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Let $T$ be a closed triangle contained, together with a neighbourhood, in the domain of a holomorphic function $f$. We prove Cauchy theorem for a triangle without assuming continuity of $f^{\prime }$.

Join the edge midpoints to subdivide $T$ into four triangles, each with half its diameter and perimeter. Their positively oriented contour integrals sum to the integral around $T$, because the internal edges cancel. At least one subtriangle has integral of modulus at least one quarter of the original. Repeat that choice to obtain nested closed triangles $T_n$ satisfyingTheir intersection is a single point $z_{\ast }$, by compactness and shrinking diameter. complex differentiability at a point at $z_{\ast }$ givesThe first two terms have polynomial primitives, hence integrate to zero around each triangle. For sufficiently large $n$, $|r|<\epsilon$ on $T_n$, so the complex line integral estimate givesCombining the bounds gives $|{\int }_{\partial T}f(z)dz|\le \epsilon d_0{P}_0$. Since $\epsilon$ is arbitrary,The orientation can be reversed with no change to the conclusion.

For a positively oriented simple closed contour, the residue theorem states that the integral of a function holomorphic on and inside the contour except at finitely many isolated interior singularities is $2\pi i$ times the sum of its residues. There must be no singularity on the contour.

Factor the denominator as $(z^2+z+1)(z^2-z+1)$. For $R>1$, the upper semicircle encloses the two simple polesThe other two poles are in the lower half-plane. Since $D^{\prime }(z)=4z^3+2z=2z(2z^2+1)$,Writing $h=\sqrt{3}/2$ and using $1/a=1/2-ih$, $1/b=-1/2-ih$, their numerator difference simplifies toThus for every $R>1$ the closed-contour integral isTo remove the arc, on $C_R$ we have $|e^{iz}|=e^{-\mathrm{Im}z}\le 1$ and $|z^4+z^2+1|\ge R^4-R^2-1$. For sufficiently large $R$, the complex line integral estimate givesThe real-line integral is absolutely convergent. Its imaginary part is zero by oddness of $\sin x/(x^4+x^2+1)$. Therefore