Complex Analysis - Series Analytic Regions

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SUMMARY

The discussion focuses on determining the analytic regions for two complex functions. For function A, \( f(z) = \sum_{n=1}^{\infty} \left[ \frac{z(z+n)}{n} \right]^n \), it converges for \( |z| < 1 \) due to its growth behavior resembling \( z^n \exp(z) \). Function B, \( f(z) = \sum_{n=1}^{\infty} 2^{-n^2 z} \cdot n^n \), can be analyzed similarly, suggesting convergence under the same condition. The participants emphasize the importance of understanding series growth rates in complex analysis.

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Homework Statement
Find the analytic regions of the next functions:

A. [tex]f(z)=\sum_{n=1}^{\infty} [ \frac{z(z+n)}{n}]^n[/tex]
B. [tex]f(z)=\sum_{n=1}^{\infty} 2^{-n^2 z} \cdot n^n[/tex]

Homework Equations


The Attempt at a Solution



In the first one: I've tried writing : [tex]f(z)= \sum \frac{z^{2n}}{n^n} + \sum z^n[/tex]
and the second element in the sum converges iff |z|<1... Is it enough?

About B: We can write this series as: [tex]\sum [ \frac{n}{2^{nz}}]^n[/tex] ... But I don't think it helps us...

Hope you'll be able to help me


TNX!
 
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WannaBe22 said:
Homework Statement
Find the analytic regions of the next functions:

A. [tex]f(z)=\sum_{n=1}^{\infty} [ \frac{z(z+n)}{n}]^n[/tex]
B. [tex]f(z)=\sum_{n=1}^{\infty} 2^{-n^2 z} \cdot n^n[/tex]
The first series ultimately grows like z^n exp(z) & hence converges for |z|<1.
The second problem can be tackled similarly.
 

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