Complex number epsilon-delta problem

[jjr]

Homework Statement

Prove that \lim_{z\rightarrow z_0} Re\hspace{1mm}z = Re\hspace{1mm} z_0

Homework 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|

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

 

[Svein]

Start with the fact that lim_z→z0z = z₀.

 

[Dick]

Homework Statement

Prove that \lim_{z\rightarrow z_0} Re\hspace{1mm}z = Re\hspace{1mm} z_0

Homework 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|

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

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

 

[jjr]

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

 

[Dick]

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

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

 

[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?

 

[Dick]

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?

Yes.

 

[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

 

[Dick]

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

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

 

[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

 

[Svein]

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

 

[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.