Prove that for closed polygons, the sum of the exponents [latex]\left(\alpha_{1} / \pi\right)-1,\left(\alpha_{2} / \pi\right)-1, \ldots,\left(\alpha_{n} / \pi\right)-1[/latex] in the Schwarz-Christoffel transformation (8.9) or (8.10), page 247 , is equal to -2 . [latex]{\frac{d w}{d z}}=A(z-x_{1})^{\alpha_{1}/\pi-1}(z-x_{2})^{\alpha_{2}/\pi-1}\cdot\cdot\cdot(z-x_{n})^{\alpha_{n}/\pi-1}[/latex] (8.9) [latex]w=A\int(z-x_{1})^{\alpha_{1}/\pi-1}(z-x_{2})^{\alpha_{2}/\pi-1}\cdot\cdot\cdot(z-x_{n})^{\alpha_{n}/\pi-1}d z+B[/latex] (8.10)

The sum of the exterior angles of any closed polygon is [latex]2 \pi[/latex]. Then [latex]\left(\pi-\alpha_{1}\right)+\left(\pi-\alpha_{2}\right)+\cdots+\left(\pi-\alpha_{n}\right)=2 \pi[/latex] and dividing by [latex]-\pi[/latex], we obtain as required, [latex]\left(\frac{\alpha_{1}}{\pi}-1\right)+\left(\frac{\alpha_{2}}{\pi}-1\right)+\cdots+\left(\frac{\alpha_{n}}{\pi}-1\right)=-2[/latex]

Question 8.18: Prove that for closed polygons, the sum of the exponents (a1...

Schaum’s Outline of Complex Variables

Prove that for closed polygons, the sum of the exponents $\left(\alpha_{1} / \pi\right)-1,\left(\alpha_{2} / \pi\right)-1, \ldots,\left(\alpha_{n} / \pi\right)-1$ in the Schwarz-Christoffel transformation (8.9) or (8.10), page 247 , is equal to -2 .

${\frac{d w}{d z}}=A(z-x_{1})^{\alpha_{1}/\pi-1}(z-x_{2})^{\alpha_{2}/\pi-1}\cdot\cdot\cdot(z-x_{n})^{\alpha_{n}/\pi-1}$ (8.9)

$w=A\int(z-x_{1})^{\alpha_{1}/\pi-1}(z-x_{2})^{\alpha_{2}/\pi-1}\cdot\cdot\cdot(z-x_{n})^{\alpha_{n}/\pi-1}d z+B$ (8.10)

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