Complex number epsilon-delta problem

• jjr
In summary: 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.
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

jjr said:

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...

jjr said:
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. $\delta = \epsilon$?

jjr said:
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

jjr said:
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

Last edited:
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.

1. What is a complex number epsilon-delta problem?

A complex number epsilon-delta problem is a mathematical problem that involves using the concepts of limits, epsilon, and delta to prove the convergence or divergence of a sequence or function involving complex numbers.

2. How do you solve a complex number epsilon-delta problem?

To solve a complex number epsilon-delta problem, you need to first define the limit of the sequence or function in terms of epsilon and delta. Then, you need to manipulate the expressions using algebraic and trigonometric identities to find a suitable value for delta that satisfies the given epsilon value.

3. What are the common applications of complex number epsilon-delta problems?

Complex number epsilon-delta problems are commonly used in calculus and analysis to prove the convergence or divergence of complex sequences and functions. They are also used in physics and engineering to solve problems involving complex variables.

4. How do you choose the value of epsilon in a complex number epsilon-delta problem?

The value of epsilon is usually chosen based on the given problem and the desired level of precision. It should be small enough to ensure that the difference between the limit and the value of the function or sequence is within the specified epsilon range.

5. What are some common challenges in solving complex number epsilon-delta problems?

Some common challenges in solving complex number epsilon-delta problems include determining the correct value for delta, manipulating complex expressions, and understanding the concept of convergence and divergence in terms of epsilon and delta. Practice and familiarity with complex numbers and their properties can help overcome these challenges.

• Calculus and Beyond Homework Help
Replies
7
Views
2K
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
2K
• Calculus and Beyond Homework Help
Replies
23
Views
2K
• Calculus and Beyond Homework Help
Replies
9
Views
2K
• Topology and Analysis
Replies
4
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
2K
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
2K
• Calculus and Beyond Homework Help
Replies
6
Views
5K