I need to show that $f(z)$ is nowhere analytic, where $$ f(z) = e^{z\operatorname{Re}(z)}.$$
Letting $z = x + yi$,
$$ f(x,y)=e^{x^2}e^{xyi}=e^{x^2}(\cos(xy)+i\sin(xy))$$
and letting $$u(x,y)=e^{x^2}\cos(xy),\quad v(x,y)=e^{x^2}\sin(xy),$$
we have:
\begin{align}u_x &= e^{x^2}(2x\cos(xy)-y\sin(xy))\\ u_y &= -e^{x^2}x\sin(xy) \\ v_x &= e^{x^2}(2x\sin(xy)+y\cos(xy))\\v_y &= e^{x^2}x\cos(xy)\end{align}
Using the Cauchy-Riemann equations $u_x = v_y$ and $u_y = -v_x$, I end up with the equations:
$$ x\cos(xy) = y\sin(xy) $$ $$ -x\sin(xy) = y\cos(xy) $$
- Is this it? Am I able to conclude using these two equations?
- If not, am I supposed to show that $f(z)$ is nowhere analytic by some other method?
Note that from the two equations, we get $$\frac{x\cos xy}{-x\sin xy}=\frac{y\sin xy}{y\cos xy}\implies -\cot xy=\tan xy\implies \tan^2 xy=-1$$ which is absurd for $x,y\in\Bbb R\setminus\{0\}$.