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

  1. Does $\lim_{z\rightarrow0}\frac{\bar{z}^2}{z^2}$ exits? If the answer is no, why?
  2. Does $\bar{z}$ represents $a-bi$?
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Do you mean $\frac{\overline{x}^{2}}{z^{2}}$, or should the $x$ be a $z$? And yes, $\overline{z}$ is the complex conjugate of $z=a+bi$, $\overline{z}=a-bi$. –  user123123 Oct 27 '12 at 17:11
    
@Peter I have changed $\bar{x}$ to $\bar{z}$ in the title. –  user17762 Oct 27 '12 at 17:14

1 Answer 1

up vote 7 down vote accepted

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.

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