The Cauchy-Goursat theorem states that if a function $f$ is analytic on, and in the interior of, a contour $C$ then: $$\int_{C} f(z) \hspace{1mm} dz = 0.$$
From here, a lot of lecture notes deduce the following.
Cor. Suppose $f$ is analytic on some simply connected region $A$. Then for any two points $a, b \in A$ we have that $$\int_{C} f(z) \hspace{1mm} dz$$ is independent of the contour $C$ in $A$ joining $a$ and $b$.
However, the 'proof' often only shows $$\int_{C_1} f(z) \hspace{1mm} dz = \int_{C_2} f(z) \hspace{1mm} dz.$$ when $C_1$ and $C_2$ are non-intersecting (because then $C := C_1\cup C_2$ is a closed contour). What is the best way to prove independence for all $C_1$ and $C_2$? I can 'see' that one can just pick a third contour $C_3$ which doesn't intersect either $C_1$ or $C_2$, and the proof follows from that. But how can I make the existence of $C_3$ rigorous without 'advanced machinery'?
If $\gamma_1$ and $\gamma_2$ are two paths going from $a$ to $b$ and if $\gamma$ is the path $\gamma_1$ followed by the inverse path of $\gamma_2$ (usually, $\gamma$ is denoted by $\gamma_2^-\vee\gamma_1$, then $\gamma$ is a closed path and therefore $\oint_\gamma f(z)\,\mathrm dz=0$. But this is the same thing as asserting that$$\int_{\gamma_1}f(z)\,\mathrm dz=\int_{\gamma_2}f(z)\,\mathrm dz.$$Whether the paths $\gamma_1$ or $\gamma_2$ intersect or not doesn't matter.