Riemann rearrangement theorem

The Riemann rearrangement theorem states that if $\sum\limits_{n=0} ^{+ \infty} a_n$ is conditionally convergent and $M \in \mathbb{R}$ then there exists a permutation $\sigma (n)$ such that $\sum\limits_{n=0}^{+ \infty} a_{\sigma(n)} \ =M$.

Could you tell me how to use this to prove a more general statement?

That if we have $\sum\limits_{n=0} ^{+ \infty} c_n$ - conditionally convergent series of complex numbers, then there exists a line $l$ on the plane such that each point of this line can be a limit of the series.

I'd appreciate any help.

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Not sure if I am right in asking this but, How did you go from a permutation in case of reals to 'line' in general case? – Bhargav Mar 2 '13 at 18:59
Reinhold Remmert, Theory of Complex Functions, page 30. It is only mentioned there, without a proof. Maybe you could tell me where I could find it. This is a very interesting result. – Hagrid Mar 2 '13 at 19:08
I feel certain this has been asked (or at least answered) here before, but I can't seem to find it. – Erick Wong Mar 2 '13 at 19:31

A complex series converges if and only if the real and imaginary parts converge, and an identical statements holds when taking absolute values. Then if $\sum_{n=1}^{\infty} c_{n}$ is convergent, so are $\sum_{n=1}^{\infty} a_{n}$ and $\sum_{n=1}^{\infty} b_{n}$ where $c_{n} = a_{n} + ib_{n}$. If $\sum_{n=1}^{\infty} |c_{n}|$ diverges, at least one of $\sum_{n=1}^{\infty} |a_{n}|$ of $\sum_{n=1}^{\infty} |b_{n}|$ diverges, and suppose without loss of generality that only one converges, and that it is the former. Then we can force the real part of our series, $\sum_{n=1}^{\infty} a_{n}$ to converge to whatever we want by the original Riemann Rearrangement Theorem. If $\sum_{n=1}^{\infty} |b_{n}|$ converges, we might be stuck with a fixed sum for the complex part, but we can still hit the entire horizontal line through $i\sum_{n=1}^{\infty} b_{n}$.
cant there be cases where both $a_{n},b_{n}$ are divergent and their sum being convergent? I am asking this because I have never read anything about sum of two divergent series – Bhargav Mar 2 '13 at 19:21
@b555 $a_n$ is not added to $b_n$, it is added to $ib_n$. Yes it could be both $a_n$ and $b_n$ are divergent and $a_n+b_n$ is convergent, but that is not what this problem deals with. – Maesumi Mar 2 '13 at 19:24
@IsaacSolomon I am not sure about the last line of your proof. The problem is as we fix a rearrangement for $a_n$ to reach some number $A$ then the arrangement for $b_n$ is fixed and we wont know what it converges to, or that it converges at all. – Maesumi Mar 2 '13 at 19:29