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In real analysis there was an easy property that converted limits to infinity in limits at zero. More precisely, $1/z_n$ converges to 0 if and only if $z_n$ diverges (this is converges to infinity).

I want to apply this property to complex numbers as follows:

Let $z_n$ a complex sequence. $1/z_n$ converges to 0 if and only if $z_n$ diverges

I've written $1/z_n = x_n/(x_n^2+y_n^2)-i(y_n/(x_n^2+y_n^2))$ where $x_n = Re(z_n)$ and $y_n = Im(z_n)$ but I can't see either of the implications.

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    $\begingroup$ Try to use the modulus $|1/z_n|$ and $|z_n|$ to see whether you can straight away use the result from real analysis. $\endgroup$ – Alan Wang Apr 12 '16 at 1:08
  • $\begingroup$ This is not true. For example if $z_n$ is an alternating sequence. Do you perhaps mean diverge to $\pm \infty$? $\endgroup$ – user223391 Apr 12 '16 at 1:24
  • $\begingroup$ Yes, I meant that $\endgroup$ – Rodrigo Apr 12 '16 at 1:30
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The proposed proposition is false. The sequence $z_n = (-1)^n$ diverges, but $1/z_n$ does not approach $0$.

Here is a somewhat modified proposition that is true:

$1/z_n$ converges to $0$ if and only if $z_n$ diverges to $\infty$.

To prove this, you can use the following facts:

  • $w_n \to \infty$ as $n\to \infty$ means $\forall M>0\ \exists N\in\mathbb N\ \forall n\ge N\ |z_n|> M$.
  • $w_n\to 0$ as $n\to\infty$ means $\forall \varepsilon>0\ \exists N\in\mathbb N\ \forall n\ge N\ |z_n| < \varepsilon$.
  • For $A,B\in\mathbb C$, $|A|<|B|$ if and only if $|1/A|>|1/B|$.

You can also try to figure out how to prove the three bulleted facts.

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I would phrase it this way:

If $z_n$ is a complex sequence then $1/z_n\to 0$ if and only if $|z_n| \to \infty$.

In terms of standard limit nomenclature, this becomes:

If $z_n$ is a complex sequence then $\left(\forall \epsilon > 0, \ \exists n(\epsilon) \text{ such that } n > n(\epsilon) \implies \dfrac1{|z_n|} < \epsilon \right)$ if and only if $\left(\forall v > 0, \ \exists n(v) \text{ such that } n > n(v) \implies |z_n| > v \right)$.

The proof becomes quite straightforward.

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This is the way I look at it now:

If $1/z_n$ converges to 0, that implies $1/|z_n|$ converges to zero (note that the implication $|z_n|$ converges to |z| does not imply that $z_n$ converges to z). Now, using the real case |zn| converges to infinity $\iff$ zn converges to infinity.

For the other side, we just need to note that the implication $|z_n|$ converges to 0 does imply that $z_n$ converges to 0 (remember that $z_n$ converges to z $\iff \ d(z_n,z)=|z_n-z|$ converges to 0)

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