Koebe Function

The function

$$f(z)\equiv {z\over (1-z)^2}.$$

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

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

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

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

References

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