# Complex number epsilon-delta problem

1. Feb 23, 2015

### jjr

1. The problem statement, all variables and given/known data
Prove that $\lim_{z\rightarrow z_0} Re\hspace{1mm}z = Re\hspace{1mm} z_0$

2. Relevant equations
It is specifically mentioned in the text that the epsilon-delta relation should be used,

$|f(z)-\omega_0| < \epsilon\hspace{3mm}\text{whenever}\hspace{3mm}0<|z-z_0|<\delta$.

Where $\lim_{z\rightarrow z_0}f(z) = \omega_0$
Other equations that might be useful are
$|z_1+z_2| \leq |z_1| + |z_2|\hspace{3mm}\text{(Triangle inequality)}$
and perhaps
$Re\hspace{1mm}z \leq |Re\hspace{1mm} z| \leq |z|$
3. The attempt at a solution
Here $\omega_0 = Re\hspace{1mm}z_0$ and $f(z) = Re\hspace{1mm} z$,
so we want to find a delta $|z-z_0| < \delta$ such that $|Re\hspace{1mm}z - Re\hspace{1mm}z_0| < \epsilon$.

I am honestly not sure how to approach this. Any clues would be very helpful.

Thanks,
J

2. Feb 23, 2015

### Svein

3. Feb 23, 2015

### Dick

Think about it. Which is larger $|z-z_0|$ or $|Re(z)-Re(z_0)|$?

4. Feb 23, 2015

### jjr

Well, the distance between two complex points is definitely larger or equal to the distance between the real parts of the same points...

5. Feb 23, 2015

### Dick

Ok, so how small shall we make $|z-z_0|$ to make sure that $|Re(z)-Re(z_0)| \lt \epsilon$?

6. Feb 23, 2015

### jjr

Hmm. I suppose that if we set $|z-z_0|$ smaller than $\epsilon$, then $|Re(z)-Re(z_0)|$ would have to be smaller than $\epsilon$. I.e. $\delta = \epsilon$?

7. Feb 23, 2015

### Dick

Yes.

8. Feb 23, 2015

### jjr

Thank you. Not too hard to see really, but I'm still not sure how to show it. Is the statement I made sufficient? I feel like it's lacking some rigor

9. Feb 23, 2015

### Dick

You just need to show $|Re(z)| \le |z|$.

10. Feb 24, 2015

### jjr

Okay, I think I've got my head wrapped around it. I have another one, if you would care to have a look:

To be shown:
$\lim_{z\rightarrow z_0} \bar{z}=\bar{z_0}$

So
$|\bar{z}-\bar{z_0}|<\epsilon \to |z-z_0|<\delta$

My work:
I'm trying to find some relation between $z,z_0$ and their conjugates, so I tried setting
$\bar{z}=z-2Im(z)$
$\to |\bar{z}-\bar{z_0}| = |z-z_0-2Im(z)+2Im(z)|$.
Here I tried invoking the triangle inequality, but the expression I ended up with didn't help much.
Suggestions?

Thanks,
J

edit: typo

Last edited: Feb 24, 2015
11. Feb 24, 2015

### Svein

Hint: $\lvert z-z_{0}\rvert ^{2}=(z-z_{0})(\bar{z}-\bar{z_{0}})=\lvert \bar{z}-\bar{z_{0}}\rvert ^{2}$

12. Feb 24, 2015

### jjr

Thank you. I suppose one could take the square root from both sides and obtain $|z-z_0| = |\bar{z}-\bar{z_0}|$, which in turn enables us to set $\delta = \epsilon$? Thinking about this again I realize that this makes sense, that the distance between two points $z, z_0$ should be the same as the distance between the same two points mirrored about the imaginary axis.