Question

prove f(z) =cos2z is analytic​

Answer 1

if $\large\rm { f(z)}$ is a complex valued function, we can write it as

$\large\rm{ f( x + \imath y) = u(x,y) + iv(x,y)}$

where u and v are real valued functions.

having u_x=v_y and u_y=−v_x at some point z is not enough to conclude that f is holomorphic there. However, if we add that u and v have continuous partial derivatives at z, then we can conclude that f is holomorphic at z. (Alternatively, we could add that the mapping :

$\large\rm { (x,y) \mapsto (u,(x,y) , v(x,y))}$ is differentiable as a $\large\rm { \mathbb{R}^{2} \longrightarrow \mathbb{R}^{2}}$ function to conclude f is holomorphic at x+y.)

after evaluation we have

$\large\rm { u(x,y) = \cos \ x \ \cosh \ y}$

$\large\rm { v(x,y) = - \sin \ x \ \sinh \ y}$ We can check that the Cauchy Riemann equations hold everywhere, and furthermore, that all four partial derivatives are continuous everywhere. This is enough to conclude that f is holomorphic.so it is analytic.