1. Find the harmonic conjugate of $u(x,y) = x^3 - 3xy^2$. The harmonic conjugate $v$ satisfies the Cauchy-Riemann equations: $$u_x = v_y, \quad u_y = -v_x$$ Calculate partial derivatives: $$u_x = 3x^2 - 3y^2, \quad u_y = -6xy$$ From $u_x = v_y$, we get: $$v_y = 3x^2 - 3y^2$$ Integrate w.r.t. $y$: $$v = 3x^2 y - y^3 + h(x)$$ From $u_y = -v_x$: $$-6xy = -v_x \implies v_x = 6xy$$ Calculate $v_x$ from above $v$: $$v_x = 6xy + h'(x)$$ Equate: $$6xy + h'(x) = 6xy \implies h'(x) = 0$$ So $h(x)$ is constant, ignore it. Thus, $$v = 3x^2 y - y^3 + c$$ Answer: option c. 2. Find the analytic function whose real part is $e^x (x \cos y - y \sin y)$. Let $f(z) = u + iv$ be analytic with real part $u = e^x (x \cos y - y \sin y)$. Note that $z = x + iy$ and $e^z = e^x (\cos y + i \sin y)$. Rewrite $u$: $$u = \text{Re}(z e^z)$$ So the analytic function is: $$f(z) = z e^z + c$$ Answer: option b. 3. Find the orthogonal trajectory of family of curves $\cos x \cosh y = c$. Differentiate implicitly: $$-\sin x \cosh y + \cos x \sinh y \frac{dy}{dx} = 0$$ Solve for $\frac{dy}{dx}$: $$\frac{dy}{dx} = \frac{\sin x \cosh y}{\cos x \sinh y}$$ Orthogonal trajectories satisfy: $$\frac{dy}{dx} = -\frac{\cos x \sinh y}{\sin x \cosh y}$$ Rewrite: $$-\sin x \sinh y = c$$ Answer: option a. 4. Find the imaginary part of the analytic function whose real part is $r^2 \cos 2\theta - r \sin \theta$. Using polar form, the harmonic conjugate $v$ satisfies Cauchy-Riemann equations in polar coordinates. The harmonic conjugate of $r^2 \cos 2\theta$ is $r^2 \sin 2\theta$. The harmonic conjugate of $-r \sin \theta$ is $r \cos \theta$. So, $$v = r^2 \sin 2\theta + r \cos \theta + c$$ Answer: option c. 5. Evaluate $\int_{y=x} (x^2 - i y) dz$ where $z = x + iy$. Parametrize along $y = x$, so $z = x + i x = x(1 + i)$. Then, $$dz = (1 + i) dx$$ Substitute $y = x$: $$\int (x^2 - i x) dz = \int (x^2 - i x)(1 + i) dx$$ Expand: $$(x^2 - i x)(1 + i) = x^2 + i x^2 - i x - i^2 x = x^2 + i x^2 - i x + x = (x^2 + x) + i (x^2 - x)$$ Integrate from 0 to 1: $$\int_0^1 (x^2 + x) dx + i \int_0^1 (x^2 - x) dx = \left[ \frac{x^3}{3} + \frac{x^2}{2} \right]_0^1 + i \left[ \frac{x^3}{3} - \frac{x^2}{2} \right]_0^1 = \left( \frac{1}{3} + \frac{1}{2} \right) + i \left( \frac{1}{3} - \frac{1}{2} \right) = \frac{5}{6} + i \left( -\frac{1}{6} \right) = \frac{5}{6} - \frac{i}{6}$$ Answer: option c.