Differentiability of $\sin ^2(x+y)+i\cos ^2(x+y)$ I want to find the set of all points $z\in\mathbb{C}$ such that $f:\mathbb{C}\to\mathbb{C}$ given by $f(x+iy)=\sin ^2(x+y)+i\cos ^2(x+y)$ is differentiable at $z$.
Isn't it true that  $f$ is differentiable at all points of $\mathbb{C}$? Because $\sin$ and $\cos$ are differentiable... Am I wrong?
Thanks.

The total differential is straightforward:
$$ df(z) = 2 \sin(x+y) \cos(x+y) (dx + dy) - 2 i \cos(x+y) \sin(x+y) (dx + dy) $$
Also,
$$ dx = \frac{dz + d\bar{z}}{2} \qquad \qquad dy = \frac{dz - d\bar{z}}{2i} $$
Checking differentiability is the same thing as checking that the coefficient on $d\bar{z}$ is zero, if you write it in the $\{ dz, d\bar{z} \}$ basis. Equivalently, you set $dz=0$ (so that $dx + i dy = 0$) and check if the total differential is zero.
