Proving limits of complex numbers

Question

Prove $\displaystyle\lim\limits_{z\mathop\to 0}\frac{\overline z^2}{z^2}=1$ and $\displaystyle\lim\limits_{z\mathop\to 0}\frac{\overline z^2}{z}=0 $

Write $z = r e^{i\theta}$. But I don't see how the first identity can be true unless you specify how $z \to 0$. — Stephen Montgomery-Smith, Apr 01 '14 at 03:59

André Nicolas · Accepted Answer · 2014-04-01T04:19:06.303

4

The first is not correct. For example, let $z_n=\dfrac{1}{n}(1+i)$. Then $\dfrac{\bar{z_n}^2}{z_n^2}=-1$ for all $n$.

For the second, go to polar form, letting $z=re^{i t}$. Then our ratio is $re^{-3it}$, which has limit $0$ as $r\to 0$.

edited Apr 01 '14 at 04:19

answered Apr 01 '14 at 04:02

André Nicolas

http://www.wolframalpha.com/input/?i=lim+z+to+0+z*^2%2Fz^2 Wolfram Alpha says it is though ? – Superbus Apr 01 '14 at 04:03
Interesting. I am barely acquainted with Wolfram Alpha. It seems highly useful. Somewhat imperfect. – André Nicolas Apr 01 '14 at 04:08
So does the limit not exist for the first* one? – Superbus Apr 01 '14 at 04:16
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@LuciusTarquiniusSuperbus Be wary of using Wolfram Alpha for complex limits. I have seen quite a few examples of it giving wrong answers when the limit is dependent on the path (hence does not exist). For instance, look at this. – Viktor Vaughn Apr 01 '14 at 04:16
+1 $i^{1/2}$ was the example that came to mind when I saw the question. – robjohn Apr 01 '14 at 04:23
@LuciusTarquiniusSuperbus: the first limit does not exist. – robjohn Apr 01 '14 at 04:24

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