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

the radio= 1

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

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up vote 1 down vote accepted

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 $1$.

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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

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