Continuity of a complex valued function and Cauchy Riemann equation checking.

Question

$f(z) = \left\{ \begin{array}{ll} 0 &, z = 0 \ \\ \exp(-1/z^4) & ,z \neq 0\ \end{array} \right.$ 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?

Cauchy Riemann equations only tell about differentiablity. But a continuous function needs not to be differentiable. — Frank Lu, Jan 22 '17 at 14:10
BTW, I don't see why you need to make a transformation. As $z\to 0$, we have $\frac{1}{\exp(z^4)}\to 1$ — Frank Lu, Jan 22 '17 at 14:19
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. — BAYMAX, Jan 22 '17 at 14:21
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 $(x_0,y_0)$. — Frank Lu, Jan 22 '17 at 14:26
I very strongly suspect that the function should be $\exp\bigl( - \frac{1}{z^4}\bigr)$ for $z\neq 0$. — Daniel Fischer, Jan 22 '17 at 15:04
You compute the partial derivatives, and see that they come out to be $0$. You probably know the real version with $\exp(-1/x^2)$? — Daniel Fischer, Jan 22 '17 at 15:20
Some posts about the same function - which might be useful too: Is $f(z)=\exp (-\frac{1}{z^4})$ holomorphic? and Is this function holomorphic at 0? — Martin Sleziak, Jan 23 '17 at 18:42
And this might also help you: Does the limit of $e^{-1/z^4}$ as $z\to 0$ exist? Found using Approach0. — Martin Sleziak, Jan 23 '17 at 19:10

score 1 · Accepted Answer · answered Jan 22 '17 at 15:20

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 $z=0$.

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. — BAYMAX, Jan 24 '17 at 01:21

Continuity of a complex valued function and Cauchy Riemann equation checking.

1 Answers1