Let's solve the problem step by step as requested: --- ## **Step 1: Understanding and Sketching the Path \(\alpha(t)\)** Given: \[ \alpha(t) = e^{i3\pi t} + 2e^{i12\pi t}, \quad \leq t \leq 1 \] Let's analyze each part: - \(e^{i3\pi t}\): This term traces out a circle of radius 1, starting at \(1\) when \(t=\) and ending at \(e^{i3\pi} = -1\) when \(t=1\). It traces half a circle (since \(3\pi\) is 1.5 turns). - \(2e^{i12\pi t}\): This term traces out a circle of radius 2, starting at 2 when \(t=\) and ending at \(2e^{i12\pi} = 2\) when \(t=1\) (since \(12\pi\) is 6 full turns, so it ends where it started). **Combined Path:** The path \(\alpha(t)\) is a combination of these two motions. As \(t\) goes from to 1, the point moves along a path in the complex plane starting from: \[ \alpha() = 1 + 2 = 3 \] and ending at: \[ \alpha(1) = -1 + 2 = 1 \] - The path starts at \(z=3\), ends at \(z=1\), and is not a simple arc or straight line, but a spiral-like motion due to the different frequencies in the exponentials. **Explanation:** - The first term rotates slowly (half a circle). - The second term rotates quickly (6 circles) around the origin, but with twice the radius. **Sketch:** If you were to sketch, you would start at \(z=3\), spiral around the origin (due to the fast \(2e^{i12\pi t}\)), while also slowly moving from \(z=3\) to \(z=1\) due to the first term. --- ## **Step 2: Evaluating the Path Integrals** ### **a) \(\int_{\alpha} \frac{dz}{z}\)** This integral is the logarithmic change in \(z\) as we move along \(\alpha\): \[ \int_{\alpha} \frac{dz}{z} = \log(\alpha(1)) - \log(\alpha()) \] since the path does not encircle the origin (as verified from the sketch). \[ \alpha(1) = 1, \quad \alpha() = 3 \] \[ \Rightarrow \int_{\alpha} \frac{dz}{z} = \log(1) - \log(3) = - \log(3) = -\log(3) \] --- ### **b) \(\int_{\alpha} \frac{dz}{z+2}\)** Let’s consider the function \(f(z) = 1/(z+2)\). \[ \int_{\alpha} \frac{dz}{z+2} = \log(\alpha(1) + 2) - \log(\alpha() + 2) \] \[ \alpha(1) + 2 = 1 + 2 = 3 \] \[ \alpha() + 2 = 3 + 2 = 5 \] \[ \Rightarrow \int_{\alpha} \frac{dz}{z+2} = \log(3) - \log(5) = -\log\left(\frac{5}{3}\right) \] --- ### **c) \(\int_{\alpha} \frac{dz}{z^2 + 2z}\)** Factor \(z^2 + 2z = z(z+2)\). Use partial fractions: \[ \frac{1}{z(z+2)} = \frac{A}{z} + \frac{B}{z+2} \] Solving, \(A=1/2, B=-1/2\): \[ \int_{\alpha} \frac{dz}{z^2 + 2z} = \int_{\alpha} \left(\frac{1}{2z} - \frac{1}{2(z+2)}\right) dz = \frac{1}{2} \int_{\alpha} \frac{dz}{z} - \frac{1}{2} \int_{\alpha} \frac{dz}{z+2} \] Plug in previous results: \[ = \frac{1}{2}(-\log 3) - \frac{1}{2}(-\log(5/3)) = -\frac{1}{2} \log 3 + \frac{1}{2} \log(5/3) \] \[ = \frac{1}{2} \log\left(\frac{5/3}{3}\right) = \frac{1}{2} \log\left(\frac{5}{9}\right) \] --- ## **Final Answers** \[ \boxed{ \begin{aligned} &\int_{\alpha} \frac{dz}{z} = -\log(3) \\ &\int_{\alpha} \frac{dz}{z+2} = -\log\left(\frac{5}{3}\right) \\ &\int_{\alpha} \frac{dz}{z^2 + 2z} = \frac{1}{2} \log\left(\frac{5}{9}\right) \end{aligned} } \] **Summary of Steps:** - Step 1: Analyze and sketch the path \(\alpha\) in \(\mathbb{C}\). - Step 2: Use properties of logarithmic integrals and partial fractions to compute each path integral using start and end points.