If $f(z)$ is not analytic in $\mathbb C$, then $e^{f(z)}$ is also not analytic in $\mathbb C$. Prove or disprove.
It seems very intuitive that $e^{f(z)}$ will not be analytic in $\mathbb C$ if $f(z)$ is not analytic in $\mathbb C$. Consider $f(z) = u(x,y)+iv(x,y)$, then we can write $e^{f(z)} = U(x,y)+iV(x,y)$, where $U(x,y)=e^{u(x,y)}$cos$(v(x,y))$ and $V(x,y)=e^{u(x,y)}$sin$(v(x,y))$.
Assuming that $e^{f(z)}$ is analytic in $\mathbb C$, $U(x,y)$ and $V(x,y)$ should satisfy Cauchy-Riemann equations. Assuming that the partial derivatives of $u(x,y)$ and $v(x,y)$, with respect to $x$ and $y$, exist, we get the contradiction that $f(z)$ is analytic in $\mathbb C$.
But are we correct in assuming that the the partial derivatives of $u(x,y)$ and $v(x,y)$, with respect to $x$ and $y$, will always exist? If not, then what is the best way to approach this problem?
Example.
Let $f(z) = z$ if $\mathrm{Re}\; z$ is rational and $f(z) = 2\pi i + z$ otherwise. Then $f(z)$ is continuous nowhere, so certainly not analytic. But $e^{f(z)} = e^z$ is analytic.