# Homework Help: Is my proof correct for lim_(n-> infty) |z_n| = |z| ? Complex Analysis

1. May 11, 2007

### laura_a

Is my proof correct for lim_(n-> infty) |z_n| = |z| ??? Complex Analysis

1. The problem statement, all variables and given/known data
Show that if lim_{n-> infty} z_n = z

then

lim_{n-> infty} |z_n| = |z|

2. Relevant equations

3. The attempt at a solution

Is this correct:

lim_{n-> infty} |z_n| = |z|

iff

Assume that the conditions hold;
lim_{n-> infty} |x_n| = |x| and lim_{n-> infty} |y_n| = |y|

According to these conditions there exist, for each positive number \epsilon, positive integers n_1 and n_2 such that:

||x_n| - |x|| < epsilon/2 whenever n > n_1

and

||y_n| - |y|| < epsilon/2 whenever n > n_2

Hence, if n_0 is the larger of the two integers n_1 and n_2

||x_n| - |x|| < epsilon/2 and ||y_n| - |y|| < epsilon/2 whenever n > n_0

Since

|(|x_n| + i|y_n|) - (|x| + i|y|) =
|(|x_n| - |x|) + i(|y_n|-|y|) <= ||x_n| - |x|| + ||y_n| - |y||

Then

||z_n| - |z|| < epislon/2 + epsilon/2 = epsilong whenver n > n_0

Thus it holds that

lim_(n-> infty) |z_n| = |z| because for every epsilon > 0 there exists N > 0 such that | |z_n| - |z|| < epsilon

I am using a proof from the text book to draw this from so just wanted to check to see if I've left anything out and if it makes sence!

Thanks

2. May 11, 2007

### NateTG

Well, being lazy, I'd be tempted to use the triangle inequality for complex numbers:
$$|z-z_n| \geq ||z|-|z_n||$$
Then take the limit of both sides as $n \rightarrow \infty$

What you seem to be doing is splitting the original convergence into the convergence of the real part, and the imaginary part, and then using those to show the convergence of the norm, which is also works. You could make it clearer by making that explicit:

Given $\epsilon > 0$ there is $N$ so that $n>N \Rightarrow |z-z_n| < \frac{\epsilon}{2}$ from the hypothesis that $z_n$ converges to $z$.

Then we know that
$$|\rm{Real}(z)-\rm{Real}(z_n)|<\frac{\epsilon}{2}$$
and
$$|\rm{Imaginary}(z)-\rm{Imaginary}(z)|<\frac{\epsilon}{2}$$
and the triangle inequality gives:
$$||z|-|z_n|| \leq |\rm{Real}(z)-\rm{Real}(z_n)|+|\rm{Imaginary}(z)-\rm{Imaginary}(z_n)|< |\frac{\epsilon}{2}+\frac{\epsilon}{2}|=\epsilon$$
so
$$n>N \Rightarrow ||z|-|z_n|| < \epsilon$$

Last edited: May 11, 2007