Let $C$ be a closed staircase contour in $D$, not necessarily simple. If we show that $\displaystyle \oint_C \map f z \rd z = 0$, then the result follows from Zero Staircase Integral Condition for Primitive. The staircase contour $C$ is a concatenation of $C_1, \ldots, C_n$, where the image of each $C_k$ is a line segment parallel with either the real axis or the imaginary axis. Denote the parameterization of $C$ as $\gamma: \closedint a b \to \C$, where $\closedint a b$ is a closed real interval. Denote the parameterization of $C_k$ as $\gamma_k: \closedint {a_k} {b_k \to \C$. === Lemma === {{:Zero Simple Staircase Integral Condition for Primitive/Lemma}} === Splitting up the Contour === The lemma shows that given a staircase contour $C$, we can assume that for $k \in \set {1, \ldots, n - 1}$, the intersection of the images of $C_k$ and $C_{k + 1}$ is equal to their common end point $\map {\gamma_k} {b_k}$. This means that in order to intersect itself, $C$ must be a concatenation of at least $4$ directed smooth curves. Now, we prove the main requirement for Zero Staircase Integral Condition for Primitive, that $\displaystyle \oint_C \map f z \rd z = 0$. The proof is by induction over $n \in \N$, the number of directed smooth curves that $C$ is a concatenation of. === Basis for the Induction === For $n = 1$, $C$ can only be a closed staircase contour if $\gamma$ is constant, so: {{begin-eqn}} {{eqn | l = \oint_C \map f z \rd z       | r = \int_a^b \map f {\map \gamma t} \map {\gamma'} t \rd t       | c = {{Defof|Complex Contour Integral}} }} {{eqn | r = 0       | c = Derivative of Complex Polynomial: $\gamma$ is constant }} {{end-eqn}} For $n = 4$, $C$ can only be a closed staircase contour if $C$ is a simple contour. Then, $\displaystyle \oint_C \map f z \rd z = 0$ by the original assumption of this theorem. === Induction Hypothesis === For $N \in \N$, if $C$ is a closed staircase contour that is a concatenation of $n$ directed smooth curves with $n \le N$, then: :$\displaystyle \oint_C \map f z \rd z = 0$ === Induction Step === Suppose that $C$ is a closed staircase contour that is a concatenation of $n + 1$ directed smooth curves. If $C$ is a simple contour, the induction hypothesis is true by the original assumption of this theorem. Otherwise, define $t_0 = a$, and $t_3 = b$. Define $t_1 \in \closedint a b$ as the infimum of all $t \in \closedint a b$ for which $\gamma$ intersects itself. Then define $t_2 \in \hointl {t_1} b$ as the infimum of all $t \in \hointl {t_1} b$ for which $\map \gamma t = \map \gamma {t_1}$. For $k \in \set {1, \ldots, 3}$, define $\tilde C_k$ as the staircase contour with parameterization $\gamma \restriction {\closedint {t_{k - 1} } {t_k} }$. Then $\tilde C_2$ is a closed staircase contour that is a concatenation of at least $4$ directed smooth curves. Then both $\tilde C_1 \cup \tilde C_3$ and $\tilde C_2$ are a concatenation of fewer than $n + 1$ directed smooth curves, so: {{begin-eqn}} {{eqn | l = \oint_C \map f z \rd z       | r = \oint_{\tilde C_1 \cup \tilde C_2 \cup \tilde C_3} \map f z \rd z }} {{eqn | r = \oint_{\tilde C_1 \cup \tilde C_3} \map f z \rd z + \oint_{\tilde C_2} \map f z \rd z        | c = Contour Integral of Concatenation of Contours }} {{eqn | r = 0       | c = Induction Hypothesis }} {{end-eqn}} {{qed}} Category:Complex Analysis d9syvp4ve618ilyjlzt9dae82wmnqfj