For which values of $z$ is $g(z)=\cos(x)\cosh(y)+i\sin(x)\sinh(y)$ differentiable?

Question

For which values of z is $g(z)=\cos(x)\cosh(y)+i\sin(x)\sinh(y)$ differentiable?

I used the Cauchy-Riemann formulas and became that:

$\partial u / \partial x = - \sin(x) \cosh (y)$

$\partial v / \partial y = \sin(x) \cosh (y)$

$\partial u / \partial y = \cos(x) \sinh (y)$

$\partial v / \partial x = \cos(x) \sinh (y)$

Next I tried to solve:

$ - \sin(x) \cosh (y) = \sin(x) \cosh (y) $ => $x=k\pi $ with k an interger

$\cos(x) \sinh (y)= -\cos(x) \sinh (y)$ => $x=\pi/2 + k\pi$ with k an interger

But what's the conclusion?

Answer 1

I did not check your computations, but if the Cauchy-Riemann equations have no solutions, then $g$ is differentiable nowhere.

Answer 2

Taking $z = x+iy$ and because $\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b)$ then $$\cos(z) = \cos(x)\cos(iy) - \sin(x)\sin(iy)$$ also $\sin(iy) = i\sinh(y)$ and $\cos(iy) = \cosh(y)$ $$\cos(z) = \cos(x)\cosh(y) - i\sin(x)\sinh(y)$$ let's assume $\tilde{x} = -x$, if $\cos(x) = \cos(\tilde{x})$ but $\sin(x) = -\sin(\tilde{x})$ so being $z = -\tilde{x}+iy$ $$\cos(z) = \cos(\tilde{x})\cosh(y) + i\sin(\tilde{x})\sinh(y)$$ and $\cos(z)$ is differentiable $\mathbb{C}^{\infty}$