Prove this function is analytic

Question

Given $$f(z):=Re{(e^{1/z})}$$ prove it is analytic everywhere except the origin.

I wrote it like this $$f(x,y):=e^\frac{x}{x^2+y^2}\cos(\frac{y}{x^2+y^2})$$ and tried to use C-R equations with $u(x,y)=f(x,y)$ and $v(x,y)=0$. But these are obviously not true.

Answer 1

You can't apply C-R equations here as C-R equations are applicable only for complex valued functions
Here, $f(z)$ is a real valued function
so, in order to prove f(z) is analytic everywhere except the origin you need to prove: $f(z)=f(x,y)=\exp(\frac{x}{x^2+y^2}) \cos(\frac{y}{x^2+y^2})$ is differentiable $\forall (x,y)$ except $(x,y)=(0,0)$
I hope now you can prove it

Answer 2

Sorry, my earlier comment missed the point of this problem entirely. It is clear that $e^{1/z}$ is analytic except at $z=0,$ because $1/z$ is analytic except for a pole at the origin, an $e^z$ is entire. It then follows automatically as you say that $\Re(e^{1/z})$ is harmonic except at the origin. There is no need to deal with the horrid derivatives in either the C-R equations or Laplace's equation.

EDIT:

If you want to use the Cauchy-Riemann equations, you can't deal with just the real part of $f$. You have to write$$ e^{1/z} = \exp\left(\frac{x-iy}{x^2+y^2}\right) = \exp\left(\frac{x}{x^2+y^2}\right)\left(\cos\left(\frac{y}{x^2+y^2}\right)-i\sin \left(\frac{y}{x^2+y^2}\right)\right) $$ Then you can verify the Cauchy-Riemann equations with $$ \begin{align} u &= \exp\left(\frac{x}{x^2+y^2}\right)\cos\left(\frac{y}{x^2+y^2}\right)\\ v &= -\exp\left(\frac{x}{x^2+y^2}\right)\sin\left(\frac{y}{x^2+y^2}\right) \end{align}, $$ or at least you can if you have enough patience.

I think the point of the problem though is that if you wanted to prove $u$ harmonic directly, you'd have to slog through Laplace's equation, but if you just observe that it's the real part of an analytic function, you're done.

Answer 3

$\exp\left(1/z\right)\,$ is analytic in $\mathbb{C}\backslash\{0\}$, therefore $u(x,y)=\Re{\left[\exp\left(1/z\right)\right]}$ is harmonic in $\mathbb{C}\backslash\{0\}$. You can also check this by verifying the Laplace PDE $$u_{xx}+u_{yy}=0.$$

A useful theorem to keep in mind is

Theorem $\,$If a function $f(z)=u(x,y)+iv(x,y)$ is analytic in a domain $D$, then its component functions $u$ and $v$ are harmonic in $D$.

This can be easily proved by using the Cauchy-Riemann Equations.