$f(z)=\bar{z}$ has no primitive As a consequence of Goursat's Theorem, we can prove that every holomorphic function on an open disk has primitive. 
Question: Is it true that every continuous function $f\colon D\rightarrow \mathbb{C}$ has primitive? [D=open disc in $\mathbb{C}$]
The answer I think is "NO". But my explanation involves use of some important theorems. The example I thought is $f(z)=\overline{z}$. If this $f$ has primitive, then $f$ has to be holomorphic, a contradiction.
The problem I would concern here is the following: 
Problem: Can we give an elementary argument that $f(z)=\overline{z}$ has no primitive in any open disc?
(I want to avoid the theorem that "a complex function which is once differentiable is infinitely many times differentiable").
 A: The proof of @Fabian is the morally right one: indeed, to find the antiderivative of a function $f(z)$ one needs to integrate $\int^z f(\zeta) d \zeta$ since we have the complex Leibniz-Newton formula
$$F'=f \implies \int_a^b f(\zeta) d \zeta = F(b) - F(a)$$
moreover the integral must not depend on the path.
Another approach: assume that there exists $F$ with $F'(z) = f(z)$. This is a complex derivative. Now we can use the real partial derivatives and conclude:
\begin{eqnarray}
\frac{\partial F}{\partial z}  &=& f(z)\\
\frac{\partial F}{\partial \bar z} &=&0
\end{eqnarray}
From the equations above it follows right away that if $f$ is of class $C^k$ then $F$ is of class $C^{k+1}$. Say $f$ is of class $C^1$;  we obtain
$$\frac{\partial f}{\partial \bar z} = \frac{\partial }{\partial  \bar z}\frac{\partial F}{\partial  z}= \frac{\partial }{\partial   z}\frac{\partial F}{\partial \bar z}=0$$
because the operators $\frac{\partial }{\partial   z}$, $\frac{\partial }{\partial  \bar z}$ commute. 
Now notice that $\frac{\partial \bar z}{\partial   \bar z}\equiv 1$
A: Take an open disc around $z_0$. Then you can find $\varepsilon>0$ such that the closed path
$$ \gamma: \phi\in[0,2\pi] \mapsto z_0 + \varepsilon e^{i\phi}$$
is inside the disc.
Now note that 
$$\oint_\gamma \bar z dz =
i \epsilon\int_0^{2\pi} (\bar z_0 + \epsilon e^{-i\phi}) e^{i\phi} d\phi 
= i \epsilon^2 \int_0^{2\pi} d\phi = 2\pi i  \epsilon^2 \ne0.$$
However, for $\bar z$ to have a primitive this integral should be $0$ as the path is closed.
