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My complex analysis textbook says the following:

... the reader can check that the sequence $\{ z_n \}$ converges to $w$ if and only if the sequence of real and imaginary parts of $z_n$ converge to the real and imaginary parts of $w$, respectively.

Since we need to prove logical equivalence, we will need two proofs (one for either direction).

Proof 1:

I begin by assuming that the sequence $\{ z_n \}$ converges to $w$, and my aim is to show that the sequence of real and imaginary parts of $z_n$ converge to the real and imaginary parts of $w$, respectively.

$$| z_n - w | \ge || z_n | - | w || \ge | \mathcal{R} \{ z_n \} - \mathcal{R} \{ w \} |$$

(Since $|\mathcal{R} \{ z \}| \le |z|$ for $z \in \mathbb{C}$.)

Therefore, we have that

$$ \lim_{n \to \infty } || z_n | - | w || \ge \lim_{n \to \infty} | \mathcal{R} \{ z_n \} - \mathcal{R} \{ w \} | $$

And since we assumed that the sequence $\{ z_n \}$ converges to $w$ (that is, that $\lim_{n \to \infty} z_n = w$), we have that $\lim_{n \to \infty} | \mathcal{R} \{ z_n \} - \mathcal{R} \{ w \} |$ converges to $0$, since $\lim_{n \to \infty } || z_n | - | w ||$ converges to $0$.

And so we have shown that the real parts of $z_n$ converge to the real parts of $w$.

The proof of convergence for the imaginary parts is analogous.

Proof 2:

I begin by assuming that the real and imaginary parts of $z_n$ converge to the real and imaginary parts of $w$.

$$\begin{align} | z_n - w | &= | (x_n + iy_n) - (x + iy)| \\ &= | (x_n - x) + i(y_n - y)| \\ &= \sqrt{ (x_n - x)^2 + (y_n - y)^2 } \end{align}$$

We assumed that $\lim_{n \to \infty} x_n = x$ and $\lim_{n \to \infty} y_n = y$.

$$\begin{align} \therefore \lim_{n \to \infty} | z_n - w | &= \lim_{n \to \infty} \sqrt{ (x_n - x)^2 + (y_n - y)^2 } \\ &= \sqrt{ (x - x)^2 + (y - y)^2 } \\ &= 0\end{align}$$

And so we have shown that the sequence $\{ z_n \}$ converges to $w$.

I would greatly appreciate it if people could please advise me as to the correctness of my proof.

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  • $\begingroup$ First displayed formula: You have a limit on the RHS which is hypothesised to be zero. The middle term may be nonzero. The second inequality sign is the wrong way round. $\endgroup$ – Lord Shark the Unknown Jun 4 at 4:40
  • $\begingroup$ The sequence also converges if it is cauchy sequence $\endgroup$ – Vedant Chourey Jun 4 at 4:43
  • $\begingroup$ @LordSharktheUnknown Thanks. Inequality sign fixed. What else is wrong with the first proof? $\endgroup$ – The Pointer Jun 4 at 4:45
  • $\begingroup$ I recommend using the inequality $|(x_n+iy_n)-(x+y)| \leq |x_n-x|+|y_n-y|$ instead of the one you used. $\endgroup$ – Kavi Rama Murthy Jun 4 at 6:21
  • $\begingroup$ @KaviRamaMurthy But the second proof seems to be correct anyway. I think it’s the first proof that’s incorrect? $\endgroup$ – The Pointer Jun 4 at 6:51
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For your forward direction:

Suppose $z_1=1$ and $w=-1$, then $$||z_1|-|w||=0$$

but $$|\mathcal{R}\{z_1\}-\mathcal{R}\{w\}|=2$$

Hence $||z_1|-|w|| \ge |\mathcal{R}\{z_1\}-\mathcal{R}\{w\}|$ is not true.

To prove $$|z_n-w|\ge |\mathcal{R}\{z_n\}-\mathcal{R}\{w\}|$$

We have

$$|z_n-w| \ge |\mathcal{R}\{z_n-w\}|=|\mathcal{R}\{z_n\}-\mathcal{R}\{w\}|$$

where the first inequality is due to $|\mathcal{R}\{z\}|\le |z|$.

Now, we can take limit on both sides,

$$\lim_{n \to \infty}|z_n-w| \ge \lim_{n \to \infty}|\mathcal{R}\{z_n\}-\mathcal{R}\{w\}|$$

and conclude that the real part converges.

For the reverse direction:

You might like to define $x_n$, $y_n$, $x$, and $y$ explicitly and I think you meant to say

$$| (x_n + iy_n) - (x + iy)| = | (x_n - x) \color{blue}+ i(y_n - y)| $$

even though what you wrote is still correct.

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  • $\begingroup$ Thank you very much for everything. You're absolutely correct. I have one question that I am unsure about: Is $\mathcal{R}$ actually a linear operator? Because that's what it's treated as, it seems. I haven't read anything about it being an operator, but I see that people treat it like one, so I'm curious what it is? Thanks again! $\endgroup$ – The Pointer Jun 9 at 4:52
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    $\begingroup$ We need a field when we discuss linear operator. If the field is the real number then yes. We can check that $\mathcal{R}(x+y) = \mathcal{R}(x)+\mathcal{R}(y)$ and $\mathcal{R}(cx)= c\mathcal{R}(x)$ where $c \in \mathbb{R}$. $\endgroup$ – Siong Thye Goh Jun 9 at 5:00
  • $\begingroup$ But in the case of $|\mathcal{R}\{z_n-w\}|=|\mathcal{R}\{z_n\}-\mathcal{R}\{w\}|$, wouldn't the field be the complex numbers ($\mathbb{C})$? $\endgroup$ – The Pointer Jun 9 at 5:01
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    $\begingroup$ I mean the $c$ in the multiplication, it has to be real. For additions, we just deal with the vector field, which is the complex number. $\endgroup$ – Siong Thye Goh Jun 9 at 5:08
  • $\begingroup$ Oh, yes, you're right. It's been a while since I thought about fields. $\endgroup$ – The Pointer Jun 9 at 5:09
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Did you perhaps mean $\ge$ instead of $\le$ in the first proof? It is perhaps best to explain why your inequality is true; it is not immediately obvious. Your second proof looks fine.

Perhaps a tip to make the equations simpler; you could simply translate the entire complex plane to send $w$ to $0$, noting that convergence under either definition is unaffected. You would then be able to avoid the use of the triangle inequality and multiple variables.

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  • $\begingroup$ I changed the first proof. What do you think? $\endgroup$ – The Pointer Jun 4 at 7:55

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