Why does $z^{-1}$ not have an anti derivative?

$\DeclareMathOperator{\Re}{Re}\DeclareMathOperator{\Im}{Im}$Because the integral of the reciprocal about the unit circle is non-zero, the value of the integral $$ w = \int_{1}^{z} \frac{d\zeta}{\zeta} \tag{1} $$ depends not only on $z$, but on the path from $1$ to $z$. You can denote the value of the integral $w = \log z$, but doing so does not make $\log$ single-valued. The green paths below depict two "lifts" of the unit circle, starting at $(1, 0)$ and tracing counterclockwise, ending at $(1, 2\pi i)$; tracing clockwise, ending at $(1, -2\pi i)$.

For each non-zero $z$, there are infinitely many distinct values of $\log z$. Fixing a path $\gamma$ from $1$ to $z$ fixes a value of $$ \log z = \int_{\gamma} \frac{d\zeta}{\zeta}. $$ Every other choice differs by an added integer multiple of $2\pi i$, corresponding to a path that winds about the origin some number of times before following $\gamma$ to $z$.

There's no perfectly adequate picture because the graph of the relation $w = \log z$ naturally sits in the space of ordered pairs $(z, w)$ of complex numbers (a real four-dimensional space), but the projection $$ (z, w) \mapsto (\Re z, \Im z, \Re w + \Im w) $$ (which perhaps is not as widely-known as it should be), captures the qualitative "parking garage" behavior of the Riemann surface of $\log$.