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I need to prove $\lim_{n\rightarrow\infty} z_n=0 \iff \lim_{n\rightarrow\infty}|z_n|=0$.

For the forward implication, I have assumed that $\lim_{n\rightarrow\infty}z_n=0$, so I can state that for a $z_n = a_n+ib_n$, then $\lim_{n\rightarrow\infty}a_n=0$ and $\lim_{n\rightarrow\infty}b_n=0$. Next, I wrote (not sure the logic is sound here): $${\lim_{n\rightarrow\infty}|z_n|^2 = \lim_{n\rightarrow\infty}(a_n^2+b_n^2)=\lim_{n\rightarrow\infty}a_n^2 + \lim_{n\rightarrow\infty}b_n^2=0\cdot0+0\cdot0}$$ from above, and as such, $\lim_{n\rightarrow\infty}|z_n|=0$.

For the backward implication I am less sure. Can I state that if $\lim_{n\rightarrow\infty}|z_n|=0$, then $|z_n|<1$, so that $a_n^2+b_n^2 < 1$? Even if I can, I'm not sure where to go so if there is a better way to tackle this, any pointers would also be appreciated.

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  • $\begingroup$ One way to argue is to use that $|a_n|, |b_n| \le |z_n|$... Do you see how you might use that? $\endgroup$ Oct 28, 2023 at 15:44
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    $\begingroup$ Hint: the definitions are literally identical $\endgroup$
    – Lorago
    Oct 28, 2023 at 15:44
  • $\begingroup$ Begin with: what is the definition of convergence for a sequence of complex numbers? Your proof will depend on that definition. Also, mention "complex numbers" in the statement of the problem. $\endgroup$
    – GEdgar
    Oct 28, 2023 at 15:44
  • $\begingroup$ Why is $\displaystyle \lim_{n \to \infty} a_n = 0$? $\endgroup$ Oct 28, 2023 at 15:47
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    $\begingroup$ @Lorago so by using the definition of converge to state that the left-hand side reads $\forall\epsilon>0,\exists N\in\mathbb{N},\forall n\ge N:|z_n-0|=|z_n|<\epsilon$ and the right hand side read: $\forall\epsilon >0,\exists N\in\mathbb{N},\forall n\ge N:||z_n|-0|=|z_n|<\epsilon$ which are equivalent? $\endgroup$ Oct 28, 2023 at 16:18

1 Answer 1

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I thought of another approach: $"\to"$
Assume $z_n \to 0$. Then $\forall \varepsilon > 0 \ \exists N: n > N, \forall \ n \in \mathbb{N} \ \Rightarrow ||z_n| - 0| = ||z_n|| = |z_n| = |z_n - 0| < \varepsilon $
So $|z_n| \to 0.$

Same for: "$\gets$"
Assume $|a_n| \to 0$ Then $\forall \varepsilon > 0 \ \exists N: n > N, \forall \ n \in \mathbb{N} \ \Rightarrow |z_n - 0| = |z_n| = ||z_n|| = ||z_n| - 0| < \varepsilon $
So $z_n \to 0.$

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