0
$\begingroup$

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..

Please Help.

$\endgroup$
  • $\begingroup$ 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 $(x,y) \to (0,0)$? $\endgroup$ – busman Apr 13 '14 at 10:36
  • $\begingroup$ @busman Since $z$ is in the exponential, $f$ is not divided into $u+iv$ explicitly. How do you evaluate partial derivatives? $\endgroup$ – user140374 Apr 13 '14 at 10:39
  • $\begingroup$ Okay, you can try $$\log f(z) = \frac{-1}{z^4} = -\frac{(x+iy)^4}{(x^2+y^2)^4}$$ and perform implicit derivation. $\endgroup$ – busman Apr 13 '14 at 10:45
  • $\begingroup$ 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 $$\log f(z) = \left( - \frac{\bar{z}}{\| z \|} \right)^4.$$ $\endgroup$ – busman Apr 13 '14 at 10:57
  • $\begingroup$ @busman Thank you. Now it is clear that it is not continuous at 0 hence not differentiable at 0. But what about Cauchy Riemann Equation/? $\endgroup$ – user140374 Apr 13 '14 at 11:16
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.