Question

verify Cauchy Goursat formula where f(z) = z+1 and D is the region bounded by the triangle with vertices at 0,3 and 3+2i.

Answer 1

Question #72095, Math / Complex Analysis

Verify Cauchy Goursat formula where $f(z) = z + 1$ and $D$ is the region bounded by the triangle with vertices at 0,3 and $3 + 2i$.

Solution.

Cauchy Goursat formula:

$$\oint_{D} f(z) \, dz = 0$$

Then:

$$f(z) = u(x, y) + i v(x, y)$$ $$\oint_{D} f(z) \, dz = \oint_{D} u(x, y) \, dx - v(x, y) \, dy + i \oint_{D} v(x, y) \, dx + u(x, y) \, dy$$

In our case:

$$u(x, y) = x + 1; \, v(x, y) = y$$

From vertex 0 to vertex 3:

$$y = 0; \, 0 \leq x \leq 3; \, dy = 0$$ $$\int_{0,3} z \, dz = \int_{0}^{3} (x + 1) \, dx = \left(\frac{x^2}{2} + x\right) \Bigg|_{0}^{3} = \frac{9}{2} + 3 = 7.5$$

From vertex 3 to vertex $3 + 2i$:

$$x = 3; \, 0 \leq y \leq 2; \, dx = 0$$ $$\int_{3,3 + 2i} z \, dz = - \int_{0}^{2} y \, dy + i \int_{0}^{2} (3 + 1) \, dy = - \left. \frac{y^2}{2} \right|_{0}^{2} + i (4y)_{0}^{2} = -2 + 8i$$

From vertex $3 + 2i$ to vertex 0:

$$y = \frac{2}{3} x; \, 0 \leq x \leq 3; \, dy = \frac{2}{3} \, dx$$

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$$\begin{array}{l} \text{Question \#72095, Math / Complex Analysis} \\ \int_{3+2i,0} zd z = \int_{3}^{0} \left(x + 1 - \frac{2}{3} \cdot \frac{2}{3} x\right) dx + i \int_{3}^{0} \left(\frac{2}{3} x + \frac{2}{3} (x + 1)\right) dx = \\ = \left(\frac{5}{9} \cdot \frac{x^{2}}{2} + x\right) \Bigg|_{3}^{0} + i \left(\frac{4}{3} \cdot \frac{x^{2}}{2} + \frac{2}{3} x\right) \Bigg|_{3}^{0} = -\left(\frac{5}{9} \cdot \frac{9}{2} + 3\right) - i \left(\frac{4}{3} \cdot \frac{9}{2} + \frac{2}{3} \cdot 3\right) = -5.5 - 8i \end{array}$$

So far:

$$\begin{array}{l} \oint_{D} f(z) dz = \int_{0,3} zd z + \int_{3,3+2i} zd z + \int_{3+2i,0} zd z = \\ = 7.5 + (-2 + 8i) + (-5.5 - 8i) = 0 \end{array}$$

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