# How do i prove this function satisfies Cauchy-Riemann equation?

This is the first time i'm trying to apply abstract complex analysis to an explicit function . So, i need to see what a correct approach looks like.

In the class, professor had shown easy examples and i am now able to check properties of rational functions. However, i think the problem below is completely different from rational functions.

Define $f(z)=e^{-1/z^4}$ if $z\neq 0$ and $f(0)=0$

(1) Show that $f$ satisfies the Cauchy-Riemann equations everywhere

(2) Show that $f$ is holomorphic everywhere except at $z=0$

(3) Show that $f(z)$ is not countinuous at $z=0$

Since $f$ is differentiable everywhere except $z=0$, $f$ satisfies the Cauchy-Riemann equation except at $z=0$. How do i prove that it satisfies the equation at $z=0$ and is not differentiable at $z=0$?

Moreover why this is not continuous at $z=0$? I cannot find a way of approaching $0$ yields discontinuity..

• Well, recall that a function may have partial derivatives at a given point yet not be differentiable in that point. With that idea, set $f(z) = f(x+iy)$ and compute $\partial_xf$ and $\partial_yf$ around zero. What happens then if you make $(x,y) \to (0,0)$? – busman Apr 13 '14 at 10:36
• @busman Since $z$ is in the exponential, $f$ is not divided into $u+iv$ explicitly. How do you evaluate partial derivatives? – user140374 Apr 13 '14 at 10:39
• Okay, you can try $$\log f(z) = \frac{-1}{z^4} = -\frac{(x+iy)^4}{(x^2+y^2)^4}$$ and perform implicit derivation. – busman Apr 13 '14 at 10:45
• Check that the numerator has a minus, not a plus (it won't let me edit). Also check that the above expression can be re-written as $$\log f(z) = \left( - \frac{\bar{z}}{\| z \|} \right)^4.$$ – busman Apr 13 '14 at 10:57
Okay, so with the last comment taken into account, a reformulation of the Cauchy-Riemann equation is the following: $$\frac{\partial f}{\partial \overline{z}} = 0.$$ You can prove this in a rather simple fashion with the aid of the Wirtinger derivatives. Now, assume $\| z \| > 0$ and consider the expressions given in my comments. Then, $$\frac{\partial}{\partial \overline{z}} \log f(z) = \frac{\partial}{\partial \overline {z}}\left(- \frac{1}{z^4} \right) = 0.$$ Thus, $$\frac{1}{f(z)} \frac{\partial}{\partial \overline{z}}f(z)=0.$$ Since $z \neq 0$ and $\exp(z) \neq 0$ for all $z \in \mathbb{C}$, we get $$\frac{\partial}{\partial \overline{z}}f(z)=0.$$ Now, making $z \to 0$ in both sides of the equality we get $\lim_{z \to 0}\partial_{\overline{z}}f(z)=0$ and we are done.