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$f(z) = \left\{ \begin{array}{ll} 0 &, z = 0 \ \\ \exp(-1/z^4) & ,z \neq 0\ \end{array} \right.$ I could show by making a transformation of $\frac{1}{z^4}$ as $w$, that $f$ is not continuous at origin, is there any other way out to prove discontinuity at origin?

Also I am getting into complications while showing Cauchy Riemann equations hold at origin? How to solve this?

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    Cauchy Riemann equations only tell about differentiablity. But a continuous function needs not to be differentiable. – 2017-01-22
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    BTW, I don't see why you need to make a transformation. As $z\to 0$, we have $\frac{1}{\exp(z^4)}\to 1$ – 2017-01-22
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    think this is the case where CR equatons hold at origin but the function is not differentiable at origin.Just thought that $\frac{1}{z^4}$ would again be a complex number reducing complications. – 2017-01-22
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    Then for CR equations how we should proceed ? – 2017-01-22
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    By writing $z=x+iy$ where $x,y$ are the real and imaginary parts of $z$, every complex function $f(z)$ has a corresponding function $f(x,y)$ of two real variables. Then $f$ is differentiable at $z_0=x_0+iy_0$ if and only if $f(x,y)$ is differentiable (in the real variable sense) at $(x_0,y_0)$ and the CR equations hold at $(x_0,y_0)$. – 2017-01-22
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    I very strongly suspect that the function should be $\exp\bigl( - \frac{1}{z^4}\bigr)$ for $z\neq 0$. – 2017-01-22
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    @DanielFischer yes just did the edit now. – 2017-01-22
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    @DanielFischer how we could check for CR equations ? – 2017-01-22
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    You compute the partial derivatives, and see that they come out to be $0$. You probably know the real version with $\exp(-1/x^2)$? – 2017-01-22
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    Yes , that will help. – 2017-01-22
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    Some posts about the same function - which might be useful too: [Is $f(z)=\exp (-\frac{1}{z^4})$ holomorphic?](http://math.stackexchange.com/q/291233) and [Is this function holomorphic at 0?](http://math.stackexchange.com/q/484098) – 2017-01-23
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    And this might also help you: [Does the limit of $e^{-1/z^4}$ as $z\to 0$ exist?](http://math.stackexchange.com/q/1477231) Found [using Approach0](https://approach0.xyz/search/?q=%24%5Cexp(-1%2Fz%5E4)%24&p=1). – 2017-01-23

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This is a classical example (by Looman) that satisfies the Cauchy–Riemann equations everywhere but is not analytic, or even continuous, at $z = 0$.

Hint:

For the continuity issue, show that $$ \lim_{z\to 0}f(z)=0 $$ is not true. Consider the limit $$ \lim_{t\to 0+}f(t(1+i)). $$

For the CR equation at $z=0$, try to use the definition of partial derivatives. Note for instance that $$ u(x,0) = \exp(-1/x^4) \quad\text{and}\quad u(0,y) = \exp(-1/y^4)$$ where $u$ denotes the real part of $f$. Eventually you would end up with the calculation of $$ \lim_{h\to 0}\frac{\exp(-h^4)}{h}=0 $$ which implies the CR equation at $z=0$.

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    In the limit calculation for showing CR eqns to hold at origin,how can we get a $h^4$ term , please show the limit value calculation. – 2017-01-24