Prove F Z Z2 is Continuous
Prove that $f(z)=z^2$ is continuous.
Given $\varepsilon > 0$, set $\delta = \min\{1, \varepsilon/(1 + 2|z_0|)\}$. For all $z$, $|z - z_0| < \delta$ implies $|z - z_0| < 1$ (which implies $|z + z_0| < 1 + 2|z_0|$ by the triangle inequality) and $$|f(z) - f(z_0)| = |z + z_0||z - z_0| < (1 + 2|z_0|)\frac{\varepsilon}{1 + 2|z_0|} = \varepsilon.$$
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Comments
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Prove that $f(z)=z^2$ is continuous for all complex and real values of $z$.
What I've got so far is:
Given $ \epsilon >0$ and $|z-z_0|<\delta$ after some calculations (which I've checked with the answer key) $$ |f(z)-f(z_0)|<\delta(\delta+2|z_0|) $$
Beyond this things get difficult when trying to create $\epsilon$ as a function of $\delta$, the answer reads:
$$\delta(\delta+2|z_0|)\leq \frac{\epsilon}{3|z_0|}(|z_0|+2|z_0|)=\epsilon $$
and I have no clue how to get there.
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You need to find a $\delta > 0$ such that $|f(z) - f(z_0)| < \epsilon$. Which $\delta$ will certainly do the job? The one for which $\delta (\delta + 2|z_0|) \leq \epsilon$. I think the answer suggests taking any $\delta > 0$ such that $\delta \leq |z_0|$ and $\delta \leq \frac{\epsilon}{3 |z_0|}$. This makes sense for $z_0 \neq 0$. If $z_0 = 0$ then $\delta = \sqrt{\epsilon}$ is ok.
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Okay, but why $\delta \leq |z_0|$ and $\delta \leq \frac{\epsilon}{3|z_0|}$ though? I guess $\delta \leq |z_0|$ is the biggest $delta$ for $z=0$ but why the other one and how do they combine into that expression? I'm having a hard time wrapping my head around the actual "image".
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It is just a particular choice of $\delta$ for which $\ldots \leq \epsilon$ easy to check. You could alternatively just solve $\delta (\delta + 2 |z_0|) = \epsilon$ (note equality) for $\delta$.
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I think this is exactly what I'm having trouble understanding. 1.Where does $\delta \leq \frac{\epsilon}{3|z_0|}$ come from? 2.How would you solve $\delta (\delta +2|z_0|)=\epsilon$?
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1. Maybe trial and error. 2. This is a quadratic equation in $\delta$. (But 1 and 2 are alternatives and 2 will (probably) not give the same answer as 1.)
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1.What do you mean trial and error? 2. $\delta (\delta+2|z_0|)=0$ is a normal quadratic but $\delta (\delta+2|z_0|)=\epsilon $ surely is not a regular one. I don't understand your reasoning.
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How do you get that $|z-z_0|<\delta$ implies $|z-z_0|<1$? Should it read $|z_0|$?
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Since $\delta$ is the smaller of the numbers $1$ and $\varepsilon/(1 + 2|z_0|)$, the condition $|z - z_0| < \delta$ implies $|z - z_0| < 1$ and $|z - z_0| < \varepsilon/(1 + 2|z_0|)$.
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I think the elegance of the solution might have been misleading me. You are saying that the first bit comes from $|z+z_0|<(1+2|z_0|)$. The second bit comes from $|z-z_0|=\delta$ but we use $\epsilon=\frac{\epsilon}{1+2|z_0|}$ since this $\epsilon\geq \delta =|z-z_0|$ so that $\frac{\epsilon}{\delta}\geq 1$?
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No, I didn't say $|z + z_0| = 1 + 2|z_0|$, nor did I say $|z - z_0| = \delta$. I supposed $|z - z_0| \color{red}{<} \delta$, which implies, $|z - z_0| < 1$, which implies $$|z + z_0| = |(z - z_0) + 2z_0| \le |z - z_0| + |2z_0| = |z - z_0| + 2|z_0| < 1 + 2|z_0|$$ Also, $|z - z_0| < \delta$ implies $|z - z_0| < \varepsilon/(1 + 2|z_0|)$. Since $|z + z_0| < 1 + 2|z_0|$ and $|z - z_0| < \varepsilon/(1 + 2|z_0|)$, we have $$|z + z_0||z - z_0| < (1 + 2|z_0|)\frac{\varepsilon}{1 + 2|z_0|} = \varepsilon$$
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I see, the only thing I'm still confused about is how $|z-z_0|<\delta$ implies $|z-z_0|<\epsilon/(1+2|z_0|)$
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Sure, let's try again. Recall the following property of order: if $a < b$ and $b \le c$, then $a < c$. I defined $\delta$ to be the smaller of $1$ and $\varepsilon/(1 + 2|z_0|)$. So $\delta \le 1$ and $\delta \le \varepsilon/(1 + 2|z_0|)$. In particular, since $\delta \le \varepsilon/(1 + 2|z_0|)$, $|z - z_0| < \delta$ implies $|z - z_0| < \varepsilon/(1 + 2|z_0|)$.
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Thank you! It's obvious now, what we want is to replace our $\delta$ with $\epsilon$ using the assigned property of $|z-z_0|<\delta$. I find the 1 you used instead of $|z_0|$ as my textbook used made it easier to decipher.
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@kobe How do you get this values of delta?
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@GermánLópez we note $|z^2 - z_0^2| = |z + z_0| | z - z_0|$. To make this product of terms less than $\epsilon$, I first consider $|z - z_0| < 1$; the inequality implies $|z + z_0| < 1 + 2|z_0|$ by the triangle inequality, so that $|z^2 - z_0^2| < (1 + 2|z_0|)|z - z_0|$. The last expression is less than $\epsilon$ provided $|z - z_0| < \epsilon/(1 + 2|z_0|)$. So I choose $\delta$ to be the smaller of the two numbers $1$ and $\epsilon/(1 + 2|z_0|)$.
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@kobe umm ok, but how to you know that you need to consider $|z - z_{0}| < 1$
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@GermánLópez it doesn't have to be $|z - z_0| < 1$. If I wanted to, I could have considered $|z - z_0| < 10$ instead. The main point is that I need to bound $|z + z_0|$, which I can do by working in some open interval about $z_0$. I happened to use $|z - z_0| < 1$ for simplicity.
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@kobe Ahh! In short, you selected it in this way so you can find a bound to $|z + z_{0}|$ since you need it on your proof. It is constructed in such a way you can cancel at the end....
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Source: https://9to5science.com/prove-that-f-z-z-2-is-continuous
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