# $\sum\limits_{n=1}^\infty\left(\frac{z-1}{z+1}\right)^n$ is locally uniformly convergent

I'm taking an introductory course on complex analysis and I've need some hints with homework.

The exercise I'm currently trying is:

Show that the series $\displaystyle\sum_{n=1}^\infty\left(\frac{z-1}{z+1}\right)^n$ is locally uniformly convergent in the semi-plane $\mathrm{Re}(z)>0$ and find the series of the sum.

The way I see it, why would this series be convergent only on $\mathrm{Re}(z)>0$? And how would I go about manually finding the series of this sum? I can't seem to find any clever way to rewrite this. Using Wolfram Alpha, I managed to find that the series converges to $\frac{z-1}{2}$

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If you set $q=(z-1)/(z+1)$, does the sum look familiar? –  Phira Nov 5 '11 at 16:57
Well, yes. I think that solves the summation part. And I think I'm starting to see the whole $Re(z)>0$ thing... Thanks. –  Leonardo Fontoura Nov 5 '11 at 17:07

For each $z$, this is a geometric series, so it converges where $\left|\frac{z-1}{z+1}\right|<1$, and it converges uniformly on any set where there is a fixed $r<1$ such that $\left|\frac{z-1}{z+1}\right|\leq r$, e.g., by the Weierstrass M test. You can show that $Re(z)>0$ if and only if $\left|\frac{z-1}{z+1}\right|<1$. The existence of the fixed $r<1$ as above bounding $\left|\frac{z-1}{z+1}\right|$ in a neighborhood of each point in the right-half plane follows from continuity, and you can apply the formula for the value of a geometric series to get the result shown by Wolfram Alpha.
It's interesting to note that the mapping $\frac{z-1}{z+1}$ is the well-known mapping from the unit disk to the right half-plane. –  Ｊ. Ｍ. Nov 5 '11 at 17:09
@J.M. I think you mean from the right half plane to the unit disk. Yes, it is interesting. It's also $z\mapsto iz$ followed by the Cayley transform. –  Jonas Meyer Nov 5 '11 at 17:12