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Suppose $f$ is continuous on $A = \{z : \text{Im}(z)\geq0\}$ and analytic on $\Omega=\{z : \text{Im}(z) > 0\}$. Let $T$ be a triangle in $A$ with one side on the $x$-axis. Prove that $\int_Tf(z) dz = 0$.

I am totally stuck on how to do this. I know there will be some approximation to $T$ be triangles in the upperhalf plane.

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For $\epsilon > 0$ let $T_\epsilon = T + i\epsilon$, then $\int_{T_\epsilon} f(z)\,dz = 0$. On the other hand \begin{align*} \left|\int_{T_\epsilon} f(z)\,dz - \int_T f(z)\,dz\right| &= \left|\int_{T} f(z+i\epsilon)\,dz - \int_T f(z)\,dz\right|\\ &\le \int_T \left|f(z+i\epsilon) - f(z)\right|\,|dz|\\ &\le C \cdot \sup_{z \in T} \left|f(z+i\epsilon) - f(z)\right|\\ \end{align*} where $C$ denotes the circumference of $T$. As $f$ is uniformly continuous on $T + [0,1]i$, this converges to $0$ for $\epsilon \to 0$.

Thus $\int_T f(z)\,dz = 0$.

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This is what I was writing. –  ncmathsadist Jun 20 '12 at 19:45

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