# Is My Epsilon-Delta Proof Correct?

## Homework Statement

Using the Epsilon-Delta Definition of a limit, show that

$$\lim_{z\to z_0} \overline{z} = \overline{z_0}$$

where $\overline{z}$ is the conjugate of z.

## Homework Equations

$$|\overline{z}|=|z|$$
$$\overline{z_1}-\overline{z_2} = \overline{z_1-z_2}$$

## The Attempt at a Solution

$$\forall \varepsilon > 0, \exists \delta > 0: 0 < |z-z_0| < \delta \implies |\overline{z} - \overline{z_0}| < \varepsilon$$
$$|\overline{z}-\overline{z_0}| = |\overline{z-z_0}| = |z-z_0|$$
Choose $\delta = \varepsilon$

$$|\overline{z-z_0}| = |z-z_0| < \delta = \varepsilon$$

I'm currently teaching myself Complex Variables using Churchill and Brown's text, but do not have access to any instructors to comment on my work (summer before university). This is the first time that I'm seeing $\varepsilon-\delta$ proofs, so I want to make sure that my proof is correct.

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SammyS
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Homework Helper
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You've got the main part sort of backwards. What you've shown is what I call the scratch work that you do on the side before writing up the formal proof. It's how you decide what relationship δ should have to ε so that when z is within the 'δ - neighborhood' of z0, then your function of z, in this case $\bar{z}$, is within the 'ε - neighborhood' of the limit L, in this case $\bar{z}_0$.

Yes, you start with ε > 0 as you did. Then you set δ, in this case you said δ = ε.

Next show that for any z such that 0 < | z - z0 | < δ, then it follows (by algebra, etc.) that $\left|\bar{z}-\bar{z}_0\right|<\epsilon\,.$

Okay, thank you. I was wondering if I had the right order of everything.
So is this right, now?

Let $\varepsilon > 0$ and choose $\delta = \varepsilon$. Then
$$0 < |z-z_0| < \delta \implies |\overline{z}-\overline{z_0}| = |\overline{z-z_0}| = |z-z_0| < \delta = \varepsilon$$

Last edited:
SammyS
Staff Emeritus
Homework Helper
Gold Member
Okay, thank you. I was wondering if I had the right order of everything.
So is this right, now?

Let $\varepsilon > 0$ and choose $\delta = \varepsilon$. Then
$$|z-z_0|<\delta \implies |\overline{z}-\overline{z_0}| = |\overline{z-z_0}| = |z-z_0| < \delta = \varepsilon$$
That's much better.

Probably should say $0<|z-z_0|<\delta \implies |\overline{z}-\overline{z_0}\,| = \dots$

You don't want (or need) z = z0.

That's much better.

Probably should say $0<|z-z_0|<\delta \implies |\overline{z}-\overline{z_0}\,| = \dots$

You don't want (or need) z = z0.
Oops, oh yeah, I did forget the greater than 0 part. Thank you very much for your help.