Is $e^{x+y}(1+xy)^{2}$ analytic?? Is the function defined by $g(z) = e^{x+y} \cdot (1+xy)^{2}$ analytic? If $f(z) = u(x,y) + i\cdot v(x,y)$ then one can use the Cauchy Riemann equation to check, whether $f$ is analytic or not. But how does one proceed here??
 A: HINT:
If $f$ is analytic, then we must have
$$\frac{\partial f}{\partial \bar z}=\frac12 \left(\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}\right)=0$$
SPOLIER ALERT:  Scroll over the highlighted area to reveal the solution

The Cauchy-Riemann Equations can be written in compact notation as $$\frac{\partial f}{\partial \bar z}=\frac12 \left(\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}\right)=0 \tag 1$$Then, for $f(x,y)=e^{x+y}(1+xy)^2$, neither partial derivative is identically zero, and both partial derivatives are real, then $(1)$ cannot hold.  Therefore, $f$ cannot be analytic. 

A: If $f(x,y) = u(x,y)+iv(x,y)$ with $v(x,y)=0$, then $u$ (and therefore $f$) must be constant. This can be easily verified by the Cauchy-Riemann equations.
The same holds if $f$ is purely imaginary.
A: Any holomorphic function from $\mathbb C$ to $\mathbb R$ is constant.
Proof: By Cauchy Riemann equations, if $f = u + iv$, then $f= u$. Also, $\frac{\partial v}{\partial x} = 0$ and $\frac {\partial v}{\partial y} = 0$.
Thus the partials in $u$ are also $0$, and your function is constant.
Your map is not constant. So it is not holomorphic.
