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I am trying to prove that the $$\lim_{z \rightarrow z_0} z^2+c = z_0^2+c,$$ where $c$ is some complex constant.

I was taught that typically it is a good idea to begin from the end of the proof and work backwards. Therefore, I will start with $$|z^2-z^2_0| < \epsilon$$ $$\Rightarrow |z+z_0|\cdot|z-z_0| < \epsilon.$$ Now, if $|z-z_0|<|z+z_0|$, then it follows that $$|z-z_0|^2<|z-z_0| \cdot |z+z_0| < \epsilon$$ $$\Rightarrow |z-z_0| < \sqrt\epsilon.$$ In that case, we would be able to set $\delta = \sqrt\epsilon$, and the proof would follow as such: $$$$ Suppose $\epsilon > 0$. We will define $\delta = \sqrt\epsilon$. Since $\epsilon > 0$, it follows that $\delta > 0$. Now, $\forall z \in \mathbb{C}$, the expression $$0 < |z-z_0| < \delta$$ implies $$0 < |z-z_0| < \sqrt\epsilon.$$ It then follows $$0 < |z+z_0| \cdot |z-z_0| < |z+z_0| \cdot \sqrt\epsilon.$$ $$$$ I now find myself in another predicament because I feel like I need to make the assumption that $|z+z_0|<|z-z_0|$ in order to continue with the proof. By making this assumption, I will then know that $|z+z_0| < \sqrt\epsilon$. $$\Rightarrow |z^2-z_0^2| < |z+z_0| \cdot \sqrt\epsilon < \sqrt\epsilon \cdot \sqrt\epsilon = \epsilon$$ $$\Rightarrow |z^2-z_0^2| < \epsilon.$$ $$$$ This proof seems to depend on the assumption that $|z+z_0| < |z-z_0|$, which confuses me because when working backwards, I made the assumption that $|z+z_0| > |z-z_0|$. However, I don't believe that irregularity poses a threat to the proof itself. What does ruin the proof is that gaping assumption that $|z+z_0| < |z-z_0|$. I have tried doing the proof by cases, where I first assume that $|z+z_0| < |z-z_0|$, then I assume that $|z+z_0| \geq |z-z_0|$. I will now show the progress I have made in the second case, which is close to none. $$$$ Suppose $\epsilon > 0$. We will define $\delta = \sqrt\epsilon$. Since $\epsilon > 0$, it follows that $\delta > 0$. Now, $\forall z \in \mathbb{C}$, the expression $$0 < |z-z_0| < \delta$$ implies $$0 < |z-z_0| < \sqrt\epsilon.$$ It then follows $$0 < |z+z_0| \cdot |z-z_0| < |z+z_0| \cdot \sqrt\epsilon.$$ $$$$ That's about as far as I've figured. However, what I have found is that if we assume that $$0 < |z-z_0| < \sqrt\epsilon$$ which we have been, it follows that $$(x-x_0)^2+(y-y_0)^2 < \epsilon,$$ by definition of the definition of the modulus ($|z-z_0| = \sqrt{(x-x_0)^2+(y-y_0)^2}$), where $z=x+iy$ and $z_0=x_0+iy_0$. Then, $$x^2-2xx_0+x_0^2+y^2-2yy_0+y_0^2 < \epsilon$$ $$\Rightarrow x^2+2xx_0+x_0^2+y^2+2yy_0+y_0^2 < \epsilon + 4xx_0 + 4yy_0$$ $$\Rightarrow |z+z_0| < \sqrt{\epsilon + 4xx_0 + 4yy_0}.$$ $$$$ I don't really see this helping either however. Thanks for the help!

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2 Answers 2

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You want to show that you can find a $\delta > 0$ such that for all $z \in U_{\delta}(z_0) := \{z \in \mathbb{C}: |z-z_0| < \delta \}$, you have that $z \in U_{\epsilon}(z_0)$. Now for any $\delta > 0$, there exists a constant $C_\delta$ such that $|z+z_0| < C_{\delta}$ for all $z \in U_{\delta}(z_0)$:

$$|z+z_0| \leq |z| + |z_0| \leq 2|z_0| + \delta =: C_{\delta}, \quad z \in U_{\delta}(z_o).$$ Note that $C_\delta$ decreases as $\delta$ decreases, which is good. We see that for all $\delta < 1$, we have $|z+z_0| < C := 2|z_0| + 1$ for all $z \in U_{\delta}(z_0)$. Therefore, if $\delta < \text{min}(1,\epsilon/C)$ we find that for $z \in U_{\delta}(z_0)$ we have $$ |z^2 - z_0^2| = |z+z_0||z-z_0| \leq C|z-z_0| \leq C\cdot \frac{1}{C}\epsilon = \epsilon.$$

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If I understood correctly, you have to find for every $\epsilon>0$ a $\delta>0$ with the following property: $|z-z_0|<\delta\Rightarrow|(z^2+c)-(z_0^2+c)|<\epsilon$.

Looking at your way of solving the problem, I noticed, you are stuck with the idea that $\delta:=\sqrt{\epsilon}$. This leads to your deadlock argument.

Try the following approach: Define $z=z_0+t$, then $|z-z_0|=|t|<\delta$. Now, take a look at $|z^2+c-(z_0^2+c)|$. Try to define your $\delta$ this way.

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