Is $f(z)=e^{z^2+z+1}+e^{Im(z)}$ Holomorphic?

Question

$$z=x+iy$$ 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!

Answer 1

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

$$v(x,y)=e^{x^2-y^2+x+1}\sin(2xy+y)$$

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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.

Answer 2

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.

Answer 3

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.

Answer 4

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.