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Please, could you help me to prove the following limit by definition?

$\lim_{z \rightarrow 1-i} |\bar{z}^{2}-2| = 2 \sqrt{2}$

Here is my try:

Let $\delta > 0$ such thtat $0 < |z - (1-i)| < \delta$. Note htat $0< ||z|-|1-i|| \leq |z-(1-i)| \leq \delta$. Then $0 \leq ||z| - \sqrt{2}|\leq \delta$. Also:

\begin{align} |\bar{z}^{2} - 2 - 2\sqrt{2}| &= |(\bar{z}-\sqrt{2})(\bar{z}+\sqrt{2}) - 2\sqrt{2}| \\ &=|\overline{(z-\sqrt{2})(z+\sqrt{2}) - 2\sqrt{2}}| \\ &=|(z-\sqrt{2})(z+\sqrt{2}) - 2\sqrt{2}| \\ &=|(z-|1-i|)(z+|1-i|) - 2|1-i|| \\ \end{align}

After that I am no sure how to proceed. Any suggestion?

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    $\begingroup$ Use continuity. $\endgroup$
    – anomaly
    Dec 7, 2023 at 0:22

2 Answers 2

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If your intention is to solve it through $\epsilon - \delta$ definition, then observe that:

$$|(z^2-2)-((1-i)^2-2)| =|z^2-(1-i)^2|.$$

Now,

$$|z^2-(1-i)^2| = |z-(1-i)||z+(1-i)|.$$

That is

$$|z^2-(1-i)^2| \leq |z-(1-i)|(|z|+|1-i|).$$

In otherword,

$$|z^2-(1-i)^2| \leq |z-(1-i)|(|z|+\sqrt{2}). \tag{1}$$ Suppose we fix $\delta <1$, then as per your calculation:

$$|z| \leq \delta+|1-i| \leq 1+\sqrt{2}.$$

Now, choose $\epsilon$ as small as possible such that $\frac{\epsilon}{1+2\sqrt{2}}<1$, and fix $\delta = \frac{\epsilon}{1+2\sqrt{2}}$. Then from equation (1), we infer that, whenever $0<|z-(1-i)| <\frac{\epsilon}{1+2\sqrt{2}}$, we will have

$$|(z^2-2)-((1-i)^2-2)| = |z^2-(1-i)^2| < \epsilon.$$

Now, from $||a|-|b|| \leq |a-b|$, we can infer that whenever $0<|z-(1-i)| <\frac{\epsilon}{1+2\sqrt{2}}$,

$$||z^2-2|-|(i-1)^2-2|| < \epsilon.$$

That is

$$||z^2-2|-2\sqrt{2}| <\epsilon,$$ whenever $0<|z-(1-i)| <\frac{\epsilon}{1+2\sqrt{2}}$.

Now, seeing all this calculation drudgery, we must avoid these arguments by resorting to arguments which involve some standard properties of functions. One such is continuity of functions.

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    $\begingroup$ Everything is clear for me. Thank you! $\endgroup$
    – Irbin B.
    Dec 8, 2023 at 2:09
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Too long for a comment.

I will try with $x\&y$ version. Given $\epsilon$, let $\epsilon'=\min\{\epsilon,\frac12\}$.

$$0<|z-(1-i)|<\delta\iff \\ 0<(x-1)^2+(y+1)^2<\delta^2\tag1$$ and $$||\bar z^2-2|-2\sqrt2|<\epsilon\iff\\ 2\sqrt2-\epsilon<\sqrt{(x^2-y^2-2)^2+4x^2y^2}<2\sqrt2+\epsilon\tag2$$ Due to $(1)$, we will choose $\delta<1$ so that $(2)$ is comfortably satified.

From right: $$((1-\delta)^2-(-1-\delta)^2-2)^2+4(1+\delta)^4<8+4\sqrt2\epsilon$$ $$(-4\delta-2)^2+4(1+\delta)^4<8+4\sqrt2\epsilon$$ $$(16+16)\delta+4(4+6+4+1)\delta<4\sqrt2\epsilon$$ From left: $$8-4\sqrt2\epsilon'+\epsilon'^2<((1+\delta)^2-(-1+\delta)^2-2)^2+4(1-\delta)^4$$ $$-(4\sqrt2-1)\epsilon'<-8\delta+4(-4-4)\delta$$ Hence $$\delta=\min\{\frac{\sqrt2\epsilon}{23},\frac{(4\sqrt2-1)\epsilon'}{40}\}$$

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