I'm trying to find a compex power series centered at $i$ with convergence radius $\sqrt{2}$ which converges for $z=1$ but not for $z=-1$. Any help?
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The function $\log(1+z)$ is a good starting-point. – Winther Jan 03 '15 at 22:46
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Consider the series: $$\sum_{n=1}^\infty \frac{z^n}{n}.$$
- It has radius of convergence $R=1$ about $z=0$.
- It diverges when $z=1$ (Harmonic series).
- It converges conditionally for all other $z$ of modulus $1$ (use the Dirichlet Test).
How might one alter this series to obtain one with the properties you list?
Mike F
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I could write it like this $\sum_{n=1}^\infty \frac{(z-i)^n}{n(1+i)}$. It is centered at $i$, it has radius $\sqrt{2}$, it diverges for $z=-1$ but what happens for $z=1$? – Stavros Cohen Jan 03 '15 at 23:08
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You're sort of on the right track. You want a change of variables such that the centre $z=i$ becomes $z=0$ and the point of divergence $z=-1$ becomes $z=1$. The substitution $z \mapsto \frac{z-i}{-1-i}$ does the trick. Plugging that into the series gives $\sum_{n=1}^\infty \frac{1}{n (-1-i)^n} (z-i)^n$ I think. – Mike F Jan 04 '15 at 01:02