Solution
= Solution
Choose
$$
\boxed{f(z)=z},
$$
or add any constant. Then $f'(z)=1$, so the <vector field> is exactly $\mathbf r$ and necessarily has the same <flux> as in part (a). The formula from part (b)(i) confirms this:
$$
2\pi(z_0+1)+\pi z_0(1-z_0^2)
=\pi(2+3z_0-z_0^3).
$$