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For$\lim_{z\rightarrow0}\frac{\bar{z}^2}{z^2}$

  1. Does $\lim_{z\rightarrow0}\frac{\bar{z}^2}{z^2}$ exist? If the answer is no, why?
  2. Does $\bar{z}$ represents $a-bi$?
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    @Peter I have changed $\bar{x}$ to $\bar{z}$ in the title.2012-10-27

1 Answers 1

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If $z = a + bi$, then $\bar{z} = a - bi$. Equivalently, if $z = r e^{i\theta}$, $\bar{z} = r e^{-i \theta}$.

Take $z = re^{i\theta}$ and as $z \to 0$, we have $r \to 0$.

Plug in $z = r^{i\theta}$ in $\dfrac{\bar{z}^2}{z^2}$. Now, let $r \to 0$ and see what happens for different $\theta$'s.