# Exercise complex variable, series.

Get the power series expansion centered at the origin of the function f, and calculate the radius of convergence of the corresponding series in each of the following cases:

$f(z)=\frac{z^2}{(z+1)^2}$, $\forall z \in C\backslash\{-1\}$.

Solution: Ok I don't know if I do the exercise well.

the serie= $\frac{1}{4}+\sum(-1+z)^n(-1)^n(2^(-2-n)(-3+n))$

Please, help. The sum is of n=1 to $\infty$.

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Note the function is not analytic at $z=-1$. Hence, you can only hope that the radius of convergence is at most $1$.
Further recall that for $\vert z \vert <1$, we have $$\dfrac1{1+z} = 1 - z + z^2 - z^3 \pm$$ Differentiate the above to get that $$-\dfrac1{(1+z)^2} = -1+2z-3z^2 + 4z^3 - 5z^4 \pm$$ Multiplying by $-z^2$, we then get the power series expansion of $\dfrac{z^2}{(1+z)^2}$ as $$\dfrac{z^2}{(1+z)^2} = z^2-2z^3 + 3z^4 - 4z^5+ 5z^6 \mp$$ Hence, the radius of convergence is $1$.
Thank ^^ and as we is commenting this type of exercise... and if $f(z)= arctan z$? Solution? – Rafael Jiménez Guerra Jan 10 '13 at 4:19
@RafaelJiménezGuerra: If $f(z) = \arctan z$, then $f'(z) = 1/(1+z^2)$. Use the geometric series to find an expression for $f'$ and integrate termwise. – mrf Jan 10 '13 at 9:44