or $C \sim [0,1]$. – GEdgar Jan 22 '13 at 1:51

By reverse triangle inequality we get following inequailty: $$|t-z_n|\ge |t-z_0|-|z_n-z_0|.$$ So $$\left|\frac{1}{t-z_n} -\frac{1}{t-z_0} \right|=\left|\frac{z_0-z_n}{(t-z_n)(t-z_0)}\right|\le \frac{|z_n-z_0|}{|t-z_0|(|t-z_0|-|z_n-z_0|)}$$ for arbitary large $n$.
Since $z_0$ does not contained closed set (on $\mathbb{C}$) $[0,1]$, there exists $c>0$ such that $|z_0-t|\ge c$ for all $t\in[0,1]$. So $$\frac{|z_n-z_0|}{|t-z_0|(|t-z_0|-|z_n-z_0|)}\le \frac{|z-z_0|}{c(c-|z_n-z_0|)}.$$ So we get upper bound of $\left|\frac{1}{t-z_n} -\frac{1}{t-z_0} \right|$ converge at 0 and does not dependent on $t$.