1. **Problem 8:** Prove continuity of \( f(z) = \frac{x^3(1+i) - y^3(1-i)}{x^2 + y^2} \) for \( z \neq 0 \) and \( f(0) = 0 \), and check Cauchy-Riemann equations at origin. Step 1: Define \( u(x,y) = \frac{x^3 - y^3}{x^2 + y^2} \) and \( v(x,y) = \frac{x^3 + y^3}{x^2 + y^2} \). Step 2: Check \( \displaystyle \lim_{(x,y) \to (0,0)} u(x,y) \) and \( \lim_{(x,y) \to (0,0)} v(x,y) \). Use polar coordinates: \( x = r \cos \theta, y = r \sin \theta \). Then \( u = \frac{r^3(\cos^3 \theta - \sin^3 \theta)}{r^2} = r(\cos^3 \theta - \sin^3 \theta) \to 0 \) as \( r \to 0 \). Similarly, \( v = r(\cos^3 \theta + \sin^3 \theta) \to 0 \). Hence, \( f(z) \) is continuous at origin. Step 3: Compute partial derivatives at origin: \( u_x = \lim_{h \to 0} \frac{u(h,0)-u(0,0)}{h}, u_y, v_x, v_y \). At \( y=0 \), \( u(h,0) = \frac{h^3}{h^2} = h \to 0 \), so \( u_x(0,0) = \lim_{h \to 0} \frac{h - 0}{h} = 1 \). Similarly, \( u_y(0,0) = \lim_{k \to 0} \frac{u(0,k) - 0}{k} = \lim_{k \to 0} \frac{-k - 0}{k} = -1 \). For \( v_x(0,0) \), at \( y=0 \), \( v(h,0) = \frac{h^3}{h^2} = h \), so \( v_x(0,0) = 1 \). For \( v_y(0,0) \), at \( x=0 \), \( v(0,k) = \frac{k^3}{k^2} = k \), so \( v_y(0,0) = 1 \). Step 4: Check Cauchy-Riemann equations: \( u_x = v_y \) and \( u_y = -v_x \). From above, \( u_x = 1, v_y = 1, u_y = -1, v_x = 1 \), so \( u_x = v_y \) true, \( u_y = -v_x \) true. Therefore, Cauchy-Riemann equations hold at origin. --- 2. **Problem 9:** Find \( a, b \) such that \( f(z) = (x^2 - 2xy + ay^2) + i(bx^2 - y^2 + 2xy) \) is analytic. Step 1: Set \( u = x^2 - 2xy + ay^2, v = bx^2 - y^2 + 2xy \). Step 2: Compute partial derivatives: \( u_x = 2x - 2y, u_y = -2x + 2ay \) \( v_x = 2bx + 2y, v_y = -2y + 2x \) Step 3: According to Cauchy-Riemann, \( u_x = v_y \) gives \( 2x - 2y = -2y + 2x \) which holds for all \( x,y \). \( u_y = -v_x \) gives \( -2x + 2ay = - (2bx + 2y) \Rightarrow -2x + 2ay = -2bx - 2y \). Equate coefficients: For \( x \): \( -2 = -2b \Rightarrow b = 1 \). For \( y \): \( 2a = -2 \Rightarrow a = -1 \). Step 4: Therefore, \( a = -1, b = 1 \). Step 5: Express \( f(z) \) in terms of \( z \): Recall \( z = x+iy, z^2 = (x+iy)^2 = x^2 - y^2 + 2ixy \). Note that \( u = x^2 - 2xy - y^2 = (x^2 - y^2) - 2xy \) and \( v = x^2 - y^2 + 2xy \) because \( b=1, a=-1 \). Rewrite \( f(z) = u + iv = (x^2 - y^2 - 2xy) + i(x^2 - y^2 + 2xy) \). Observe that \( f(z) = (x^2 - y^2) + i(x^2 - y^2) + (-2xy + 2ixy) = (1 + i)(x^2 - y^2 + 2ixy) \). Note that \( x^2 - y^2 + 2ixy = (x + iy)^2 = z^2 \). Hence \( f(z) = (1 + i) z^2 \). --- 3. **Problem 10(a):** Find analytic function with real part \( e^x \cos y \). Step 1: Let \( u = e^x \cos y \), need to find \( v \) such that Cauchy-Riemann hold. \( u_x = e^x \cos y, u_y = -e^x \sin y \) By Cauchy-Riemann, \( v_x = -u_y = e^x \sin y \) \( v_y = u_x = e^x \cos y \) Step 2: Integrate \( v_y \) w.r.t \( y \): \( v = \int e^x \cos y dy = e^x \sin y + h(x) \) Step 3: Differentiate with respect to \( x \): \( v_x = e^x \sin y + h'(x) \) Equate to \( v_x \) found before: \( e^x \sin y + h'(x) = e^x \sin y \Rightarrow h'(x) = 0 \Rightarrow h(x) = C \). Step 4: Therefore, \( v = e^x \sin y + C \). Step 5: The analytic function is $$f(z) = u + iv = e^x \cos y + i e^x \sin y = e^{x + i y} = e^z.$$ --- 4. **Problem 10(b):** Find the analytic function whose imaginary part is \( \frac{x - y}{x^2 + y^2} \). Step 1: Let \( v = \frac{x - y}{x^2 + y^2} \), and seek \( u \) such that \( f = u + iv \) is analytic. Step 2: Verify \( v \) is harmonic where \( z \neq 0 \) (can be shown). Step 3: Using Cauchy-Riemann: \( u_x = v_y, u_y = -v_x \). Calculate derivatives: \( v_x = \frac{(1)(x^2 + y^2) - (x - y)(2x)}{(x^2 + y^2)^2} = \frac{x^2 + y^2 - 2x(x - y)}{(x^2 + y^2)^2} \). Simplify numerator: \( x^2 + y^2 - 2x^2 + 2xy = -x^2 + y^2 + 2xy \). Similarly, \( v_y = \frac{(-1)(x^2 + y^2) - (x - y)(2y)}{(x^2 + y^2)^2} = \frac{-x^2 - y^2 - 2y(x - y)}{(x^2 + y^2)^2} \). Simplify numerator: \( -x^2 - y^2 - 2xy + 2y^2 = -x^2 + y^2 - 2xy \). Step 4: Then \( u_x = v_y = \frac{-x^2 + y^2 - 2xy}{(x^2 + y^2)^2}, \quad u_y = -v_x = - \frac{-x^2 + y^2 + 2xy}{(x^2 + y^2)^2} = \frac{x^2 - y^2 - 2xy}{(x^2 + y^2)^2} \). Step 5: Find \( u \) by integrating \( u_x \) w.r.t \( x \) or \( u_y \) w.r.t \( y \). Alternatively, note imaginary part matches \( \text{Im}\left(-\frac{1+i}{z}\right) \), since \( \frac{1+i}{z} = \frac{1+i}{x + iy} = \frac{(1+i)(x - iy)}{x^2 + y^2} = \frac{(x + y) + i(y - x)}{x^2 + y^2} \). Hence imaginary part is \( \frac{y - x}{x^2 + y^2} \), close but with opposite sign. So the analytic function whose imaginary part is \( v = \frac{x - y}{x^2 + y^2} \) is $$ f(z) = \frac{1 - i}{z} + C $$ --- 5. **Problem 11(a):** Find analytic function with real part \( u = x^3 - 3xy^2 \). Step 1: Recognize \( u = \text{Re}(z^3) \) because \[ z^3 = (x + iy)^3 = x^3 + 3ix^2 y - 3x y^2 - i y^3 = (x^3 - 3 x y^2) + i (3 x^2 y - y^3) \]. So analytic function is $$ f(z) = z^3. $$ --- 6. **Problem 11(b):** Analytic function with imaginary part \( v = e^x (x \sin y + y \cos y) \). Step 1: Cauchy-Riemann equations: \( u_x = v_y \), \( u_y = -v_x \). Step 2: Compute derivatives: \( v_x = e^x(x \sin y + y \cos y) + e^x \sin y = v + e^x \sin y \), \( v_y = e^x (x \cos y + \cos y - y \sin y) \). Step 3: Integrate \( u_x = v_y \) to find \( u \). Step 4: Recognize expression resembles real part of \( f(z) = z e^z \), since \( z e^z = (x + iy) e^{x + iy} = (x + iy) e^x (\cos y + i \sin y) \). Real part: \( u = e^x [x \cos y - y \sin y] \), imaginary part: \( v = e^x [x \sin y + y \cos y] \), matching given \( v \). Hence, analytic function is $$ f(z) = z e^z. $$ --- 7. **UNIT V: Complex Variable Integration** 1.(a) Line integral of \( f \) along curve \( C \): $$ \int_C f(z) dz = \int_a^b f(z(t)) z'(t) dt, $$ where \( z(t) \) parametrizes \( C \). 1.(b) Cauchy's integral theorem: For \( f \) analytic in simply connected domain \( D \), $$ \oint_C f(z) dz = 0 $$ for every closed contour \( C \) in \( D \). 1.(c) Cauchy Integral formula: $$ f^{(n)}(a) = \frac{n!}{2 \pi i} \oint_C \frac{f(z)}{(z - a)^{n+1}} dz $$ if \( f \) analytic inside and on \( C \). 1.(d) Taylor series of \( e^z \) about \( z=3 \): $$ e^z = e^3 \sum_{n=0}^\infty \frac{(z - 3)^n}{n!}. $$ 1.(e) Cauchy Residue theorem: $$ \oint_C f(z) dz = 2 \pi i \sum \text{Res}(f, a_k), $$ where \( a_k \) are singularities inside \( C \). --- 8. **Problem 2(a):** Evaluate \( \int_{(0,0)}^{(1,3)} 3x^2 y dx + (x^3 - 3y^2) dy \) along \( y=3x \). Step 1: Parametrize \( y=3x \), \( x=t, y=3t, t \in [0,1] \). Step 2: Express \( dx = dt, dy = 3 dt \). Step 3: Substitute inside integral: $$ \int_0^1 \left[3 t^2 (3 t) \cdot 1 + (t^3 - 3 (3t)^2) \cdot 3 \right] dt = \int_0^1 [9 t^3 + 3 (t^3 - 27 t^2)] dt = \int_0^1 [9 t^3 + 3 t^3 - 81 t^2] dt = \int_0^1 (12 t^3 - 81 t^2) dt. $$ Step 4: Integrate: $$ \left[3 t^4 - 27 t^3 \right]_0^1 = 3 - 27 = -24. $$ --- 9. **Problem 2(b):** Evaluate \( \int_{0}^{1+i} (x^2 - i y) dz \) along \( y = x \). Step 1: Parametrize: \( z = x + i x = x(1 + i), x \in [0,1] \). Step 2: Then \( dz = (1 + i) dx \). Step 3: Expression inside integral: \( x^2 - i y = x^2 - i x \). Step 4: Integral: $$ \int_0^1 (x^2 - i x) (1 + i) dx = (1 + i) \int_0^1 (x^2 - i x) dx. $$ Step 5: Integrate inside: $$ \int_0^1 x^2 dx = \frac{1}{3}, \quad \int_0^1 x dx = \frac{1}{2}. $$ Step 6: So $$ \int_0^1 (x^2 - i x) dx = \frac{1}{3} - i \frac{1}{2} = \frac{1}{3} - \frac{i}{2}. $$ Step 7: Multiply by \( 1 + i \): $$ (1 + i) \left( \frac{1}{3} - \frac{i}{2} \right) = \frac{1}{3}(1 + i) - \frac{i}{2}(1 + i) = \frac{1}{3} + \frac{i}{3} - \frac{i}{2} - \frac{i^2}{2} = \frac{1}{3} + \frac{i}{3} - \frac{i}{2} + \frac{1}{2}. $$ Step 8: Simplify real and imaginary parts: Real: \( \frac{1}{3} + \frac{1}{2} = \frac{5}{6} \), Imaginary: \( \frac{1}{3}i - \frac{1}{2}i = - \frac{1}{6} i \). Final answer: $$ \frac{5}{6} - \frac{1}{6} i. $$ --- 10. **Problem 3:** Evaluate \( \int_0^{3+i} z^2 dz \) (i) along \( y = x/3 \), (ii) along parabola \( x = 3 y^2 \). Step 1: For any path between two points and analytic integrand, integral depends only on endpoints. Step 2: \( \int_a^b z^2 dz = \frac{z^3}{3} \Big|_0^{3+i} = \frac{(3+i)^3}{3}. \) Step 3: Calculate \( (3+i)^3 \): \( (3+i)^2 = 9 + 6i + i^2 = 9 + 6i -1 = 8 + 6i \). \( (3+i)^3 = (3+i)(8 + 6i) = 24 + 18i + 8i + 6i^2 = 24 + 26 i - 6 = 18 + 26 i \). Step 4: Therefore: $$ \int_0^{3+i} z^2 dz = \frac{18 + 26 i}{3} = 6 + \frac{26}{3} i. $$ Same for both paths. --- 11. **Problem 4:** Show \( \oint_c (z + 1) dz = 0 \) where \( c \) is square with vertices 0, 1, 1+i, i. Step 1: Because \( f(z) = z + 1 \) is entire (analytic everywhere), By Cauchy's integral theorem, $$ \oint_c (z+1) dz = 0. $$ --- 12. **Problem 5:** Evaluate \( \oint_c \frac{e^{2z}}{(z-1)(z-2)} dz \) where \( c: |z|=3 \). Step 1: Singularities at \( z=1 \), \( z=2 \), both inside circle \( |z|=3 \). Step 2: Residues: \[ \text{Res}_{z=1} = \lim_{z \to 1} (z-1) \frac{e^{2z}}{(z-1)(z-2)} = \frac{e^{2 \cdot 1}}{1 - 2} = \frac{e^2}{-1} = -e^2, \] \[ \text{Res}_{z=2} = \lim_{z \to 2} (z-2) \frac{e^{2z}}{(z-1)(z-2)} = \frac{e^{4}}{2 - 1} = e^4. \] Step 3: Sum of residues = \( -e^2 + e^4 \). Step 4: By residue theorem: $$ \oint_c \frac{e^{2z}}{(z-1)(z-2)} dz = 2 \pi i ( e^4 - e^2 ). $$