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Show that $\lim\ z^3=z_0^3$ as $z$ approaches $z_0$.

Should I go about this where $z^n=r^n(cos(n\theta)+i(sin(n\theta)))$?

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    Whatever proof you know on the reals will work here. – 2017-02-22

2 Answers 2

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Let $\epsilon >0$. Take $\delta = \min\{1,\frac{\epsilon}{3(1+|z_{0}|)^{2}}\}$. If $|z-z_{0}|<\delta$, then $|z|\leq|z-z_{0}|+|z_{0}|< \delta+|z_{0}|$. So, $$|z^{3}-z_{0}^{3}|=|z-z_{0}||z^{2}+zz_{0}+z_{0}^{2}|< \delta((\delta+|z_{0}|)^{2}+(\delta+|z_{0}|)||z_{0}|+|z_{0}|^{2})\leq3\delta (1+|z_{0}|)^{2}\leq\epsilon$$

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Let $f(z)=z^3-z_0^3$. Then, we can write $f(z)$ as

$$f(z)=(z-z_0)^3+3z_0(z-z_0)^2+3z_0^2(z-z_0)$$

Hence from $(1)$, if $|z-z_0|<1$, then given $\epsilon>0$

$$\begin{align} f(z)&=|z^3-z_0^3|\\\\ &=|(z-z_0)^3+3z_0(z-z_0)^2+3z_0^2(z-z_0)|\\\\ &\le |z-z_0|\left(1+3|z_0|+3|z_0|^2\right)\\\\ &<\epsilon \end{align}$$

whenever $|z-z_0|<\delta=\min\left(1,\frac{\epsilon}{1+3|z_0|+3|z_0|^2}\right)$.

And we are done!

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    Please let me know how I can improve this answer too. As always, I really want to give you the best answer I can. If the answer was not useful, I am happy to delete it. -Mark – 2017-04-12