Finding Domain of Convergence for Complex Series

elimenohpee
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Homework Statement



I need to find the domain of absolute convergence of the following series:

^{\infty}_{1}\sum(z+3)^{2n}/(2n)!

Homework Equations



Ratio test?

The Attempt at a Solution



I'm not really sure how to handle the complex variable z within the series. I attempted to use the ratio test and simplified down to this:

lim (n->\infty) |(z+3)^{2}/(2n+1)(2n+2)|

I'm assuming I simplified this down to this point correctly of course. Can someone nudge me in the right direction? I just need to know how to handle z.
 
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z really doesn't need any 'handling'. It's just some fixed complex number. What's the limit as n -> infinity?
 
It would be zero correct?

I can't find any examples with complex series, but from my text it states they are almost identical to real valued series. If the series ended up being non-zero, is the answer just expressed in terms of z?
 
It would be zero.

(z + 3)^2 = 8 + 6z = 8 + 6(-1)^0.5, which is a constant.

So the limit becomes,

|8+6z| * the limit

And the limit approaches zero.
 
elimenohpee said:
It would be zero correct?

I can't find any examples with complex series, but from my text it states they are almost identical to real valued series. If the series ended up being non-zero, is the answer just expressed in terms of z?

If the limit depends on z, then you need to express it in terms of z. Here it doesn't.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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