# How to prove the statement: $\lim_{n\rightarrow\infty} z_n=0 \iff \lim_{n\rightarrow\infty}|z_n|=0$ for a sequence of complex numbers

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.

• One way to argue is to use that $|a_n|, |b_n| \le |z_n|$... Do you see how you might use that? Oct 28, 2023 at 15:44
• Hint: the definitions are literally identical Oct 28, 2023 at 15:44
• 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. Oct 28, 2023 at 15:44
• Why is $\displaystyle \lim_{n \to \infty} a_n = 0$? Oct 28, 2023 at 15:47
• @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? Oct 28, 2023 at 16:18

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 $$z_n \to 0.$$