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Koebe Function

\begin{figure}\begin{center}\BoxedEPSF{KoebeFunction.epsf scaled 900}\end{center}\end{figure}

The function

\begin{displaymath}
f(z)\equiv {z\over (1-z)^2}.
\end{displaymath}

It has a Minimum at $z=-1$, where

\begin{displaymath}
f'(z)=-{1+z\over(z-1)^3}=0,
\end{displaymath}

and an Inflection Point at $z=-2$, where

\begin{displaymath}
f''(z)={2(2+z)\over(z-1)^4}=0.
\end{displaymath}


References

Stewart, I. From Here to Infinity: A Guide to Today's Mathematics. Oxford, England: Oxford University Press, pp. 164-165, 1996.




© 1996-9 Eric W. Weisstein
1999-05-26