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My teacher said without explaining that $$\lim\limits_{z\to 0}\frac{z^4}{\left|z\right|^4}$$ does not exist, why is this?

Note: $z$ is a complex number.

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    Write $z$ in polar form. You'll see it then. – 2017-01-06
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    Check my answer to this question: http://math.stackexchange.com/questions/2084512/how-to-show-that-fz-fracz5z4-satisfies-the-cauchy-riemann-equatio/2084521#2084521 – 2017-01-06
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    Do you know why the limit of $\frac{x}{|x|}$ as $x\to 0$ doesn't exist for real numbers $x$? Because if you approach zero from the right it gives an answer of one and if you approach from the left the answer is negative one. Since these are unequal, the limit doesn't exist. For your specific example, a similar issue occurs (*though not from the positive,negative, or even imaginary axes*). Approaching zero from two different directions (say $\theta=0$ and $\theta=\frac{\pi}{4}$) will give different answers. – 2017-01-06

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Let $$z=re^{i\theta}.$$then

$$\frac{z^4}{|z|^4}=\frac{r^4e^{4i\theta}}{r^4}=e^{4i\theta}.$$ so the limit depends on which direction $z$ takes when it tends to $0$.

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    Does this occur for any $\frac{z^n}{|z|^{n}},n\in\mathbb{N}$ or just for $n=4$? – 2017-01-06
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    @Isaac For each $n\in \Bbb N$. – 2017-01-06
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    @Isaac try it yourself. What is $\frac{z^n}{|z|^n}$ equal to in terms of $r,\theta,n$ if you rewrite $z$ in polar form? – 2017-01-06
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    @SalahFatima not for $n=0$ (which some may consider included in $\mathbb{N}$) – 2017-01-06
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$$\lim_{r \rightarrow 0}\frac{r^4\exp(4\theta i)}{r^4}=\exp(4\theta i)=\cos(4\theta)+i\sin(4\theta)$$

If $\theta=0$, $\cos(4\theta)+i\sin(4\theta)=1$

If $\theta=\frac{\pi}{4}$, $\cos(4\theta)+i\sin(4\theta)=-1$

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$$\lim\limits_{z\to 0}\frac{z^4}{\left|z\right|^4}$$ Using polar coordinates, we have $$\lim\limits_{r\to 0^+}\frac{r^4 e^{4i\phi}}{r^4}=e^{4i\phi}$$ This limit is dependent on $\phi$. Therefore $$\lim\limits_{z\to 0}\frac{z^4}{\left|z\right|^4}\ \text{does not exist}$$