It's still related to my previous question about proving a function is analytic except at its singularities.
I've found a preposition that said
if $f(z)$ is differentiable at $z$, then it satisfies the Cauchy-Riemann equations
And it also said the converse isn't true. But i also found another thread said that the converse is also true. Here is the link. I'm confused now. But since I've Sen the counterexample, I believe we can't use Cauchy Riemann to prove this. Here I'm thinking how do we know the function is analytic except the singularity if the function itself I don't know whether it's differentiable except the singularities or not.
By the way, this is the function that I'm talking about:
$$\frac{e^z}{z-2}$$
Anyway, I've been trying to show whether this satisfies Cauchy Riemann equations or not and the answer is yes, it satisfies and of course I'm assuming the function is differentiable except $z=2$.
But again I can't conclude that the function is analytic except at simple pole $2$ since as far as I know the Cauchy Riemann is just a necessary condition for a function to be analytic.
Conclusion: how to prove this problem? Should I show that the function can be written as power series and to support this proof I will also include the fact that that function satisfies Cauchy Riemann? Help me please. I know this is unusual since I'm not talking about how to prove/disprove this function is "differentiable everywhere".
A complex function is complex differentiable at a point if and only if it is real differentiable at that point and satisfies the CR equations at that point. It is analytic on an open set if and only if it is complex differentiable at every point in the set. A function is real differentiable at a point if its partial derivatives exist in a neighborhood and are continuous at the point.
That said, the easiest way to prove that your given function is analytic is by using the fact that sums, differences, products and quotients (as long as the denominator doesn't vanish) of complex differentiable functions are complex differentiable. So your function is complex differentiable everywhere except where the denominator vanishes, since it's composed of everywhere complex differentiable functions. So it's complex differentiable on $\mathbb C\backslash\{2\}$, which is open, so your function is analytic on that open set. CR equations + real differentiability are overkill here.