Let's say we have the following complex function
$$f(z)=k\cdot\frac{e^z}{z}$$
Is it analytic in some domain? When substituting with $z=x+iy$, using trigonometry and deriving the function on the form $f(x,y)=u(x,y)+iv(x,y)$ as follows;
$$f(x,y)=k\cdot\frac{e^x(\cos y + i \sin y)(x-iy)}{x^2+y^2}$$
it seems to fail, at least the first of, the Cauchy-Riemann's equations: $$\frac{\partial u(x,y)}{\partial x} = \frac{\partial v(x,y)}{\partial y} \text{and} \frac{\partial u(x,y)}{\partial y} = -\frac{\partial v(x,y)}{\partial x}$$
But, intuitively it feels like the function $f(z)$ should be analytic in $\mathbb C / \{ z = 0 \} $. Is this correct? How should one go about to think about analyticity in a domain and how to show that?
You can apply the Cauchy-Reimann's test on $e^z$ and ${1\over z}$ separately and then multiply them together:$$e^z=e^x\cos y+ie^x\sin y\implies {u=e^x\cos y\\v=e^x\sin y}\\{\partial u\over\partial x}=e^x\cos y\\{\partial u\over\partial y}=-e^x\sin y\\{\partial v\over\partial x}=e^x\sin y\\{\partial v\over\partial y}=e^x\cos y$$therefore$${\partial u\over\partial x}={\partial v\over\partial y}\\{\partial u\over\partial y}=-{\partial v\over\partial x}$$for all $x,y\in \Bbb R$. Similarly$${1\over z}={x\over x^2+y^2}-i{y\over x^2+y^2}$$so we have$${\partial u\over\partial x}={\partial v\over\partial y}={y^2-x^2\over(x^2+y^2)^2}\\{\partial u\over\partial y}=-{\partial v\over\partial x}=-{2xy\over(x^2+y^2)^2}$$as long as $(x,y)\ne(0,0)$. Since $e^z$ and $1\over z$ are analytic over $\Bbb C$ and $\Bbb C-\{0+i0\}$ then their product, ${e^z\over z}$ is analytic over $\Bbb C-\{0\}$.