Complex Integration & Analytic Functions

$$ \begin{aligned} \text{Let } I &= \int_C \frac{3z^2+7z+1}{z+1} \, dz. \\ \text{The singular point is } z = -1. \\ \text{The given region C is } |z| = \frac{1}{2} \text{ (Circle with center 0, radius 0.5)}. \\ \text{Distance of singularity from center: } |-1| = 1. \\ \text{Since } 1 > 0.5, \text{ the point } z = -1 \text{ lies OUTSIDE the circle } C. \\ \text{By Cauchy's Integral Theorem, if a function is analytic inside and on C, the integral is 0.} \\ \therefore \int_C \frac{3z^2+7z+1}{z+1} \, dz = 0. \end{aligned} $$

$$ \begin{aligned} \text{Let } f(z) &= z. \text{ The integral is of the form } \int \frac{f(z)}{z-a} dz. \\ \text{Singularity at } z &= 2. \\ \text{Region C: } |z-2| &= 1.5 \text{ (Circle with center at 2, radius 1.5)}. \\ \text{The point } z=2 \text{ is the center, so it clearly lies INSIDE C.} \\ \text{By Cauchy's Integral Formula: } \int_C \frac{f(z)}{z-a} dz &= 2\pi i f(a). \\ I &= 2\pi i f(2). \\ \text{Since } f(z) = z, \implies f(2) = 2. \\ I &= 2\pi i (2) = 4\pi i. \end{aligned} $$

$$ \begin{aligned} \text{Poles are at } z=1 \text{ and } z=2. \\ \text{Region C is } |z|=3. \text{ Both } |1|<3 \text{ and } |2|<3 \text{ lie INSIDE C.} \\ \text{Using Cauchy's Residue Theorem: } I = 2\pi i (\text{Sum of Residues}). \\ f(z) &= \frac{\sin \pi z^2 + \cos \pi z^2}{(z-1)(z-2)} \\ \text{Res}(z=1) &= \lim_{z \to 1} (z-1)f(z) = \frac{\sin \pi + \cos \pi}{1-2} = \frac{0 - 1}{-1} = 1. \\ \text{Res}(z=2) &= \lim_{z \to 2} (z-2)f(z) = \frac{\sin 4\pi + \cos 4\pi}{2-1} = \frac{0 + 1}{1} = 1. \\ I &= 2\pi i (1 + 1) = 4\pi i. \end{aligned} $$

$$ \begin{aligned} \text{Poles: } z=0, z=1, z=2. \\ \text{Region: } |z|=1.5. \\ \text{Inside: } z=0 \text{ and } z=1. \quad \text{Outside: } z=2. \\ \text{Rewriting integrand using partial fractions or Residue theorem directly.} \\ \text{Let } g(z) &= \frac{4-3z}{z(z-1)(z-2)}. \\ \text{Res}(z=0) &= \lim_{z \to 0} \frac{4-3z}{(z-1)(z-2)} = \frac{4}{(-1)(-2)} = \frac{4}{2} = 2. \\ \text{Res}(z=1) &= \lim_{z \to 1} \frac{4-3z}{z(z-2)} = \frac{4-3}{1(1-2)} = \frac{1}{-1} = -1. \\ \text{Integral } &= 2\pi i (\text{Sum of residues inside}) \\ &= 2\pi i (2 - 1) = 2\pi i. \end{aligned} $$

$$ \begin{aligned} \text{Taylor Series: } f(z) &= \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (z-a)^n, \quad a = \frac{\pi}{4}. \\ f(z) &= \sin z \implies f(\pi/4) = \frac{1}{\sqrt{2}} \\ f'(z) &= \cos z \implies f'(\pi/4) = \frac{1}{\sqrt{2}} \\ f''(z) &= -\sin z \implies f''(\pi/4) = -\frac{1}{\sqrt{2}} \\ f'''(z) &= -\cos z \implies f'''(\pi/4) = -\frac{1}{\sqrt{2}} \\ \text{Substituting values:} \\ \sin z &= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}(z-\frac{\pi}{4}) - \frac{1}{\sqrt{2}}\frac{(z-\frac{\pi}{4})^2}{2!} - \frac{1}{\sqrt{2}}\frac{(z-\frac{\pi}{4})^3}{3!} + \dots \\ &= \frac{1}{\sqrt{2}} \left[ 1 + \left(z-\frac{\pi}{4}\right) - \frac{\left(z-\frac{\pi}{4}\right)^2}{2} - \frac{\left(z-\frac{\pi}{4}\right)^3}{6} + \dots \right] \end{aligned} $$

$$ \begin{aligned} \text{Partial Fractions: } \frac{1}{(z+2)(z+3)} &= \frac{1}{z+2} - \frac{1}{z+3}. \\ \text{Region } |z|<2 \implies |z/2|<1 \text{ and } |z/3|<1. \\ \text{Expand in powers of } z: \\ \text{Term 1: } \frac{1}{z+2} &= \frac{1}{2(1+z/2)} = \frac{1}{2}(1 + \frac{z}{2})^{-1} = \frac{1}{2} \sum_{n=0}^{\infty} (-1)^n (\frac{z}{2})^n. \\ \text{Term 2: } \frac{1}{z+3} &= \frac{1}{3(1+z/3)} = \frac{1}{3}(1 + \frac{z}{3})^{-1} = \frac{1}{3} \sum_{n=0}^{\infty} (-1)^n (\frac{z}{3})^n. \\ f(z) &= \sum_{n=0}^{\infty} (-1)^n z^n \left[ \frac{1}{2^{n+1}} - \frac{1}{3^{n+1}} \right]. \end{aligned} $$

$$ \begin{aligned} \text{Since numerator degree } \ge \text{ denominator, divide first:} \\ \frac{z^2-1}{z^2+5z+6} &= 1 + \frac{-5z-7}{(z+2)(z+3)}. \\ \text{Partial fractions for remainder: } \frac{-5z-7}{(z+2)(z+3)} = \frac{A}{z+2} + \frac{B}{z+3}. \\ \text{Solving: } A=3, \quad B=-8. \\ f(z) &= 1 + \frac{3}{z+2} - \frac{8}{z+3}. \\ \text{Region } 2 < |z| < 3: \\ 1. \quad |z| > 2 \implies |\frac{2}{z}| < 1. \quad \frac{3}{z+2} = \frac{3}{z(1+2/z)} = \frac{3}{z} \sum (-1)^n (\frac{2}{z})^n. \\ 2. \quad |z| < 3 \implies |\frac{z}{3}| < 1. \quad \frac{8}{z+3} = \frac{8}{3(1+z/3)} = \frac{8}{3} \sum (-1)^n (\frac{z}{3})^n. \\ \text{Result: } f(z) &= 1 + \sum_{n=0}^{\infty} \frac{3(-1)^n 2^n}{z^{n+1}} - \sum_{n=0}^{\infty} \frac{8(-1)^n z^n}{3^{n+1}}. \end{aligned} $$

(i) \( f(z) = z^2 \):

$$ \begin{aligned} f(z) &= (x+iy)^2 = x^2 - y^2 + i(2xy). \\ u &= x^2-y^2, \quad v = 2xy. \\ u_x &= 2x, \quad v_x = 2y. \\ u_y &= -2y, \quad v_y = 2x. \\ \text{CR Equations: } u_x = v_y \implies 2x=2x \text{ (True)}. \\ u_y = -v_x \implies -2y = -(2y) \text{ (True)}. \\ \text{Partial derivatives are continuous, CR eqns satisfied. } f(z) \text{ is Analytic.} \end{aligned} $$

(ii) \( f(z) = e^z \):

$$ \begin{aligned} f(z) &= e^{x+iy} = e^x(\cos y + i \sin y). \\ u &= e^x \cos y, \quad v = e^x \sin y. \\ u_x &= e^x \cos y, \quad v_x = e^x \sin y. \\ u_y &= -e^x \sin y, \quad v_y = e^x \cos y. \\ \text{Check CR: } u_x = v_y \text{ and } u_y = -v_x. \text{ (Both hold)}. \\ f'(z) &= u_x + i v_x = e^x \cos y + i e^x \sin y = e^z. \end{aligned} $$

(i) Finding Constants:

$$ \begin{aligned} u &= x+ay, \quad v = bx+cy. \\ u_x &= 1, \quad u_y = a. \\ v_x &= b, \quad v_y = c. \\ \text{CR Eqns: } u_x = v_y \implies 1 = c. \\ u_y = -v_x \implies a = -b. \\ \text{Answer: } c = 1, b = -a. \end{aligned} $$

(ii) Proof:

$$ \begin{aligned} \text{Let } \nabla^2 &= \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} = 4 \frac{\partial^2}{\partial z \partial \bar{z}}. \\ |f(z)|^2 &= f(z) \overline{f(z)} = f(z) \bar{f}(\bar{z}). \\ \nabla^2 (f(z)\bar{f}(\bar{z})) &= 4 \partial_z \partial_{\bar{z}} (f(z)\bar{f}(\bar{z})) \\ &= 4 \partial_z [ f(z) \bar{f}'(\bar{z}) ] \\ &= 4 f'(z) \bar{f}'(\bar{z}) \\ &= 4 |f'(z)|^2. \quad \text{(Hence Proved)} \end{aligned} $$

$$ \begin{aligned} \text{Milne Thomson Method:} \\ u_x &= 2\cos 2x \cosh 2y - 2\sin 2x \\ u_y &= 2\sin 2x \sinh 2y \\ \text{Put } y=0, x=z: \\ \phi_1(z,0) &= u_x(z,0) = 2\cos 2z \cosh(0) - 2\sin 2z = 2\cos 2z - 2\sin 2z. \\ \phi_2(z,0) &= u_y(z,0) = 2\sin 2z \sinh(0) = 0. \\ f'(z) &= \phi_1(z,0) - i\phi_2(z,0) = 2\cos 2z - 2\sin 2z. \\ f(z) &= \int (2\cos 2z - 2\sin 2z) dz \\ f(z) &= \sin 2z + \cos 2z + C. \end{aligned} $$

$$ \begin{aligned} u &= \frac{1}{2} \log(x^2+y^2) = \log (\sqrt{x^2+y^2}) = \log r. \\ \text{In polar,} \nabla^2 u &= \frac{1}{r} \frac{\partial}{\partial r} (r \frac{\partial u}{\partial r}) + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2}. \\ u_r &= \frac{1}{r}, \quad r u_r = 1, \quad \frac{\partial}{\partial r}(1) = 0. \quad u_{\theta\theta} = 0. \\ \therefore \nabla^2 u &= 0 \implies \text{Harmonic.} \\ \text{Milne Thomson for } f(z): \\ u_x &= \frac{x}{x^2+y^2}, \quad u_y = \frac{y}{x^2+y^2}. \\ \phi_1(z,0) &= \frac{z}{z^2} = \frac{1}{z}, \quad \phi_2(z,0) = 0. \\ f'(z) &= \frac{1}{z} \implies f(z) = \log z + C. \\ f(z) &= \log(re^{i\theta}) = \log r + i\theta. \\ \text{Conjugate } v &= \theta = \tan^{-1}(y/x). \end{aligned} $$

(i) \( (z_1, z_2, z_3) = (0, -1, i) \to (w_1, w_2, w_3) = (i, 0, \infty) \)

$$ \begin{aligned} \text{Cross Ratio: } \frac{w-w_1}{w-w_2} &= \frac{z-z_1}{z-z_2} \cdot \frac{z_3-z_2}{z_3-z_1} \\ \text{Since } w_3 = \infty \text{, we omit terms with } w_3. \\ \frac{w-i}{w-0} &= \frac{z-0}{z-(-1)} \cdot \frac{i-(-1)}{i-0} \\ \frac{w-i}{w} &= \frac{z}{z+1} \cdot \frac{1+i}{i} \\ \frac{w-i}{w} &= \frac{z(1-i)}{z+1} \cdot \frac{1}{i(1-i)} \cdot (1+i) \dots \text{Simplify: } \frac{1+i}{i} = 1-i. \\ 1 - \frac{i}{w} &= \frac{z(1-i)}{z+1}. \\ \text{Solving yields: } w &= -i \left( \frac{z+1}{z+1 - z(1-i)} \right) = \dots \implies w = \frac{z+1}{z-i} \cdot i \dots \text{result check: } \\ \text{Standard form result: } w &= \frac{-(z+1)}{i(z-i)}. \end{aligned} $$

(ii) \( (1, i, -1) \to (0, 1, \infty) \)

$$ \begin{aligned} \frac{w-0}{1-0} &= \frac{z-1}{z-(-1)} \cdot \frac{i-(-1)}{i-1} \\ w &= \frac{z-1}{z+1} \cdot \frac{i+1}{i-1} \\ \text{Constant: } \frac{i+1}{i-1} \times \frac{i+1}{i+1} &= \frac{2i}{-2} = -i. \\ w &= -i \left( \frac{z-1}{z+1} \right). \end{aligned} $$

$$ \begin{aligned} \text{Denominator roots: } z^2+2z+5 = 0 \implies z = \frac{-2 \pm \sqrt{4-20}}{2} = -1 \pm 2i. \\ \text{Poles: } \alpha = -1+2i, \quad \beta = -1-2i. \\ \text{Center of C is } -1-i. \text{ Radius } 2. \\ \text{Distance to } \alpha: |(-1+2i) - (-1-i)| = |3i| = 3 > 2 \text{ (Outside)}. \\ \text{Distance to } \beta: |(-1-2i) - (-1-i)| = |-i| = 1 < 2 \text{ (Inside)}. \\ \text{Residue at } \beta (-1-2i): \lim_{z \to \beta} (z-\beta) \frac{z+4}{(z-\alpha)(z-\beta)} = \frac{\beta+4}{\beta-\alpha}. \\ \beta+4 = 3-2i. \quad \beta-\alpha = (-1-2i) - (-1+2i) = -4i. \\ \text{Res} &= \frac{3-2i}{-4i}. \\ \text{Integral } &= 2\pi i \left( \frac{3-2i}{-4i} \right) = -\frac{\pi}{2}(3-2i) = \frac{\pi}{2}(2i-3). \end{aligned} $$

$$ \begin{aligned} \text{Roots of } z^2+2z+4=0 \text{ are } z = -1 \pm i\sqrt{3}. \\ z_1 = -1 + 1.732i, \quad z_2 = -1 - 1.732i. \\ \text{Center: } -1-i (-1 - 1i). \\ \text{Dist } z_1: |(-1+1.732i) - (-1-i)| = |2.732i| > 2 \text{ (Out)}. \\ \text{Dist } z_2: |(-1-1.732i) - (-1-i)| = |-0.732i| < 2 \text{ (In)}. \\ \text{Pole } z_2 \text{ is of order 2.} \\ f(z) &= \frac{z+1}{(z-z_1)^2 (z-z_2)^2} \implies \phi(z) = \frac{z+1}{(z-z_1)^2}. \\ \text{Integral } &= 2\pi i \cdot \phi'(z_2). \\ \phi'(z) &= \frac{(z-z_1)^2(1) - (z+1)2(z-z_1)}{(z-z_1)^4} = \frac{(z-z_1) - 2(z+1)}{(z-z_1)^3} = \frac{-z -z_1 - 2}{(z-z_1)^3}. \\ \text{Substitute } z_2 \text{ and solve.} \end{aligned} $$

$$ \begin{aligned} f(z) &= \frac{1}{(z-1)(z-2)} = \frac{1}{z-2} - \frac{1}{z-1}. \\ \text{Region } 1 < |z| < 2: \\ \text{Since } |z|>1 \implies |1/z|<1. \quad \text{Use } \frac{1}{z-1} = \frac{1}{z(1-1/z)} = \frac{1}{z}\sum (\frac{1}{z})^n. \\ \text{Since } |z|<2 \implies |z/2|<1. \quad \text{Use } \frac{1}{z-2} = \frac{-1}{2(1-z/2)} = -\frac{1}{2}\sum (\frac{z}{2})^n. \\ f(z) &= -\frac{1}{2}\sum_{n=0}^{\infty} \left(\frac{z}{2}\right)^n - \sum_{n=0}^{\infty} \frac{1}{z^{n+1}}. \end{aligned} $$