complex analysis - Prove that $\int_Bg(z)dz=\int_{B_1}g(z)dz+\int_{B_2}g(z)dz$ - Mathematics Stack Exchange

let $b$,$b_{1}$ , $b_{2}$ non-interesecfting simple closed curves, such $b_{1}$ , $b_{2}$ contained inside $b$. let $g$ analytic in region between these curves, , on curves. want prove using cauchy-goursat theorem , basic properties of integrals $$\int_bg(z)dz=\int_{b_1}g(z)dz+\int_{b_2}g(z)dz$$

my attempt @ proof; first let direction along $b$ anti-clockwise, since $g$ satisfies conditions cauchy-goursat theorem have $$\int_{b}g(z)dz=0$$ let direction along $b_{1}$ , $b_{2}$ clockwise, cauchy goursat theorem again have $$\int_{-b_{1}}g(z)dz+\int_{-b_{2}}g(z)dz=0\implies -\int_{b_1}g(z)dz-\int_{b_2}g(z)dz=0$$ $$\implies \int_{b_1}g(z)dz+\int_{b_2}g(z)dz=0=\int_{b}g(z)dz=\sum_{i=1}^{2}\int_{b_{i}}g(z)dz$$ required

could tell me if incorrect (and how fix if is) , if correct?

this proposition stated not hold. $b_1$ or $b_2$ allowed inside of other, resulting in $$\int_bg(z)dz=\int_{b_1}g(z)dz=\int_{b_2}g(z)dz$$

now add condition neither of $b_1$ , $b_2$ resides inside of other, $g$ analytic in (open , connected) domain $d$ contains union of following 2 sets. first set intersection of, outerior of $b_1$ , $b_2$, , interior of $b$. second $b\cup b_1\cup b_2$. produce desired equation.

proof: transform domain homeomorphically $b\cup b_1\cup b_2$ becomes circles depicted in graph below. take contour $c$ shown in graph below running counter-clockwise on large circle, clockwise on smaller circles, letting distance between gap zero.

take contour $\gamma$ image of $c$ on original domain, apply cauchy-goursat theorem $g$ on $d$ , contour $\gamma$ $$0=\int_\gamma g(z)dz=\int_b g(z)dz-\int_{b_1}g(z)dz-\int_{b_2}g(z)dz+0+0$$ last 2 $0$'s signifies integration on parallel sides of straight "streets" in graph in opposite directions results in cancellation of integration on parts of $\gamma$.

now have obtained desired equation.