Continuity of a complex function Show that $f(z)=xy+iy$ is everywhere continuous but is not analytic. We have $\displaystyle w=f(z)=u(x,y)+iv(x,y)$ and $\displaystyle z=x+iy$. Do I just say that as both u and v here are polynomials, so there are continuous everywhere?? How to show that some function is continuous everywhere? Second part is to check the C-R equations, which are not satisfied, so the function is not analytic. Anything to add/substract here?

Observing that they are each polynomials is enough to see that the function is continuous in each variable separately but not necessarily that it is continuous. (You probably saw examples in Calc III where a function was not continuous at the origin even though it was continuous in each variable separately at the origin. That is, where $\lim_{x \to 0} f(x,0)$ and $\lim_{y \to 0} f(0,y)$ both exist but $\lim_{(x,y) \to (0,0)} f(x,y)$ does not exist.)

An easier way is to write $f(z)=y(x+i)$, then it is a product of two polynomials, each of which are continuous, therefore it is continuous.