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If the Taylor series of 2 complex functions $f,g$ have radii of convergence $r_f, r_g$ respectively, does it follow that the radius of convergence of their composition has radius of convergence equal to $\min\{r_f, r_g\}$?

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    $\begingroup$ No: $f=0$, $g=\Sigma n^n.z^n$, $f \circ g=0$. More down to earth : $f(z)=\frac {1}{z-1}$, $g(z)=\frac {1}{z-2}$, $f\circ g(z)=\frac {z-2}{3-z}$ $\endgroup$ – Georges Elencwajg Feb 19 '12 at 11:51
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The series for $1/(2-z)$ converges out to $2$, that for $e^z$ to infinity, but the composition $1/(2-e^z)$ can't converge beyond the singularity at $\log2$.

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