Is $f(z)=e^{z^2+z+1}+e^{Im(z)}$ Holomorphic? $$z=x+iy$$
I can write the function $f$ in this form:
$$f(z)=u(x,y)+iv(x,y)$$
 
$$e^{x^2-y^2+2ixy+x+iy+1}+e^y$$

$$v(x,y)=e^{x^2-y^2+2ixy+x+iy+1}$$
$$u(x,y)=e^y$$
I need to check these conditions:
\begin{cases}
\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y} \\
\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}
\end{cases}
$$\frac{\partial u}{\partial x}=0 \ne (-2y+2ix+i) \ \ e^{x^2-y^2+2ixy+x+iy+1}=\frac{\partial v}{\partial y}$$

$f$ is not holomorphic

Is it correct? 
Thanks!
 A: Suppose $f(z) = e^{z^2+z+1} + e^y$ is holomorphic in some open connected set $U\subset \mathbb C.$ Because $e^{z^2+z+1}$ is holomorphic in $U,$ we then have $e^y = f(z) - e^{z^2+z+1}$ holomorphic in $U.$ But every real-valued holomorphic function in $U$ is constant (from the C-R equations or the open mapping theorem). Cleary $e^y$ is not constant on $U,$ contradiction.
A: This is not correct, the real and imaginary parts of $f$ ($u$ and $v$) are not given correctly:
\begin{align}
f(x,y)& =\exp\left[(x+iy)^2+x+iy+1 \right] +\exp(y) \\
&=\exp\left[ (x^2-y^2+x+1)+i(2xy+y) \right]+\exp(y) \\
&=\exp\left[x^2-y^2+x+1\right]\left(\cos(2xy+y)+i\sin(2xy+y)\right)+\exp(y) \\
&=\left[\exp(x^2-y^2+x+1)\cos(2xy+y)+\exp(y) \right]+i\left[\exp(x^2-y^2+x+1)\sin(2xy+y)\right].
\end{align}
From here, $u$ and $v$ are clear.  
A: A simpler approach is to notice that $e^{1+z+z^2}$ is holomorphic while $g(z)=\exp\text{Im}(z)$ is not, since the integral of $g(z)$ on the boundary of the square $[-1,1]\times[-1,1]$, counter-clockwise oriented, is not zero.
A: The analysis in the OP is not correct.  While it is true that
$$f(z)=e^{z^2+z+1}+e^y=e^{(x^2-y^2+x+1)+i(2xy+y)}+e^y$$
the real part of $f(z)$, $u(x,y)$, is given by
$$u(x,y)=e^{x^2-y^2+x+1}\cos(2xy+y)+e^y$$
and the imaginary party of $f(z)$, v(x,y)$ is given by
$$v(x,y)=e^{x^2-y^2+x+1}\sin(2xy+y)$$


An easy way to proceed to test analyticity is to note that $e^{\text{Im}(z)}=e^{-i(z-\bar z)/2}$ and
$$\frac{\partial e^{-i(z-\bar z)/2}}{\partial \bar z}\ne 0 \tag 1$$
which is equivalent to showing that the Cauchy-Riemann equations are not satisfied.

