Let $f(x,y) = u+iv = e^y\cos(x)+ie^y\sin(x)$.
Then
$$\bar{f}(z,y) = e^y\cos(x)-ie^y\sin(x)$$ $$u(x,y)=e^y\cos(x)$$ $$v(x,y)=e^y\sin(x)$$
The given function represents a locally sourceless and irrotational flow since $\bar{f}$ is analytic on $\mathbb{C}$.
Note that
$$\frac{\partial{v}}{\partial{x}} = e^y\cos(x)$$ $$\frac{\partial{u}}{\partial{y}} = e^y\cos(x)$$ $$\frac{\partial{v}}{\partial{x}} - \frac{\partial{u}}{\partial{y}} = 0$$
This shows that $f$ is globally irrotational.
$$\frac{\partial{u}}{\partial{x}} = -e^y\sin(x)$$ $$\frac{\partial{v}}{\partial{y}} = e^y\sin(x)$$ $$\frac{\partial{u}}{\partial{x}} + \frac{\partial{v}}{\partial{y}} = 0$$
This shows that $f$ is globally sourceless.
So $f$ is both globally sourceless and globally irrotational.
