1. **Problem Statement:** Find the contour integral $$\int_C f(z) \, dz$$ where $$f(z) = \pi \exp(\pi \overline{z})$$ and $$C$$ is the boundary of the square with vertices $$0, 1, 1+i, i$$ traced counterclockwise. 2. **Understanding the function and contour:** The function involves the conjugate $$\overline{z}$$, which is not holomorphic (complex differentiable) because it depends on $$\overline{z}$$, not just $$z$$. 3. **Parameterizing the contour:** The contour $$C$$ is the square with vertices $$0 \to 1 \to 1+i \to i \to 0$$. We can split the integral into four line integrals over each side: - Side 1: from $$0$$ to $$1$$ along the real axis, parameterize as $$z = t$$, $$t \in [0,1]$$. - Side 2: from $$1$$ to $$1+i$$ along vertical line, parameterize as $$z = 1 + it$$, $$t \in [0,1]$$. - Side 3: from $$1+i$$ to $$i$$ along horizontal line, parameterize as $$z = 1 - t + i$$, $$t \in [0,1]$$. - Side 4: from $$i$$ to $$0$$ along vertical line, parameterize as $$z = i - it$$, $$t \in [0,1]$$. 4. **Expressing $$f(z)$$ on each side:** Recall $$f(z) = \pi \exp(\pi \overline{z})$$. Since $$\overline{z}$$ is the complex conjugate, for $$z = x + iy$$, $$\overline{z} = x - iy$$. 5. **Compute each integral:** **Side 1:** $$z = t$$, $$\overline{z} = t$$, $$dz = dt$$. $$\int_0^1 \pi e^{\pi t} dt = \pi \int_0^1 e^{\pi t} dt = \pi \left[ \frac{e^{\pi t}}{\pi} \right]_0^1 = e^{\pi} - 1$$ **Side 2:** $$z = 1 + it$$, $$\overline{z} = 1 - it$$, $$dz = i dt$$. Integral: $$\int_0^1 \pi e^{\pi (1 - it)} i dt = \pi i e^{\pi} \int_0^1 e^{-i \pi t} dt = \pi i e^{\pi} \left[ \frac{e^{-i \pi t}}{-i \pi} \right]_0^1 = e^{\pi} (1 - e^{-i \pi})$$ Since $$e^{-i \pi} = -1$$, this becomes: $$e^{\pi} (1 - (-1)) = 2 e^{\pi}$$ **Side 3:** $$z = 1 - t + i$$, $$\overline{z} = 1 - t - i$$, $$dz = -dt$$. Integral: $$\int_0^1 \pi e^{\pi (1 - t - i)} (-dt) = -\pi e^{\pi (1 - i)} \int_0^1 e^{-\pi t} dt = -\pi e^{\pi (1 - i)} \left[ \frac{e^{-\pi t}}{-\pi} \right]_0^1 = e^{\pi (1 - i)} (1 - e^{-\pi})$$ **Side 4:** $$z = i - it$$, $$\overline{z} = -i + it$$, $$dz = -i dt$$. Integral: $$\int_0^1 \pi e^{\pi (-i + it)} (-i) dt = -\pi i e^{-i \pi} \int_0^1 e^{i \pi t} dt = -\pi i e^{-i \pi} \left[ \frac{e^{i \pi t}}{i \pi} \right]_0^1 = - e^{-i \pi} (e^{i \pi} - 1)$$ Since $$e^{-i \pi} = -1$$ and $$e^{i \pi} = -1$$, this becomes: $$-(-1)(-1 - 1) = 2$$ 6. **Sum all integrals:** $$I = (e^{\pi} - 1) + 2 e^{\pi} + e^{\pi (1 - i)} (1 - e^{-\pi}) + 2$$ Simplify terms: $$I = e^{\pi} - 1 + 2 e^{\pi} + e^{\pi} e^{-i \pi} (1 - e^{-\pi}) + 2$$ Recall $$e^{-i \pi} = -1$$, so: $$I = 3 e^{\pi} - 1 + (- e^{\pi})(1 - e^{-\pi}) + 2 = 3 e^{\pi} - 1 - e^{\pi} + e^{\pi} e^{-\pi} + 2$$ Since $$e^{\pi} e^{-\pi} = 1$$: $$I = 2 e^{\pi} - 1 + 1 + 2 = 2 e^{\pi} + 2$$ 7. **Final answer:** $$\boxed{\int_C f(z) \, dz = 2 e^{\pi} + 2}$$