If we have a complex function
$f(x+iy) = \cos(x)\cos(y)-i\sin(x)\sin(y)$ for $x,y ∈ {\rm I\!R}$
Find the set of points at which f:$\mathbb{C}$ $\Rightarrow$ $\mathbb{C}$ are holomorphic and compute $f '(z)$.
I understand how to differentiate the function however i'm unsure as to go about proving at which points the function is holomorphic. And how to show it.
Holomorphic functions $f:\Omega\to \mathbb{C}$ satisfy a pair of partial differential equations called the Cauchy-Riemann Equations. If we write $f(x,y)=u(x,y)+iv(x,y)$ then the Cauchy-Riemann Equations read $$ \begin{cases} u_x=v_y\\ u_y=-v_x. \end{cases}$$ Compute the set of $(x,y)\in \mathbb{C}$ so that $f$ satisfies these equations. Then, to differentiate, observe that $$ f'(z)=\lim_{h\to 0}\frac{f(z+h)-f(z)}{h}.$$ However, because $h$ can tend to $0$ from whichever direction we wish, $f'(z)=u_x+iv_x$. In particular, the complex derivative is equal to either of the $x$ or $y$ partial derivatives.