2
$\begingroup$

enter image description here

I am talking about part (i). We know it's convergent so it converges to some $\large z_s \in \mathbb{C}$ , don't know how to show it's equivalent to the series of the conjugates though.

$\endgroup$

1 Answer 1

2
$\begingroup$

You can prove this with the definition of convergence of a series.

Say $\sum_{k=0}^\infty z_k = z$. That means that for any $\epsilon > 0$, there is an $N$ such that $$ \left| \sum_{k=0}^n z_k - z \right| < \epsilon \; \text{ for all } n \geq N.$$

But then

$$\left| \; \sum_{k=0}^n \bar{z_k} - \bar{z} \; \right | = \left| \; \overline{ \sum_{k=0}^n z_k - z } \; \right | = \left| \sum_{k=0}^n z_k - z \right| < \epsilon \; \text{ for all } n \geq N$$

since conjugation behaves well with absolute values and finite sums.

$\endgroup$
3
  • $\begingroup$ it makes sense, however it seems like the problem wants me to show explicitly that the sum of $z_k$ bar is bar of the sum of $z_k$. I'm not sure if what you have is enough. $\endgroup$ Mar 2, 2014 at 4:49
  • $\begingroup$ how can I do part (ii)? $\endgroup$ Mar 2, 2014 at 6:59
  • $\begingroup$ I don't know how you would the sums are equal more explicitly - they are infinite, so defining them at all requires this sort of limit. For (ii), you could use a similar argument as the above, or use the fact that $2\text{Re }z = z + \bar{z}$ and $2\text{Im }z = z - \bar{z}$. $\endgroup$
    – ec92
    Mar 3, 2014 at 14:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .