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$$ \lim_{z \rightarrow 3+i} z^2 = 8 + 6i $$

Is this correct? I am not sure how to go further.

$$ 0 < |z-(3+i)| < \delta $$ $$ |z^2-(8+6i)| < \epsilon $$ $$ \delta = \epsilon $$

A possible way forward, but I am not sure.

$$ |z-(3+i)(z+(3+i)| < \epsilon $$ $$ |6+2i| < \epsilon $$ $$ |3+i| < \frac{\epsilon}{2} $$ $$ \delta = \frac{\epsilon}{2} $$

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    $\begingroup$ This is a no re-inventing of the wheel moment. Assuming that your problem is from a math book or class in Complex Analysis, there should have been some Theorems, worked examples, or previously solved problems that are pertinent here. You should try to build on these previous results. ...see next comment $\endgroup$ Sep 25 at 15:01
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    $\begingroup$ I would suggest that you write in the question the definition of continuity using $\epsilon$-$\delta$. And that you explain what you're supposed to do with it. And then start to write equations... you're jumping by far too quick on equations. $\endgroup$ Sep 25 at 15:02
  • $\begingroup$ The solution that is considered valid will depend on what previous results that you are permitted to use. For example: let $f(z) = z^2 = (x + iy)^2 = (x^2 - y^2) + i(2xy).$ Then, you have that $f(z) = u(z) + iv(z)$ where $u(z) = u(x + iy) = (x^2 - y^2), v(z) = v(x + iy) = 2(xy).$ There is a theorem that says that if $f(z) = u(z) + iv(z)$ and if $\lim_{z \to z_1}u(z) = L_1$ and if $\lim_{z \to z_1}v(z) = L_2,$ then $\lim_{z \to z_1} f(z) = L_1 + iL_2.$ Are you permitted to use this result? $\endgroup$ Sep 25 at 15:07
  • $\begingroup$ @user2661923 The title of the question requests to use $\epsilon$-$\delta$ definition of the limit. $\endgroup$ Sep 25 at 15:10
  • $\begingroup$ I actually have a hard time reading the equations. Can you explain what's your specific goal for each equation written? $\endgroup$
    – Lab
    Sep 25 at 15:11
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For any given $\epsilon > 0$, choose $\delta = \min\left\{1,\frac{\epsilon}{1+2\sqrt{10}}\right\}$. Then \begin{align} 0 < |z-(3+i)| < \delta \quad\Rightarrow\quad |z^2 - (8+6i)| &= |z-(3+i)||z+(3+i)| \\ &< \delta\big[|z-(3+i)| + 2|3+i|\big] \\ &< \delta\big(\delta + 2\sqrt{10}\big) \\ &< \delta\big(1+ 2\sqrt{10}\big) \\ &< \epsilon \end{align}

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We will show that for any $z_0\in \mathbb{C}$, we have

$$\lim_{z\to z_0}z^2=z_0^2$$

We will restrict $z$ such that $|z-z_0|\le 1$ so that $|z-z_0|^2\le |z-z_0|$. Then, we have for any $\varepsilon>0$

$$\begin{align} |z^2-z_0^2|&=|z-z_0|\,|z+z_0|\\\\ &=|z-z_0|\,|z-z_0+2z_0|\\\\ &\le |z-z_0|\left(|z-z_0|+2|z_0|\right)\\\\ &\le |z-z_0| \left(1+2|z_0|\right)\\\\ &<\varepsilon \end{align}$$

whenever $|z-z_0|<\delta =\min\left(1,\varepsilon/\left(1+2|z_0|\right)\right)$.


For the case $z_0=3+i$, $|z_0|=\sqrt{10}$, we have for any $\varepsilon>0$

$$|z^2-(3+i)^2|<\varepsilon$$

whenever $|z-(3+i)|<\delta =\min\left(1, \varepsilon/(1+2\sqrt{10})\right)\le\min\left(1,\varepsilon/7\right)$.

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Recall that $\lim\limits_{z \to a}f(z) = L$ by definition means: $$ \forall \epsilon > 0 \; \exists \delta > 0: 0 < |z-a|<\delta \implies |f(x)-f(a)| < \epsilon.$$

Make sure you fully understand the definition before looking at the approach below.


  • Observe that to make $|z^2 - (8+6i)| < \epsilon$ one may factor LHS to $|z-(3+i)||z+(3+i)| < \epsilon$.

  • We can make $|z-(3+i)|$ arbitrarily close to $0$ and noted that intuitively $|z+(3+i)|$ is bigger than a constant as $|z-(3+i)|$ being arbitrarily small(the proof is below).

(a.) The idea is $|z-(3+i)||z+(3+i)| < \epsilon$ is equivalent to $$ |z-(3+i)| < \frac{\epsilon}{|z+(3+i)|}$$ where $\frac{\epsilon}{|z+(3+i)|}$ should be greater than$\frac{\epsilon}{C}$, where $C$ is a constant for $z$ we care(why? See below).

(b.) Noted that $|z+(3+i)| = |z-(-3-i)| \geq |(3+i)-(-3-i)|-|z-(3+i)|$ (by triangle inequality. Draw the triangle on x-y plane if you gets confused here, this might help a lot). If we only consider $\delta$ we choose that is $< 1$, then $|z-(3+i)|<1$, then $|(3+i)-(-3-i)|-|z-(3+i)| > 4\sqrt20 - 1$.

  • Combine (a) (b) we shall only consider $$ |z-(3+i)| < \frac{\epsilon}{|z+(3+i)|} < \frac{\epsilon}{|4\sqrt20 - 1|}.$$ For any $\epsilon > 0$ take $\delta < \frac{\epsilon}{|4\sqrt20 - 1|}$, then we are done.

  • Conclusively, take $\delta = \min\{1, \frac{\epsilon}{|4\sqrt20 - 1|}\}$ should end the proof.

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