<s>[INST] estimating absolute value of complex function Let $f(z) = \frac{1}{z^2 - 2z + 4 }$  where $|z| = R > 1 $. How can I estimate $|f(z)|$ to get an upper bound of the form $\frac{ M}{|z|^p }$ where $M > 0$ and $p>1$ ? I know 
$$ |f(z)| = \frac{1}{|z^2 - 2z + 4| } \leq \frac{1}{|z^2 - 2z|} = \frac{1}{|z||z-2|}$$
and here is where I got Stuck. Any ideas?
 [/INST] Clearly $|z^2 -2z+4|=|z^2-2(z-2)|=|2(z-2)-z^2|\geq ||2z-4|-|z^2||$.
$\Rightarrow$
$$ |f(z)| = \frac{1}{|z^2 - 2z + 4| } \leq \frac{1}{||2z-4|-|z^2| |}$$
So by triangle inequality we have that  $|2z-4|\geq |2z|-|4|$
Since $|z|=R>1$ , there exists $a \in \mathbb R^+$ such that $4=a|z|$ and $a>1$ .
Therefore $$ |f(z)| = \frac{1}{|z^2 - 2z + 4| } \leq \frac{1}{||2z|-|4|-|z^2| |}$$
$$ |f(z)| = \frac{1}{|z^2 - 2z + 4| } \leq \frac{1}{\left | |2z|-a|z|-|z^2| \right |}$$
$$ |f(z)| = \frac{1}{|z^2 - 2z + 4| } \leq \frac{1}{|(2-a)|z|-|z^2|| }$$
So $$ |f(z)| = \frac{1}{|z^2 - 2z + 4| } \leq \frac{1}{|(2-a)|z|-|z^2|| }=\frac{1}{||z^2|-(2-a)|z|| }$$
Since $|z|>1$ , $|z|^2 >|z|$.
So $||z^2|-(2-a)|z|| \geq ||z^2|-(2-a)|z^2||=||z^2|(a-1)||$
Now put $M= \frac{1}{a-1} >0$ as $a>1$ .
Then $$ |f(z)| = \frac{1}{|z^2 - 2z + 4| } \leq \frac{M}{|z|^2 }$$
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