= Solution
{c}
Orient the contour so that the plus side is on the left; for a positively oriented simple closed contour this is its interior. If $\zeta(s)$ is a smooth arclength parametrization with $\zeta(0)=t$, define the <Cauchy principal value> by deleting a symmetric interval $|s|<\eta$ and taking $\eta\downarrow0$. This pairs the two singular sides at equal local arclength.
Near $t$, write $f(\zeta)=f(t)+[f(\zeta)-f(t)]$. The second term divided by $\zeta-t$ is regular because $f$ is analytic there. For the constant term, indent the contour on either side of $t$ by a small semicircle. Its limiting contribution to the Cauchy integral is $\pm f(t)/2$, while the remaining contour tends to its principal value. This proves the <Sokhotski–Plemelj theorem>:
$$
w_\pm(t)=\pm\frac12 f(t)+\frac1{2\pi i}
\operatorname{PV}\int_\Gamma\frac{f(\zeta)}{\zeta-t}\,d\zeta.
$$
In particular $w_+-w_-=f$ and $w_++w_-=g$ for the given integral equation.
Define $W=w$ inside and $W=-w$ outside. Its jump is $W_+-W_-=w_++w_-=g$, and it vanishes at infinity. Therefore
$$
W(z)=\frac1{2\pi i}\int_\Gamma\frac{g(t)}{t-z}\,dt.
$$
Indeed, the difference between the two sides has no jump, extends analytically across the smooth contour, and vanishes at infinity; <Liouville's theorem> makes it zero. Applying the <Sokhotski–Plemelj theorem> again,
$$
f=W_++W_-=\boxed{\frac1{\pi i}
\operatorname{PV}\int_\Gamma\frac{g(t)}{t-z}\,dt}.
$$
Thus the normalized closed-contour singular integral operator is its own inverse.
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