2

So I previously made a post Showing a complex function is NOT holomorphic

I am needing to ask about if my methods are correct, and I also have a few general questions that are easiest to ask with an example. So I have a few equations and I want to show they are holomorphic, however, in the examples and solutions provided, I am confused.

I have in my notes for example, show $g(z)=\sin(z)-\frac{z^2}{z+1}$ is holomorphic.

So, I thought I would have to write everything in terms of x and y and do partial derivatives to check Cauchy Riemann. But then I also have in my notes that if g has a derivative everywhere then it is holomorphic. So the solution given was simply, it is holomorphic for $z \neq -1$ because $$g'(z)=cos(z)-\frac{(z+1)(2z)-(z^2)}{(z+1)^{2}}$$

Consider,

$$f(z)=\frac{\cos(z)}{z^2+1}$$

and say we wanted to show it was holomorphic. Is really all I have to do is say $$f'(z)=\frac{(z^2+1)(-\sin(z))-(\cos(z))(2z)}{(z^2+1)^2}$$

Just doing it for the first time it feels like I must be wrong because it almost seems to easy. Is there something I am not realizing?

Community
  • 1
PersonaA
  • 1,049
  • I told you, forget $d/dx,d/dy$, all you have to show is that $f(z) = f(z_0)+(z-z_0) f'(z_0)+o(|z-z_0|)$ as $z\to z_0$ from any direction of the complex plane (note how this is different from being real differentiable). Using that, it is easy to show the composition, quotient, multiplication, addition of holomorphic functions is holomorphic, and that your function is. – reuns Sep 22 '16 at 23:43
  • @user1952009 Can you please explain that notation, and what you are saying by this? id like to be able to understand but I'm not sure – PersonaA Sep 25 '16 at 03:02
  • As $z \to z_0$, the $o(|z-z_0|)$ term $\to 0$ faster than the $(z-z_0)f'(z_0)$ term, so that $\lim_{z \to z_0} \frac{f(z)-f(z_0)}{z-z_0} =\lim_{z \to z_0} \frac{(z-z_0)f'(z_0)+o(|z-z_0|)}{z-z_0} = f'(z_0)$. So this is equivalent to the usual definition of complex differentiability at $z_0$, and from it you can easily show that the sum/product/quotient/composition of complex differentiable functions are complex differentiable. – reuns Oct 14 '16 at 20:35
  • And holomorphic means complex differentiable not only at one point but on an open set $U$ (for example $U = {z \in \mathbb{C},|z-z_0| < 1}$). In the same way, analytic at only one point doesn't exist, a function is analytic on an open set. – reuns Oct 14 '16 at 20:35

3 Answers3

1

A function that is COMPLEX differentiable in a region is already holomorphic there!

So, yes the existence of the first (complex) derivation is actually enough.

Peter
  • 86,576
1

The strategy is correct, and so is the computation.

A function $f(z)$ is differentiable on $\mathbb{C}$ if and only if it is holomorphic (analytic). This can be seen from the Cauchy-Riemann criteria. If the partial derivatives exist AND are themselves continuous, then the function has a derivative $f'(z)$ at every point the criteria hold. The converse is true as well.Review the statement of the thereom. What's more that the existence of one derivative implies smoothness for $f(z)$, i.e. derivatives of all orders exist. This is not true for functions on the real line.

In other words showing that a function $f(z)$ satisfies Cauchy-Riemann implies the derivative exists everywhere that Cauchy-Riemann criteria holds. This is often advantageous to computing the derivative using the formal definition of $f'(z)$

Kernel_Dirichlet
  • 1,783
1

The sum of g(z) and f(z) both holomorph in D results in a function holomorph in D. In this case sin(z) is holomorph in C and $\frac{z^2}{z+1}$ is holomorph in C(z=-1) so the sum is holomorph in C(z=-1)

Pedro
  • 39