Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\frac{z-1}{2}$

share|improve this question
2  
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
add comment

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer
    
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
    
Oops, I was thinking of the negative reciprocal, sorry. Thanks. –  J. M. Nov 5 '11 at 17:17
    
Thanks. I knew this was supposed to be easy... I wasn't thinking straight. –  Leonardo Fontoura Nov 5 '11 at 17:32
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.