$\sum_{n=1}^\infty x_n + iy_n$ converges $\iff$ $\sum_{n=1}^\infty x_n$ and $\sum_{n=1}^\infty y_n$ converge

Question

How can I prove that $$\sum_{n=1}^\infty x_n + iy_n$$ converges $\iff$ $\sum_{n=1}^\infty x_n$ and $\sum_{n=1}^\infty y_n$ converge and, in addition, we have: if $$\sum_{n=1}^\infty x_n + iy_n = S = A + iB,$$ then $$\sum_{n=1}^\infty x_n=A \ \text{ and } \ \sum_{n=1}^\infty y_n=B?$$

Answer 1

The series converges iff the sequence of partial sums, i.e. $S_n = \sum_{k=1}^n x_k + iy_k$, converges, which is iff there is $L \in \Bbb C$ such that for each $\epsilon > 0$, there exists some $N$ such that for $n\ge N$, $|S_n - L| < \epsilon$, i.e. if say $L = x +iy$, then $\left|\left(\sum_{k=1}^n x_ k - x\right) + i\left(\sum_{k=1}^n y_k - y \right) \right| < \epsilon$. We know that for $z\in \Bbb C$, $|\Re z| \le |z|$ and $|\Im z| \le |z|$. Conclude.

Answer 2

Note that $$ |(x_n-x)+i(y_n-y)|^2=|x_n-x|^2+|y_n-y|^2\tag{$*$} $$ Trivially, the $=$ above can be replaced with $\leq$ then \begin{aligned} 0\leq|(x_n-x)+i(y_n-y)|^2\leq|x_n-x|^2+|y_n-y|^2 \end{aligned} so this gives one direction. Note that ($*$) also implies \begin{aligned} 0\leq|x_n-x|^2&\leq |(x_n-x)+i(y_n-y)|^2,\\ 0\leq|y_n-y|^2&\leq |(x_n-x)+i(y_n-y)|^2. \end{aligned} This gives you the other direction.